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python
Complete the code in python to solve this programming problem: Description: You are given an integer $$$x$$$ and an array of integers $$$a_1, a_2, \ldots, a_n$$$. You have to determine if the number $$$a_1! + a_2! + \ldots + a_n!$$$ is divisible by $$$x!$$$.Here $$$k!$$$ is a factorial of $$$k$$$ — the product of all positive integers less than or equal to $$$k$$$. For example, $$$3! = 1 \cdot 2 \cdot 3 = 6$$$, and $$$5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120$$$. Input Specification: The first line contains two integers $$$n$$$ and $$$x$$$ ($$$1 \le n \le 500\,000$$$, $$$1 \le x \le 500\,000$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le x$$$) — elements of given array. Output Specification: In the only line print "Yes" (without quotes) if $$$a_1! + a_2! + \ldots + a_n!$$$ is divisible by $$$x!$$$, and "No" (without quotes) otherwise. Notes: NoteIn the first example $$$3! + 2! + 2! + 2! + 3! + 3! = 6 + 2 + 2 + 2 + 6 + 6 = 24$$$. Number $$$24$$$ is divisible by $$$4! = 24$$$.In the second example $$$3! + 2! + 2! + 2! + 2! + 2! + 1! + 1! = 18$$$, is divisible by $$$3! = 6$$$.In the third example $$$7! + 7! + 7! + 7! + 7! + 7! + 7! = 7 \cdot 7!$$$. It is easy to prove that this number is not divisible by $$$8!$$$. Code: def rl(): return [int(i) for i in input().split()] def solve(): [n,x]=rl() a=rl() nax=500000+5 ct=[0 for i in range(nax)] for b in a: ct[b]+=1 for i in range(x): if # TODO: Your code here: return "No" ct[i+1]+=ct[i]/(i+1) return "Yes" print(solve())
def rl(): return [int(i) for i in input().split()] def solve(): [n,x]=rl() a=rl() nax=500000+5 ct=[0 for i in range(nax)] for b in a: ct[b]+=1 for i in range(x): if {{completion}}: return "No" ct[i+1]+=ct[i]/(i+1) return "Yes" print(solve())
ct[i]%(i+1)
[{"input": "6 4\n3 2 2 2 3 3", "output": ["Yes"]}, {"input": "8 3\n3 2 2 2 2 2 1 1", "output": ["Yes"]}, {"input": "7 8\n7 7 7 7 7 7 7", "output": ["No"]}, {"input": "10 5\n4 3 2 1 4 3 2 4 3 4", "output": ["No"]}, {"input": "2 500000\n499999 499999", "output": ["No"]}]
control_completion_005984
control_fixed
python
Complete the code in python to solve this programming problem: Description: The derby between Milan and Inter is happening soon, and you have been chosen as the assistant referee for the match, also known as linesman. Your task is to move along the touch-line, namely the side of the field, always looking very carefully at the match to check for offside positions and other offences.Football is an extremely serious matter in Italy, and thus it is fundamental that you keep very close track of the ball for as much time as possible. This means that you want to maximise the number of kicks which you monitor closely. You are able to monitor closely a kick if, when it happens, you are in the position along the touch-line with minimum distance from the place where the kick happens.Fortunately, expert analysts have been able to accurately predict all the kicks which will occur during the game. That is, you have been given two lists of integers, $$$t_1, \ldots, t_n$$$ and $$$a_1, \ldots, a_n$$$, indicating that $$$t_i$$$ seconds after the beginning of the match the ball will be kicked and you can monitor closely such kick if you are at the position $$$a_i$$$ along the touch-line. At the beginning of the game you start at position $$$0$$$ and the maximum speed at which you can walk along the touch-line is $$$v$$$ units per second (i.e., you can change your position by at most $$$v$$$ each second). What is the maximum number of kicks that you can monitor closely? Input Specification: The first line contains two integers $$$n$$$ and $$$v$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le v \le 10^6$$$) — the number of kicks that will take place and your maximum speed. The second line contains $$$n$$$ integers $$$t_1, \ldots, t_n$$$ ($$$1 \le t_i \le 10^9$$$) — the times of the kicks in the match. The sequence of times is guaranteed to be strictly increasing, i.e., $$$t_1 < t_2 < \cdots < t_n$$$. The third line contains $$$n$$$ integers $$$a_1, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the positions along the touch-line where you have to be to monitor closely each kick. Output Specification: Print the maximum number of kicks that you can monitor closely. Notes: NoteIn the first sample, it is possible to move to the right at maximum speed for the first $$$3.5$$$ seconds and stay at position $$$7$$$ until the first kick happens, and then immediately move right also at maximum speed to watch the second kick at position $$$17$$$. There is no way to monitor closely the third kick after the second kick, so at most $$$2$$$ kicks can be seen. Code: from bisect import bisect_right,bisect_left n,v = map(int,input().split()) t = [*map(int,input().split())] a = [*map(int,input().split())] res = [] for i in range(n): xi,yi = t[i]*v+a[i],t[i]*v-a[i] if(xi>=0 and yi>=0): res.append((xi,yi)) res.sort() dp = [float("inf")]*(n+3) dp[0] = 0 dp[n+2] = 0 for i in range(len(res)): pos = bisect_right(dp,res[i][1],0,n+2) dp[pos] = res[i][1] for i in range(n,-1,-1): if(dp[i]!=float("inf")): # TODO: Your code here
from bisect import bisect_right,bisect_left n,v = map(int,input().split()) t = [*map(int,input().split())] a = [*map(int,input().split())] res = [] for i in range(n): xi,yi = t[i]*v+a[i],t[i]*v-a[i] if(xi>=0 and yi>=0): res.append((xi,yi)) res.sort() dp = [float("inf")]*(n+3) dp[0] = 0 dp[n+2] = 0 for i in range(len(res)): pos = bisect_right(dp,res[i][1],0,n+2) dp[pos] = res[i][1] for i in range(n,-1,-1): if(dp[i]!=float("inf")): {{completion}}
print(i) break
[{"input": "3 2\n5 10 15\n7 17 29", "output": ["2"]}, {"input": "5 1\n5 7 8 11 13\n3 3 -2 -2 4", "output": ["3"]}, {"input": "1 2\n3\n7", "output": ["0"]}]
block_completion_001106
block
python
Complete the code in python to solve this programming problem: Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$? Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i &lt; 32768$$$). Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$. Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times. Code: n = int(input()) mod = 1 << 15 for x in map(int, input().split()): res = 16 for a in range(15): for b in range(15): if (x + a) * (1 << b) % mod == 0: # TODO: Your code here print(res)
n = int(input()) mod = 1 << 15 for x in map(int, input().split()): res = 16 for a in range(15): for b in range(15): if (x + a) * (1 << b) % mod == 0: {{completion}} print(res)
res = min(res, a + b)
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
block_completion_003353
block
python
Complete the code in python to solve this programming problem: Description: You are given an array $$$a$$$ of $$$n$$$ integers. Initially there is only one copy of the given array.You can do operations of two types: Choose any array and clone it. After that there is one more copy of the chosen array. Swap two elements from any two copies (maybe in the same copy) on any positions. You need to find the minimal number of operations needed to obtain a copy where all elements are equal. Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. Description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$. Output Specification: For each test case output a single integer — the minimal number of operations needed to create at least one copy where all elements are equal. Notes: NoteIn the first test case all elements in the array are already equal, that's why the answer is $$$0$$$.In the second test case it is possible to create a copy of the given array. After that there will be two identical arrays:$$$[ \ 0 \ 1 \ 3 \ 3 \ 7 \ 0 \ ]$$$ and $$$[ \ 0 \ 1 \ 3 \ 3 \ 7 \ 0 \ ]$$$After that we can swap elements in a way so all zeroes are in one array:$$$[ \ 0 \ \underline{0} \ \underline{0} \ 3 \ 7 \ 0 \ ]$$$ and $$$[ \ \underline{1} \ 1 \ 3 \ 3 \ 7 \ \underline{3} \ ]$$$Now let's create a copy of the first array:$$$[ \ 0 \ 0 \ 0 \ 3 \ 7 \ 0 \ ]$$$, $$$[ \ 0 \ 0 \ 0 \ 3 \ 7 \ 0 \ ]$$$ and $$$[ \ 1 \ 1 \ 3 \ 3 \ 7 \ 3 \ ]$$$Let's swap elements in the first two copies:$$$[ \ 0 \ 0 \ 0 \ \underline{0} \ \underline{0} \ 0 \ ]$$$, $$$[ \ \underline{3} \ \underline{7} \ 0 \ 3 \ 7 \ 0 \ ]$$$ and $$$[ \ 1 \ 1 \ 3 \ 3 \ 7 \ 3 \ ]$$$.Finally, we made a copy where all elements are equal and made $$$6$$$ operations.It can be proven that no fewer operations are enough. Code: from collections import Counter for li in[*open(0)][2::2]: n=len(li:=li.split()); m = max(Counter(li).values()) ans =n-m while(m<n): # TODO: Your code here print(ans)
from collections import Counter for li in[*open(0)][2::2]: n=len(li:=li.split()); m = max(Counter(li).values()) ans =n-m while(m<n): {{completion}} print(ans)
ans+=1 m=2*m
[{"input": "6\n1\n1789\n6\n0 1 3 3 7 0\n2\n-1000000000 1000000000\n4\n4 3 2 1\n5\n2 5 7 6 3\n7\n1 1 1 1 1 1 1", "output": ["0\n6\n2\n5\n7\n0"]}]
block_completion_004425
block
python
Complete the code in python to solve this programming problem: Description: A ticket is a string consisting of six digits. A ticket is considered lucky if the sum of the first three digits is equal to the sum of the last three digits. Given a ticket, output if it is lucky or not. Note that a ticket can have leading zeroes. Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of testcases. The description of each test consists of one line containing one string consisting of six digits. Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. Output "YES" if the given ticket is lucky, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer). Notes: NoteIn the first test case, the sum of the first three digits is $$$2 + 1 + 3 = 6$$$ and the sum of the last three digits is $$$1 + 3 + 2 = 6$$$, they are equal so the answer is "YES".In the second test case, the sum of the first three digits is $$$9 + 7 + 3 = 19$$$ and the sum of the last three digits is $$$8 + 9 + 4 = 21$$$, they are not equal so the answer is "NO".In the third test case, the sum of the first three digits is $$$0 + 4 + 5 = 9$$$ and the sum of the last three digits is $$$2 + 0 + 7 = 9$$$, they are equal so the answer is "YES". Code: for # TODO: Your code here: a = [int(j) for j in input()] print("YES" if sum(a[0:3])==sum(a[3:6]) else "NO")
for {{completion}}: a = [int(j) for j in input()] print("YES" if sum(a[0:3])==sum(a[3:6]) else "NO")
i in range(int(input()))
[{"input": "5\n213132\n973894\n045207\n000000\n055776", "output": ["YES\nNO\nYES\nYES\nNO"]}]
control_completion_007491
control_fixed
python
Complete the code in python to solve this programming problem: Description: Consider a hallway, which can be represented as the matrix with $$$2$$$ rows and $$$n$$$ columns. Let's denote the cell on the intersection of the $$$i$$$-th row and the $$$j$$$-th column as $$$(i, j)$$$. The distance between the cells $$$(i_1, j_1)$$$ and $$$(i_2, j_2)$$$ is $$$|i_1 - i_2| + |j_1 - j_2|$$$.There is a cleaning robot in the cell $$$(1, 1)$$$. Some cells of the hallway are clean, other cells are dirty (the cell with the robot is clean). You want to clean the hallway, so you are going to launch the robot to do this.After the robot is launched, it works as follows. While at least one cell is dirty, the robot chooses the closest (to its current cell) cell among those which are dirty, moves there and cleans it (so the cell is no longer dirty). After cleaning a cell, the robot again finds the closest dirty cell to its current cell, and so on. This process repeats until the whole hallway is clean.However, there is a critical bug in the robot's program. If at some moment, there are multiple closest (to the robot's current position) dirty cells, the robot malfunctions.You want to clean the hallway in such a way that the robot doesn't malfunction. Before launching the robot, you can clean some (possibly zero) of the dirty cells yourself. However, you don't want to do too much dirty work yourself while you have this nice, smart (yet buggy) robot to do this. Note that you cannot make a clean cell dirty.Calculate the maximum possible number of cells you can leave dirty before launching the robot, so that it doesn't malfunction. Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the number of columns in the hallway. Then two lines follow, denoting the $$$1$$$-st and the $$$2$$$-nd row of the hallway. These lines contain $$$n$$$ characters each, where 0 denotes a clean cell and 1 denotes a dirty cell. The starting cell of the robot $$$(1, 1)$$$ is clean. Output Specification: Print one integer — the maximum possible number of cells you can leave dirty before launching the robot, so that it doesn't malfunction. Notes: NoteIn the first example, you can clean the cell $$$(1, 2)$$$, so the path of the robot is $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 2)$$$.In the second example, you can leave the hallway as it is, so the path of the robot is $$$(1, 1) \rightarrow (1, 2) \rightarrow (2, 2)$$$.In the third example, you can clean the cell $$$(1, 2)$$$, so the path of the robot is $$$(1, 1) \rightarrow (2, 1) \rightarrow (2, 3) \rightarrow (2, 4) \rightarrow (1, 4)$$$.In the fourth example, the hallway is already clean. Maybe you have launched the robot earlier? Code: import sys input = lambda: sys.stdin.readline().rstrip() def solve(): N = int(input()) G = [[int(x) for x in input()] + [0] for _ in range(2)] dp = [[0] * 2 for _ in range(N + 1)] for j in range(2): dp[N - 1][j] = G[1 - j][N - 1] for i in range(N - 2, - 1, -1): for j in range(2): dp[i][j] = G[j][i + 1] + dp[i + 1][j] # can always ignore row 1 - j and proceed right if G[1 - j][i]: if # TODO: Your code here: dp[i][j] = max(dp[i][j], 1 + G[1 - j][i + 1] + G[1 - j][i + 2] + dp[i + 2][1 - j]) else: dp[i][j] = max(dp[i][j], 1 + G[1 - j][i + 1] + dp[i + 1][1 - j]) print(dp[0][0]) return solve()
import sys input = lambda: sys.stdin.readline().rstrip() def solve(): N = int(input()) G = [[int(x) for x in input()] + [0] for _ in range(2)] dp = [[0] * 2 for _ in range(N + 1)] for j in range(2): dp[N - 1][j] = G[1 - j][N - 1] for i in range(N - 2, - 1, -1): for j in range(2): dp[i][j] = G[j][i + 1] + dp[i + 1][j] # can always ignore row 1 - j and proceed right if G[1 - j][i]: if {{completion}}: dp[i][j] = max(dp[i][j], 1 + G[1 - j][i + 1] + G[1 - j][i + 2] + dp[i + 2][1 - j]) else: dp[i][j] = max(dp[i][j], 1 + G[1 - j][i + 1] + dp[i + 1][1 - j]) print(dp[0][0]) return solve()
G[j][i + 1]
[{"input": "2\n01\n11", "output": ["2"]}, {"input": "2\n01\n01", "output": ["2"]}, {"input": "4\n0101\n1011", "output": ["4"]}, {"input": "4\n0000\n0000", "output": ["0"]}, {"input": "5\n00011\n10101", "output": ["4"]}, {"input": "6\n011111\n111111", "output": ["8"]}, {"input": "10\n0101001010\n1010100110", "output": ["6"]}]
control_completion_008236
control_fixed
python
Complete the code in python to solve this programming problem: Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock. Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query. Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase. Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$. Code: (n, q) = map(int, input().split()) arr = list(map(int, input().split())) arr.sort(reverse=True) for i in range (1,n): arr[i] = arr[i] + arr[i-1] for trial in range(q): (x, y) = map(int, input().split()) if (x==y): print (arr[x-1]) else: # TODO: Your code here
(n, q) = map(int, input().split()) arr = list(map(int, input().split())) arr.sort(reverse=True) for i in range (1,n): arr[i] = arr[i] + arr[i-1] for trial in range(q): (x, y) = map(int, input().split()) if (x==y): print (arr[x-1]) else: {{completion}}
print (arr[x-1] - arr[x-y-1])
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
block_completion_000523
block
python
Complete the code in python to solve this programming problem: Description: An integer array $$$a_1, a_2, \ldots, a_n$$$ is being transformed into an array of lowercase English letters using the following prodecure:While there is at least one number in the array: Choose any number $$$x$$$ from the array $$$a$$$, and any letter of the English alphabet $$$y$$$. Replace all occurrences of number $$$x$$$ with the letter $$$y$$$. For example, if we initially had an array $$$a = [2, 3, 2, 4, 1]$$$, then we could transform it the following way: Choose the number $$$2$$$ and the letter c. After that $$$a = [c, 3, c, 4, 1]$$$. Choose the number $$$3$$$ and the letter a. After that $$$a = [c, a, c, 4, 1]$$$. Choose the number $$$4$$$ and the letter t. After that $$$a = [c, a, c, t, 1]$$$. Choose the number $$$1$$$ and the letter a. After that $$$a = [c, a, c, t, a]$$$. After the transformation all letters are united into a string, in our example we get the string "cacta".Having the array $$$a$$$ and the string $$$s$$$ determine if the string $$$s$$$ could be got from the array $$$a$$$ after the described transformation? Input Specification: The first line contains a single integer $$$t$$$ $$$(1 \leq t \leq 10^3$$$) — the number of test cases. Then the description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 50$$$) — the length of the array $$$a$$$ and the string $$$s$$$. The second line of each test case contains exactly $$$n$$$ integers: $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 50$$$) — the elements of the array $$$a$$$. The third line of each test case contains a string $$$s$$$ of length $$$n$$$, consisting of lowercase English letters. Output Specification: For each test case, output "YES", if we can get the string $$$s$$$ from the array $$$a$$$, and "NO" otherwise. You can output each letter in any case. Notes: NoteThe first test case corresponds to the sample described in the statement.In the second test case we can choose the number $$$50$$$ and the letter a.In the third test case we can choose the number $$$11$$$ and the letter a, after that $$$a = [a, 22]$$$. Then we choose the number $$$22$$$ and the letter b and get $$$a = [a, b]$$$.In the fifth test case we can change all numbers one by one to the letter a. Code: from sys import stdin from collections import deque lst = stdin.read().split() _s = 0 def inp(n=1): global _s ret = lst[_s:_s + n] _s += n return ret def inp1(): return inp()[0] t = int(inp1()) for _ in range(t): n = int(inp1()) a = list(map(int, inp(n))) s = list(inp1()) d = {} ok = True for i in range(n): if a[i] not in d: d[a[i]] = s[i] elif d[a[i]] != s[i]: # TODO: Your code here print("YES" if ok else "NO")
from sys import stdin from collections import deque lst = stdin.read().split() _s = 0 def inp(n=1): global _s ret = lst[_s:_s + n] _s += n return ret def inp1(): return inp()[0] t = int(inp1()) for _ in range(t): n = int(inp1()) a = list(map(int, inp(n))) s = list(inp1()) d = {} ok = True for i in range(n): if a[i] not in d: d[a[i]] = s[i] elif d[a[i]] != s[i]: {{completion}} print("YES" if ok else "NO")
ok = not ok break
[{"input": "7\n\n5\n\n2 3 2 4 1\n\ncacta\n\n1\n\n50\n\na\n\n2\n\n11 22\n\nab\n\n4\n\n1 2 2 1\n\naaab\n\n5\n\n1 2 3 2 1\n\naaaaa\n\n6\n\n1 10 2 9 3 8\n\nazzfdb\n\n7\n\n1 2 3 4 1 1 2\n\nabababb", "output": ["YES\nYES\nYES\nNO\nYES\nYES\nNO"]}]
block_completion_004085
block
python
Complete the code in python to solve this programming problem: Description: This is the hard version of the problem. The only difference between the two versions is that the harder version asks additionally for a minimum number of subsegments.Tokitsukaze has a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones, $$$n$$$ is even.Now Tokitsukaze divides $$$s$$$ into the minimum number of contiguous subsegments, and for each subsegment, all bits in each subsegment are the same. After that, $$$s$$$ is considered good if the lengths of all subsegments are even.For example, if $$$s$$$ is "11001111", it will be divided into "11", "00" and "1111". Their lengths are $$$2$$$, $$$2$$$, $$$4$$$ respectively, which are all even numbers, so "11001111" is good. Another example, if $$$s$$$ is "1110011000", it will be divided into "111", "00", "11" and "000", and their lengths are $$$3$$$, $$$2$$$, $$$2$$$, $$$3$$$. Obviously, "1110011000" is not good.Tokitsukaze wants to make $$$s$$$ good by changing the values of some positions in $$$s$$$. Specifically, she can perform the operation any number of times: change the value of $$$s_i$$$ to '0' or '1' ($$$1 \leq i \leq n$$$). Can you tell her the minimum number of operations to make $$$s$$$ good? Meanwhile, she also wants to know the minimum number of subsegments that $$$s$$$ can be divided into among all solutions with the minimum number of operations. Input Specification: The first contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 10\,000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 2 \cdot 10^5$$$) — the length of $$$s$$$, it is guaranteed that $$$n$$$ is even. The second line contains a binary string $$$s$$$ of length $$$n$$$, consisting only of zeros and ones. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. Output Specification: For each test case, print a single line with two integers — the minimum number of operations to make $$$s$$$ good, and the minimum number of subsegments that $$$s$$$ can be divided into among all solutions with the minimum number of operations. Notes: NoteIn the first test case, one of the ways to make $$$s$$$ good is the following.Change $$$s_3$$$, $$$s_6$$$ and $$$s_7$$$ to '0', after that $$$s$$$ becomes "1100000000", it can be divided into "11" and "00000000", which lengths are $$$2$$$ and $$$8$$$ respectively, the number of subsegments of it is $$$2$$$. There are other ways to operate $$$3$$$ times to make $$$s$$$ good, such as "1111110000", "1100001100", "1111001100", the number of subsegments of them are $$$2$$$, $$$4$$$, $$$4$$$ respectively. It's easy to find that the minimum number of subsegments among all solutions with the minimum number of operations is $$$2$$$.In the second, third and fourth test cases, $$$s$$$ is good initially, so no operation is required. Code: for _ in range(int(input().strip())): n = int(input().strip()) arr = input() ans = 0 t = [] for i in range(0, len(arr), 2): if arr[i] != arr[i + 1]: ans += 1 else: # TODO: Your code here seg = 1 for i in range(0, len(t) - 1): if t[i] != t[i + 1]: seg += 1 print(ans, seg)
for _ in range(int(input().strip())): n = int(input().strip()) arr = input() ans = 0 t = [] for i in range(0, len(arr), 2): if arr[i] != arr[i + 1]: ans += 1 else: {{completion}} seg = 1 for i in range(0, len(t) - 1): if t[i] != t[i + 1]: seg += 1 print(ans, seg)
t.append(arr[i])
[{"input": "5\n10\n1110011000\n8\n11001111\n2\n00\n2\n11\n6\n100110", "output": ["3 2\n0 3\n0 1\n0 1\n3 1"]}]
block_completion_008095
block
python
Complete the code in python to solve this programming problem: Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$? Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i &lt; 32768$$$). Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$. Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times. Code: n, s = open(0) for x in map(int, s.split()): # TODO: Your code here
n, s = open(0) for x in map(int, s.split()): {{completion}}
print(min(15-i+-x % 2**i for i in range(16)))
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
block_completion_003354
block
python
Complete the code in python to solve this programming problem: Description: We have an array of length $$$n$$$. Initially, each element is equal to $$$0$$$ and there is a pointer located on the first element.We can do the following two kinds of operations any number of times (possibly zero) in any order: If the pointer is not on the last element, increase the element the pointer is currently on by $$$1$$$. Then move it to the next element. If the pointer is not on the first element, decrease the element the pointer is currently on by $$$1$$$. Then move it to the previous element.But there is one additional rule. After we are done, the pointer has to be on the first element.You are given an array $$$a$$$. Determine whether it's possible to obtain $$$a$$$ after some operations or not. Input Specification: The first line contains a single integer $$$t$$$ $$$(1\le t\le 1000)$$$  — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ $$$(1\le n\le 2 \cdot 10^5)$$$  — the size of array $$$a$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — elements of the array. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$. Output Specification: For each test case, print "Yes" (without quotes) if it's possible to obtain $$$a$$$ after some operations, and "No" (without quotes) otherwise. You can output "Yes" and "No" in any case (for example, strings "yEs", "yes" and "Yes" will be recognized as a positive response). Notes: NoteIn the first test case we can obtain the array after some operations, but the pointer won't be on the first element.One way of obtaining the array in the second test case is shown below.$$$\langle \underline{0}, 0, 0, 0\rangle \to \langle 1, \underline{0}, 0, 0 \rangle \to \langle \underline{1}, -1, 0, 0\rangle \to \langle 2, \underline{-1}, 0, 0\rangle \to \langle 2, 0, \underline{0}, 0\rangle \to \langle 2, \underline{0}, -1, 0\rangle \to \langle \underline{2}, -1, -1, 0\rangle$$$ Code: from sys import stdin t = int(stdin.readline()) for h in range(t): n = int(stdin.readline()) a = list(map(int,stdin.readline().split(' '))) b = 0 v = True for i in range(n): b += a[i] if b<0: v = False break elif b==0: for j in range(i+1,n): if a[j] != 0: # TODO: Your code here break if v and sum(a) == 0: print('YES') else: print('NO')
from sys import stdin t = int(stdin.readline()) for h in range(t): n = int(stdin.readline()) a = list(map(int,stdin.readline().split(' '))) b = 0 v = True for i in range(n): b += a[i] if b<0: v = False break elif b==0: for j in range(i+1,n): if a[j] != 0: {{completion}} break if v and sum(a) == 0: print('YES') else: print('NO')
v = False break
[{"input": "7\n2\n1 0\n4\n2 -1 -1 0\n4\n1 -4 3 0\n4\n1 -1 1 -1\n5\n1 2 3 4 -10\n7\n2 -1 1 -2 0 0 0\n1\n0", "output": ["No\nYes\nNo\nNo\nYes\nYes\nYes"]}]
block_completion_000425
block
python
Complete the code in python to solve this programming problem: Description: Let's call a string $$$s$$$ perfectly balanced if for all possible triplets $$$(t,u,v)$$$ such that $$$t$$$ is a non-empty substring of $$$s$$$ and $$$u$$$ and $$$v$$$ are characters present in $$$s$$$, the difference between the frequencies of $$$u$$$ and $$$v$$$ in $$$t$$$ is not more than $$$1$$$.For example, the strings "aba" and "abc" are perfectly balanced but "abb" is not because for the triplet ("bb",'a','b'), the condition is not satisfied.You are given a string $$$s$$$ consisting of lowercase English letters only. Your task is to determine whether $$$s$$$ is perfectly balanced or not.A string $$$b$$$ is called a substring of another string $$$a$$$ if $$$b$$$ can be obtained by deleting some characters (possibly $$$0$$$) from the start and some characters (possibly $$$0$$$) from the end of $$$a$$$. Input Specification: The first line of input contains a single integer $$$t$$$ ($$$1\leq t\leq 2\cdot 10^4$$$) denoting the number of testcases. Each of the next $$$t$$$ lines contain a single string $$$s$$$ ($$$1\leq |s|\leq 2\cdot 10^5$$$), consisting of lowercase English letters. It is guaranteed that the sum of $$$|s|$$$ over all testcases does not exceed $$$2\cdot 10^5$$$. Output Specification: For each test case, print "YES" if $$$s$$$ is a perfectly balanced string, and "NO" otherwise. You may print each letter in any case (for example, "YES", "Yes", "yes", "yEs" will all be recognized as positive answer). Notes: NoteLet $$$f_t(c)$$$ represent the frequency of character $$$c$$$ in string $$$t$$$.For the first testcase we have $$$t$$$$$$f_t(a)$$$$$$f_t(b)$$$$$$a$$$$$$1$$$$$$0$$$$$$ab$$$$$$1$$$$$$1$$$$$$aba$$$$$$2$$$$$$1$$$$$$b$$$$$$0$$$$$$1$$$$$$ba$$$$$$1$$$$$$1$$$ It can be seen that for any substring $$$t$$$ of $$$s$$$, the difference between $$$f_t(a)$$$ and $$$f_t(b)$$$ is not more than $$$1$$$. Hence the string $$$s$$$ is perfectly balanced.For the second testcase we have $$$t$$$$$$f_t(a)$$$$$$f_t(b)$$$$$$a$$$$$$1$$$$$$0$$$$$$ab$$$$$$1$$$$$$1$$$$$$abb$$$$$$1$$$$$$2$$$$$$b$$$$$$0$$$$$$1$$$$$$bb$$$$$$0$$$$$$2$$$ It can be seen that for the substring $$$t=bb$$$, the difference between $$$f_t(a)$$$ and $$$f_t(b)$$$ is $$$2$$$ which is greater than $$$1$$$. Hence the string $$$s$$$ is not perfectly balanced.For the third testcase we have $$$t$$$$$$f_t(a)$$$$$$f_t(b)$$$$$$f_t(c)$$$$$$a$$$$$$1$$$$$$0$$$$$$0$$$$$$ab$$$$$$1$$$$$$1$$$$$$0$$$$$$abc$$$$$$1$$$$$$1$$$$$$1$$$$$$b$$$$$$0$$$$$$1$$$$$$0$$$$$$bc$$$$$$0$$$$$$1$$$$$$1$$$$$$c$$$$$$0$$$$$$0$$$$$$1$$$It can be seen that for any substring $$$t$$$ of $$$s$$$ and any two characters $$$u,v\in\{a,b,c\}$$$, the difference between $$$f_t(u)$$$ and $$$f_t(v)$$$ is not more than $$$1$$$. Hence the string $$$s$$$ is perfectly balanced. Code: for # TODO: Your code here: string=tuple(input().strip()) k=len(set(string)) print("NO" if any([string[i]!=string[i%k] for i in range (len(string))]) else "YES")
for {{completion}}: string=tuple(input().strip()) k=len(set(string)) print("NO" if any([string[i]!=string[i%k] for i in range (len(string))]) else "YES")
_ in range(int(input()))
[{"input": "5\naba\nabb\nabc\naaaaa\nabcba", "output": ["YES\nNO\nYES\nYES\nNO"]}]
control_completion_004715
control_fixed
python
Complete the code in python to solve this programming problem: Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock. Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query. Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase. Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$. Code: f=open(0) R=lambda:map(int,next(f).split()) n,q=R();p=[0] for w in sorted(R()): p+=p[-1]+w, for # TODO: Your code here: x, y=R();print(p[n-x+y]-p[n-x])
f=open(0) R=lambda:map(int,next(f).split()) n,q=R();p=[0] for w in sorted(R()): p+=p[-1]+w, for {{completion}}: x, y=R();print(p[n-x+y]-p[n-x])
_ in " "*q
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
control_completion_000504
control_fixed
python
Complete the code in python to solve this programming problem: Description: You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0&lt;x_1&lt;x_2&lt;\ldots&lt;x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$)  — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 &lt; x_2 &lt; \ldots &lt; x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. Output Specification: For each test case, output a single integer  — the minimum cost to conquer all kingdoms. Notes: NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. Code: for _ in range(int(input())): n,a,b=map(int, input().split()) w=[int(x) for x in input().split()] fb=sum(w)*b fa=0 ans = fb cap = 0 cur = n for x in w: fb -= x * b cur -= 1 if # TODO: Your code here: ans += (x - cap) * a ans -= (x - cap) * cur * b cap = x #print(cap) print(ans)
for _ in range(int(input())): n,a,b=map(int, input().split()) w=[int(x) for x in input().split()] fb=sum(w)*b fa=0 ans = fb cap = 0 cur = n for x in w: fb -= x * b cur -= 1 if {{completion}}: ans += (x - cap) * a ans -= (x - cap) * cur * b cap = x #print(cap) print(ans)
(x - cap) * a + fb - (x - cap) * cur * b < fb
[{"input": "4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030", "output": ["173\n171\n75\n3298918744"]}]
control_completion_008538
control_fixed
python
Complete the code in python to solve this programming problem: Description: A ticket is a string consisting of six digits. A ticket is considered lucky if the sum of the first three digits is equal to the sum of the last three digits. Given a ticket, output if it is lucky or not. Note that a ticket can have leading zeroes. Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of testcases. The description of each test consists of one line containing one string consisting of six digits. Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. Output "YES" if the given ticket is lucky, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer). Notes: NoteIn the first test case, the sum of the first three digits is $$$2 + 1 + 3 = 6$$$ and the sum of the last three digits is $$$1 + 3 + 2 = 6$$$, they are equal so the answer is "YES".In the second test case, the sum of the first three digits is $$$9 + 7 + 3 = 19$$$ and the sum of the last three digits is $$$8 + 9 + 4 = 21$$$, they are not equal so the answer is "NO".In the third test case, the sum of the first three digits is $$$0 + 4 + 5 = 9$$$ and the sum of the last three digits is $$$2 + 0 + 7 = 9$$$, they are equal so the answer is "YES". Code: for c in [input() for i in range(int(input()))]: # TODO: Your code here
for c in [input() for i in range(int(input()))]: {{completion}}
print(('NO', 'YES')[sum(int(p) for p in (c[:3])) == sum(int(p) for p in (c[3:]))])
[{"input": "5\n213132\n973894\n045207\n000000\n055776", "output": ["YES\nNO\nYES\nYES\nNO"]}]
block_completion_007622
block
python
Complete the code in python to solve this programming problem: Description: You like the card board game "Set". Each card contains $$$k$$$ features, each of which is equal to a value from the set $$$\{0, 1, 2\}$$$. The deck contains all possible variants of cards, that is, there are $$$3^k$$$ different cards in total.A feature for three cards is called good if it is the same for these cards or pairwise distinct. Three cards are called a set if all $$$k$$$ features are good for them.For example, the cards $$$(0, 0, 0)$$$, $$$(0, 2, 1)$$$, and $$$(0, 1, 2)$$$ form a set, but the cards $$$(0, 2, 2)$$$, $$$(2, 1, 2)$$$, and $$$(1, 2, 0)$$$ do not, as, for example, the last feature is not good.A group of five cards is called a meta-set, if there is strictly more than one set among them. How many meta-sets there are among given $$$n$$$ distinct cards? Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 10^3$$$, $$$1 \le k \le 20$$$) — the number of cards on a table and the number of card features. The description of the cards follows in the next $$$n$$$ lines. Each line describing a card contains $$$k$$$ integers $$$c_{i, 1}, c_{i, 2}, \ldots, c_{i, k}$$$ ($$$0 \le c_{i, j} \le 2$$$) — card features. It is guaranteed that all cards are distinct. Output Specification: Output one integer — the number of meta-sets. Notes: NoteLet's draw the cards indicating the first four features. The first feature will indicate the number of objects on a card: $$$1$$$, $$$2$$$, $$$3$$$. The second one is the color: red, green, purple. The third is the shape: oval, diamond, squiggle. The fourth is filling: open, striped, solid.You can see the first three tests below. For the first two tests, the meta-sets are highlighted.In the first test, the only meta-set is the five cards $$$(0000,\ 0001,\ 0002,\ 0010,\ 0020)$$$. The sets in it are the triples $$$(0000,\ 0001,\ 0002)$$$ and $$$(0000,\ 0010,\ 0020)$$$. Also, a set is the triple $$$(0100,\ 1000,\ 2200)$$$ which does not belong to any meta-set. In the second test, the following groups of five cards are meta-sets: $$$(0000,\ 0001,\ 0002,\ 0010,\ 0020)$$$, $$$(0000,\ 0001,\ 0002,\ 0100,\ 0200)$$$, $$$(0000,\ 0010,\ 0020,\ 0100,\ 0200)$$$. In there third test, there are $$$54$$$ meta-sets. Code: n,k=map(int,input().split()) a=[] d={} for i in range(n): a+=[''.join(input().split())] d[a[-1]]=0 def cal(s,t): res="" for i in range(k): res+=str((9-int(s[i])-int(t[i]))%3) return res for i in range(n): for # TODO: Your code here: try: d[cal(a[i],a[j])]+=1 except: pass ans=0 for y in d.values(): ans+=(y*(y-1))//2 print(ans)
n,k=map(int,input().split()) a=[] d={} for i in range(n): a+=[''.join(input().split())] d[a[-1]]=0 def cal(s,t): res="" for i in range(k): res+=str((9-int(s[i])-int(t[i]))%3) return res for i in range(n): for {{completion}}: try: d[cal(a[i],a[j])]+=1 except: pass ans=0 for y in d.values(): ans+=(y*(y-1))//2 print(ans)
j in range(i)
[{"input": "8 4\n0 0 0 0\n0 0 0 1\n0 0 0 2\n0 0 1 0\n0 0 2 0\n0 1 0 0\n1 0 0 0\n2 2 0 0", "output": ["1"]}, {"input": "7 4\n0 0 0 0\n0 0 0 1\n0 0 0 2\n0 0 1 0\n0 0 2 0\n0 1 0 0\n0 2 0 0", "output": ["3"]}, {"input": "9 2\n0 0\n0 1\n0 2\n1 0\n1 1\n1 2\n2 0\n2 1\n2 2", "output": ["54"]}, {"input": "20 4\n0 2 0 0\n0 2 2 2\n0 2 2 1\n0 2 0 1\n1 2 2 0\n1 2 1 0\n1 2 2 1\n1 2 0 1\n1 1 2 2\n1 1 0 2\n1 1 2 1\n1 1 1 1\n2 1 2 0\n2 1 1 2\n2 1 2 1\n2 1 1 1\n0 1 1 2\n0 0 1 0\n2 2 0 0\n2 0 0 2", "output": ["0"]}]
control_completion_005229
control_fixed
python
Complete the code in python to solve this programming problem: Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once. Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$. Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$. Code: n, k = [int(i) for i in input().split()] a = [int(i) for i in input().split()] dd = [0]*(len(a)+5) add = 0 diff = 0 moves = 0 for key, i in reversed([*enumerate(a)]): add += diff i += add diff += dd[-1] dd.pop() if i > 0: # TODO: Your code here print(moves)
n, k = [int(i) for i in input().split()] a = [int(i) for i in input().split()] dd = [0]*(len(a)+5) add = 0 diff = 0 moves = 0 for key, i in reversed([*enumerate(a)]): add += diff i += add diff += dd[-1] dd.pop() if i > 0: {{completion}} print(moves)
K = min(k, key+1) dd[-K] -= (i+K-1)//K diff += (i+K-1)//K moves += (i+K-1)//K add -= K*((i+K-1)//K)
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
block_completion_003446
block
python
Complete the code in python to solve this programming problem: Description: Julia's $$$n$$$ friends want to organize a startup in a new country they moved to. They assigned each other numbers from 1 to $$$n$$$ according to the jobs they have, from the most front-end tasks to the most back-end ones. They also estimated a matrix $$$c$$$, where $$$c_{ij} = c_{ji}$$$ is the average number of messages per month between people doing jobs $$$i$$$ and $$$j$$$.Now they want to make a hierarchy tree. It will be a binary tree with each node containing one member of the team. Some member will be selected as a leader of the team and will be contained in the root node. In order for the leader to be able to easily reach any subordinate, for each node $$$v$$$ of the tree, the following should apply: all members in its left subtree must have smaller numbers than $$$v$$$, and all members in its right subtree must have larger numbers than $$$v$$$.After the hierarchy tree is settled, people doing jobs $$$i$$$ and $$$j$$$ will be communicating via the shortest path in the tree between their nodes. Let's denote the length of this path as $$$d_{ij}$$$. Thus, the cost of their communication is $$$c_{ij} \cdot d_{ij}$$$.Your task is to find a hierarchy tree that minimizes the total cost of communication over all pairs: $$$\sum_{1 \le i &lt; j \le n} c_{ij} \cdot d_{ij}$$$. Input Specification: The first line contains an integer $$$n$$$ ($$$1 \le n \le 200$$$) – the number of team members organizing a startup. The next $$$n$$$ lines contain $$$n$$$ integers each, $$$j$$$-th number in $$$i$$$-th line is $$$c_{ij}$$$ — the estimated number of messages per month between team members $$$i$$$ and $$$j$$$ ($$$0 \le c_{ij} \le 10^9; c_{ij} = c_{ji}; c_{ii} = 0$$$). Output Specification: Output a description of a hierarchy tree that minimizes the total cost of communication. To do so, for each team member from 1 to $$$n$$$ output the number of the member in its parent node, or 0 for the leader. If there are many optimal trees, output a description of any one of them. Notes: NoteThe minimal possible total cost is $$$566 \cdot 1+239 \cdot 1+30 \cdot 1+1 \cdot 2+1 \cdot 2=839$$$: Code: n=int(input()) c=[] for _ in range(n): c.append(tuple(map(int,input().split()))) prefix_sum=[[0]*(n+1) for _ in range(n+1)] for i in range(1,n+1): temp=0 for j in range(1,n+1): temp+=c[i-1][j-1] prefix_sum[i][j]+=prefix_sum[i-1][j]+temp def get_rectangel_sum(x1,y1,x2,y2): return prefix_sum[x2+1][y2+1]-prefix_sum[x1][y2+1]-prefix_sum[x2+1][y1]+prefix_sum[x1][y1] def cost(x,y): if x>y: return 0 a=get_rectangel_sum(x,0,y,x-1) if x!=0 else 0 b=get_rectangel_sum(x,y+1,y,n-1) if y!=n-1 else 0 return a+b dp=[[float("INF")]*n for _ in range(n)] best_root_for_range=[[-1]*n for _ in range(n)] for i in range(n): dp[i][i]=0 best_root_for_range[i][i]=i def get_dp_cost(x,y): return dp[x][y] if x<=y else 0 for length in range(1,n): # actual length is length+1 for i in range(n-length): j=i+length for root in range(i,j+1): temp=cost(i,root-1)+cost(root+1,j)+get_dp_cost(i,root-1)+get_dp_cost(root+1,j) if temp<dp[i][j]: # TODO: Your code here ans=[-1]*n def assign_ans(ansecstor,x,y): if x>y: return root=best_root_for_range[x][y] ans[root]=ansecstor assign_ans(root,x,root-1) assign_ans(root,root+1,y) assign_ans(-1,0,n-1) print(*[i+1 for i in ans]) # 3 # 0 1 2 # 1 0 3 # 2 3 0 # 4 # 0 1 2 3 # 1 0 5 7 # 2 5 0 4 # 3 7 4 0 # 6 # 0 100 20 37 14 73 # 100 0 17 13 20 2 # 20 17 0 1093 900 1 # 37 13 1093 0 2 4 # 14 20 900 2 0 1 # 73 2 1 4 1 0
n=int(input()) c=[] for _ in range(n): c.append(tuple(map(int,input().split()))) prefix_sum=[[0]*(n+1) for _ in range(n+1)] for i in range(1,n+1): temp=0 for j in range(1,n+1): temp+=c[i-1][j-1] prefix_sum[i][j]+=prefix_sum[i-1][j]+temp def get_rectangel_sum(x1,y1,x2,y2): return prefix_sum[x2+1][y2+1]-prefix_sum[x1][y2+1]-prefix_sum[x2+1][y1]+prefix_sum[x1][y1] def cost(x,y): if x>y: return 0 a=get_rectangel_sum(x,0,y,x-1) if x!=0 else 0 b=get_rectangel_sum(x,y+1,y,n-1) if y!=n-1 else 0 return a+b dp=[[float("INF")]*n for _ in range(n)] best_root_for_range=[[-1]*n for _ in range(n)] for i in range(n): dp[i][i]=0 best_root_for_range[i][i]=i def get_dp_cost(x,y): return dp[x][y] if x<=y else 0 for length in range(1,n): # actual length is length+1 for i in range(n-length): j=i+length for root in range(i,j+1): temp=cost(i,root-1)+cost(root+1,j)+get_dp_cost(i,root-1)+get_dp_cost(root+1,j) if temp<dp[i][j]: {{completion}} ans=[-1]*n def assign_ans(ansecstor,x,y): if x>y: return root=best_root_for_range[x][y] ans[root]=ansecstor assign_ans(root,x,root-1) assign_ans(root,root+1,y) assign_ans(-1,0,n-1) print(*[i+1 for i in ans]) # 3 # 0 1 2 # 1 0 3 # 2 3 0 # 4 # 0 1 2 3 # 1 0 5 7 # 2 5 0 4 # 3 7 4 0 # 6 # 0 100 20 37 14 73 # 100 0 17 13 20 2 # 20 17 0 1093 900 1 # 37 13 1093 0 2 4 # 14 20 900 2 0 1 # 73 2 1 4 1 0
dp[i][j]=temp best_root_for_range[i][j]=root
[{"input": "4\n0 566 1 0\n566 0 239 30\n1 239 0 1\n0 30 1 0", "output": ["2 4 2 0"]}]
block_completion_003210
block
python
Complete the code in python to solve this programming problem: Description: You are given an array of length $$$2^n$$$. The elements of the array are numbered from $$$1$$$ to $$$2^n$$$.You have to process $$$q$$$ queries to this array. In the $$$i$$$-th query, you will be given an integer $$$k$$$ ($$$0 \le k \le n-1$$$). To process the query, you should do the following: for every $$$i \in [1, 2^n-2^k]$$$ in ascending order, do the following: if the $$$i$$$-th element was already swapped with some other element during this query, skip it; otherwise, swap $$$a_i$$$ and $$$a_{i+2^k}$$$; after that, print the maximum sum over all contiguous subsegments of the array (including the empty subsegment). For example, if the array $$$a$$$ is $$$[-3, 5, -3, 2, 8, -20, 6, -1]$$$, and $$$k = 1$$$, the query is processed as follows: the $$$1$$$-st element wasn't swapped yet, so we swap it with the $$$3$$$-rd element; the $$$2$$$-nd element wasn't swapped yet, so we swap it with the $$$4$$$-th element; the $$$3$$$-rd element was swapped already; the $$$4$$$-th element was swapped already; the $$$5$$$-th element wasn't swapped yet, so we swap it with the $$$7$$$-th element; the $$$6$$$-th element wasn't swapped yet, so we swap it with the $$$8$$$-th element. So, the array becomes $$$[-3, 2, -3, 5, 6, -1, 8, -20]$$$. The subsegment with the maximum sum is $$$[5, 6, -1, 8]$$$, and the answer to the query is $$$18$$$.Note that the queries actually change the array, i. e. after a query is performed, the array does not return to its original state, and the next query will be applied to the modified array. Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 18$$$). The second line contains $$$2^n$$$ integers $$$a_1, a_2, \dots, a_{2^n}$$$ ($$$-10^9 \le a_i \le 10^9$$$). The third line contains one integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$). Then $$$q$$$ lines follow, the $$$i$$$-th of them contains one integer $$$k$$$ ($$$0 \le k \le n-1$$$) describing the $$$i$$$-th query. Output Specification: For each query, print one integer — the maximum sum over all contiguous subsegments of the array (including the empty subsegment) after processing the query. Notes: NoteTransformation of the array in the example: $$$[-3, 5, -3, 2, 8, -20, 6, -1] \rightarrow [-3, 2, -3, 5, 6, -1, 8, -20] \rightarrow [2, -3, 5, -3, -1, 6, -20, 8] \rightarrow [5, -3, 2, -3, -20, 8, -1, 6]$$$. Code: import sys input = lambda: sys.stdin.readline().rstrip() class Node: def __init__(self, seg, suf, pref, sum) -> None: self.best = seg self.suf = suf self.pref = pref self.sum = sum def merge(a, b): seg = max(a.best, b.best, a.suf + b.pref) suf = max(b.suf, b.sum + a.suf) pref = max(a.pref, a.sum + b.pref) sum = a.sum + b.sum return Node(seg, suf, pref, sum) def single(a): v = max(a, 0) return Node(v, v, v, a) def build(v, l, r): if l + 1 == r: return [single(A[l])] else: m = (l + r) // 2 vl = build(2 * v + 1, l, m) vr = build(2 * v + 2, m, r) ans = [] for _ in range(2): for # TODO: Your code here: ans.append(merge(vl[i], vr[i])) vl, vr = vr, vl return ans N = int(input()) A = list(map(int, input().split())) Q = int(input()) M = 1 << N tree = build(0, 0, M) curr = 0 for _ in range(Q): K = int(input()) curr ^= (1 << K) print(tree[curr].best)
import sys input = lambda: sys.stdin.readline().rstrip() class Node: def __init__(self, seg, suf, pref, sum) -> None: self.best = seg self.suf = suf self.pref = pref self.sum = sum def merge(a, b): seg = max(a.best, b.best, a.suf + b.pref) suf = max(b.suf, b.sum + a.suf) pref = max(a.pref, a.sum + b.pref) sum = a.sum + b.sum return Node(seg, suf, pref, sum) def single(a): v = max(a, 0) return Node(v, v, v, a) def build(v, l, r): if l + 1 == r: return [single(A[l])] else: m = (l + r) // 2 vl = build(2 * v + 1, l, m) vr = build(2 * v + 2, m, r) ans = [] for _ in range(2): for {{completion}}: ans.append(merge(vl[i], vr[i])) vl, vr = vr, vl return ans N = int(input()) A = list(map(int, input().split())) Q = int(input()) M = 1 << N tree = build(0, 0, M) curr = 0 for _ in range(Q): K = int(input()) curr ^= (1 << K) print(tree[curr].best)
i in range((r - l) // 2)
[{"input": "3\n-3 5 -3 2 8 -20 6 -1\n3\n1\n0\n1", "output": ["18\n8\n13"]}]
control_completion_008162
control_fixed
python
Complete the code in python to solve this programming problem: Description: On a beach there are $$$n$$$ huts in a perfect line, hut $$$1$$$ being at the left and hut $$$i+1$$$ being $$$100$$$ meters to the right of hut $$$i$$$, for all $$$1 \le i \le n - 1$$$. In hut $$$i$$$ there are $$$p_i$$$ people.There are $$$m$$$ ice cream sellers, also aligned in a perfect line with all the huts. The $$$i$$$-th ice cream seller has their shop $$$x_i$$$ meters to the right of the first hut. All ice cream shops are at distinct locations, but they may be at the same location as a hut.You want to open a new ice cream shop and you wonder what the best location for your shop is. You can place your ice cream shop anywhere on the beach (not necessarily at an integer distance from the first hut) as long as it is aligned with the huts and the other ice cream shops, even if there is already another ice cream shop or a hut at that location. You know that people would come to your shop only if it is strictly closer to their hut than any other ice cream shop.If every person living in the huts wants to buy exactly one ice cream, what is the maximum number of ice creams that you can sell if you place the shop optimally? Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 200\,000$$$, $$$1 \le m \le 200\,000$$$) — the number of huts and the number of ice cream sellers. The second line contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le 10^9$$$) — the number of people in each hut. The third line contains $$$m$$$ integers $$$x_1, x_2, \ldots, x_m$$$ ($$$0 \le x_i \le 10^9$$$, $$$x_i \ne x_j$$$ for $$$i \ne j$$$) — the location of each ice cream shop. Output Specification: Print the maximum number of ice creams that can be sold by choosing optimally the location of the new shop. Notes: NoteIn the first sample, you can place the shop (coloured orange in the picture below) $$$150$$$ meters to the right of the first hut (for example) so that it is the closest shop to the first two huts, which have $$$2$$$ and $$$5$$$ people, for a total of $$$7$$$ sold ice creams. In the second sample, you can place the shop $$$170$$$ meters to the right of the first hut (for example) so that it is the closest shop to the last two huts, which have $$$7$$$ and $$$8$$$ people, for a total of $$$15$$$ sold ice creams. Code: N, M = [int(x) for x in input().split()] hut = [int(x) for x in input().split()] shop = [int(x) for x in input().split()] shop = sorted([-1e9] + shop + [1e9]) events = [] j = 0 for i in range(N): while shop[j] < 100*i: j += 1 if # TODO: Your code here: d = min(100*i - shop[j-1], shop[j] - 100*i) events.append((100*i-d, hut[i])) events.append((100*i+d, -hut[i])) events.sort() cont = 0 max = 0 for a in events: cont += a[1] if cont > max: max = cont print(max)
N, M = [int(x) for x in input().split()] hut = [int(x) for x in input().split()] shop = [int(x) for x in input().split()] shop = sorted([-1e9] + shop + [1e9]) events = [] j = 0 for i in range(N): while shop[j] < 100*i: j += 1 if {{completion}}: d = min(100*i - shop[j-1], shop[j] - 100*i) events.append((100*i-d, hut[i])) events.append((100*i+d, -hut[i])) events.sort() cont = 0 max = 0 for a in events: cont += a[1] if cont > max: max = cont print(max)
shop[j] != 100 * i
[{"input": "3 1\n2 5 6\n169", "output": ["7"]}, {"input": "4 2\n1 2 7 8\n35 157", "output": ["15"]}, {"input": "4 1\n272203905 348354708 848256926 939404176\n20", "output": ["2136015810"]}, {"input": "3 2\n1 1 1\n300 99", "output": ["2"]}]
control_completion_001129
control_fixed
python
Complete the code in python to solve this programming problem: Description: $$$m$$$ chairs are arranged in a circle sequentially. The chairs are numbered from $$$0$$$ to $$$m-1$$$. $$$n$$$ people want to sit in these chairs. The $$$i$$$-th of them wants at least $$$a[i]$$$ empty chairs both on his right and left side. More formally, if the $$$i$$$-th person sits in the $$$j$$$-th chair, then no one else should sit in the following chairs: $$$(j-a[i]) \bmod m$$$, $$$(j-a[i]+1) \bmod m$$$, ... $$$(j+a[i]-1) \bmod m$$$, $$$(j+a[i]) \bmod m$$$.Decide if it is possible to sit down for all of them, under the given limitations. Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 5 \cdot 10^4$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$) — the number of people and the number of chairs. The next line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ... $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the minimum number of empty chairs, on both sides of the $$$i$$$-th person. It is guaranteed that the sum of $$$n$$$ over all test cases will not exceed $$$10^5$$$. Output Specification: For each test case print "YES" (without quotes) if it is possible for everyone to sit down and fulfil the restrictions, and "NO" (without quotes) otherwise. You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" will all be recognized as positive answers). Notes: NoteTest case $$$1$$$: $$$n&gt;m$$$, so they can not sit down.Test case $$$2$$$: the first person can sit $$$2$$$-nd and the second person can sit in the $$$0$$$-th chair. Both of them want at least $$$1$$$ empty chair on both sides, chairs $$$1$$$ and $$$3$$$ are free, so this is a good solution.Test case $$$3$$$: if the second person sits down somewhere, he needs $$$2$$$ empty chairs, both on his right and on his left side, so it is impossible to find a place for the first person, because there are only $$$5$$$ chairs.Test case $$$4$$$: they can sit in the $$$1$$$-st, $$$4$$$-th, $$$7$$$-th chairs respectively. Code: for T in range (int(input())) : n,m = map(int, input().strip().split()) a = sorted(list(map(int,input().strip().split())),reverse=True) m -= 2*a[0] + 1 cont = 0 for i in range(1,n) : if # TODO: Your code here: break m -= a[i] + 1 cont +=1 if cont == n-1 : print('YES') else : print ('NO')
for T in range (int(input())) : n,m = map(int, input().strip().split()) a = sorted(list(map(int,input().strip().split())),reverse=True) m -= 2*a[0] + 1 cont = 0 for i in range(1,n) : if {{completion}}: break m -= a[i] + 1 cont +=1 if cont == n-1 : print('YES') else : print ('NO')
m <= 0
[{"input": "6\n3 2\n1 1 1\n2 4\n1 1\n2 5\n2 1\n3 8\n1 2 1\n4 12\n1 2 1 3\n4 19\n1 2 1 3", "output": ["NO\nYES\nNO\nYES\nNO\nYES"]}]
control_completion_001003
control_fixed
python
Complete the code in python to solve this programming problem: Description: There is a grid, consisting of $$$2$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$2$$$ from top to bottom. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right.The robot starts in a cell $$$(1, 1)$$$. In one second, it can perform either of two actions: move into a cell adjacent by a side: up, right, down or left; remain in the same cell. The robot is not allowed to move outside the grid.Initially, all cells, except for the cell $$$(1, 1)$$$, are locked. Each cell $$$(i, j)$$$ contains a value $$$a_{i,j}$$$ — the moment that this cell gets unlocked. The robot can only move into a cell $$$(i, j)$$$ if at least $$$a_{i,j}$$$ seconds have passed before the move.The robot should visit all cells without entering any cell twice or more (cell $$$(1, 1)$$$ is considered entered at the start). It can finish in any cell.What is the fastest the robot can achieve that? Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of testcases. The first line of each testcase contains a single integer $$$m$$$ ($$$2 \le m \le 2 \cdot 10^5$$$) — the number of columns of the grid. The $$$i$$$-th of the next $$$2$$$ lines contains $$$m$$$ integers $$$a_{i,1}, a_{i,2}, \dots, a_{i,m}$$$ ($$$0 \le a_{i,j} \le 10^9$$$) — the moment of time each cell gets unlocked. $$$a_{1,1} = 0$$$. If $$$a_{i,j} = 0$$$, then cell $$$(i, j)$$$ is unlocked from the start. The sum of $$$m$$$ over all testcases doesn't exceed $$$2 \cdot 10^5$$$. Output Specification: For each testcase, print a single integer — the minimum amount of seconds that the robot can take to visit all cells without entering any cell twice or more. Code: def readline(): line = input() while not line: line = input() return line def main(): t = int(readline()) for _ in range(t): m = int(readline()) a = [ [], [] ] a[0] += list(map(int, readline().split())) a[1] = list(map(int, readline().split())) h = [ [None] * m, [None] * m ] h[0][m-1] = max(a[1][m-1]+1, a[0][m-1]+1+1) h[1][m-1] = max(a[0][m-1]+1, a[1][m-1]+1+1) for i in reversed(range(m-1)): h[0][i] = max(a[1][i]+1, h[0][i+1]+1, a[0][i]+(i!=0)+(m-i-1)*2+1) h[1][i] = max(a[0][i]+1, h[1][i+1]+1, a[1][i]+1+(m-i-1)*2+1) pos = (0,0) t = 0 best_total_time = 10**10 for i in range(2*m-1): if i % 2 == 0: total_time = max(t+(m-i//2-1)*2+1, h[pos[0]][pos[1]]) # print(pos, t,total_time) best_total_time = min(best_total_time, total_time) if i % 4 == 0: # abajo pos = (pos[0]+1, pos[1]) elif # TODO: Your code here: # derecha pos = (pos[0], pos[1]+1) elif (i-2) % 4 == 0: # arr pos = (pos[0]-1, pos[1]) elif (i-3) % 4 == 0: # derecha pos = (pos[0], pos[1]+1) t = max(a[pos[0]][pos[1]] + 1, t+1) # for line in h: # print(line) print(best_total_time) main()
def readline(): line = input() while not line: line = input() return line def main(): t = int(readline()) for _ in range(t): m = int(readline()) a = [ [], [] ] a[0] += list(map(int, readline().split())) a[1] = list(map(int, readline().split())) h = [ [None] * m, [None] * m ] h[0][m-1] = max(a[1][m-1]+1, a[0][m-1]+1+1) h[1][m-1] = max(a[0][m-1]+1, a[1][m-1]+1+1) for i in reversed(range(m-1)): h[0][i] = max(a[1][i]+1, h[0][i+1]+1, a[0][i]+(i!=0)+(m-i-1)*2+1) h[1][i] = max(a[0][i]+1, h[1][i+1]+1, a[1][i]+1+(m-i-1)*2+1) pos = (0,0) t = 0 best_total_time = 10**10 for i in range(2*m-1): if i % 2 == 0: total_time = max(t+(m-i//2-1)*2+1, h[pos[0]][pos[1]]) # print(pos, t,total_time) best_total_time = min(best_total_time, total_time) if i % 4 == 0: # abajo pos = (pos[0]+1, pos[1]) elif {{completion}}: # derecha pos = (pos[0], pos[1]+1) elif (i-2) % 4 == 0: # arr pos = (pos[0]-1, pos[1]) elif (i-3) % 4 == 0: # derecha pos = (pos[0], pos[1]+1) t = max(a[pos[0]][pos[1]] + 1, t+1) # for line in h: # print(line) print(best_total_time) main()
(i-1) % 4 == 0
[{"input": "4\n\n3\n\n0 0 1\n\n4 3 2\n\n5\n\n0 4 8 12 16\n\n2 6 10 14 18\n\n4\n\n0 10 10 10\n\n10 10 10 10\n\n2\n\n0 0\n\n0 0", "output": ["5\n19\n17\n3"]}]
control_completion_008132
control_fixed
python
Complete the code in python to solve this programming problem: Description: Alina has discovered a weird language, which contains only $$$4$$$ words: $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$. It also turned out that there are no spaces in this language: a sentence is written by just concatenating its words into a single string.Alina has found one such sentence $$$s$$$ and she is curious: is it possible that it consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$?In other words, determine, if it's possible to concatenate these $$$a+b+c+d$$$ words in some order so that the resulting string is $$$s$$$. Each of the $$$a+b+c+d$$$ words must be used exactly once in the concatenation, but you can choose the order in which they are concatenated. Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains four integers $$$a$$$, $$$b$$$, $$$c$$$, $$$d$$$ ($$$0\le a,b,c,d\le 2\cdot 10^5$$$) — the number of times that words $$$\texttt{A}$$$, $$$\texttt{B}$$$, $$$\texttt{AB}$$$, $$$\texttt{BA}$$$ respectively must be used in the sentence. The second line contains the string $$$s$$$ ($$$s$$$ consists only of the characters $$$\texttt{A}$$$ and $$$\texttt{B}$$$, $$$1\le |s| \le 2\cdot 10^5$$$, $$$|s|=a+b+2c+2d$$$)  — the sentence. Notice that the condition $$$|s|=a+b+2c+2d$$$ (here $$$|s|$$$ denotes the length of the string $$$s$$$) is equivalent to the fact that $$$s$$$ is as long as the concatenation of the $$$a+b+c+d$$$ words. The sum of the lengths of $$$s$$$ over all test cases doesn't exceed $$$2\cdot 10^5$$$. Output Specification: For each test case output $$$\texttt{YES}$$$ if it is possible that the sentence $$$s$$$ consists of precisely $$$a$$$ words $$$\texttt{A}$$$, $$$b$$$ words $$$\texttt{B}$$$, $$$c$$$ words $$$\texttt{AB}$$$, and $$$d$$$ words $$$\texttt{BA}$$$, and $$$\texttt{NO}$$$ otherwise. You can output each letter in any case. Notes: NoteIn the first test case, the sentence $$$s$$$ is $$$\texttt{B}$$$. Clearly, it can't consist of a single word $$$\texttt{A}$$$, so the answer is $$$\texttt{NO}$$$.In the second test case, the sentence $$$s$$$ is $$$\texttt{AB}$$$, and it's possible that it consists of a single word $$$\texttt{AB}$$$, so the answer is $$$\texttt{YES}$$$.In the third test case, the sentence $$$s$$$ is $$$\texttt{ABAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{B} = \texttt{ABAB}$$$.In the fourth test case, the sentence $$$s$$$ is $$$\texttt{ABAAB}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{AB}$$$, and one word $$$\texttt{BA}$$$, as $$$\texttt{A} + \texttt{BA} + \texttt{AB} = \texttt{ABAAB}$$$. In the fifth test case, the sentence $$$s$$$ is $$$\texttt{BAABBABBAA}$$$, and it's possible that it consists of one word $$$\texttt{A}$$$, one word $$$\texttt{B}$$$, two words $$$\texttt{AB}$$$, and two words $$$\texttt{BA}$$$, as $$$\texttt{BA} + \texttt{AB} + \texttt{B} + \texttt{AB} + \texttt{BA} + \texttt{A}= \texttt{BAABBABBAA}$$$. Code: def canmake(s,a,b,c,d): anum = s.count('A') bnum = s.count('B') cnum = s.count('AB') dnum = s.count('BA') if cnum < c or dnum < d or anum != a + c + d or bnum != b + c + d: return False n=len(s) ans=0 abls=[] bals=[] l=0 while l<n: while l<n-1 and s[l]==s[l+1]: l+=1 r=l while # TODO: Your code here: r+=1 if s[l]== s[r]=='B': ans+=(r-l+1)//2 if s[l]==s[r]=='A': ans+=(r-l+1)//2 if s[l]=='A' and s[r]=='B': abls.append((r-l+1)//2) if s[l]=='B' and s[r]=='A': bals.append((r-l+1)//2) l=r+1 abls.sort() bals.sort() for i in abls: if i<=c: c-=i else: d-=i-c-1 c = 0 for i in bals: if i<=d: d-=i else: c-=i-d-1 d = 0 return (c+d)<=ans t=int(input()) for _ in range(t): a,b,c,d=[int(x) for x in input().split()] s=input() res=canmake(s,a,b,c,d) if res: print('YES') else: print('NO')
def canmake(s,a,b,c,d): anum = s.count('A') bnum = s.count('B') cnum = s.count('AB') dnum = s.count('BA') if cnum < c or dnum < d or anum != a + c + d or bnum != b + c + d: return False n=len(s) ans=0 abls=[] bals=[] l=0 while l<n: while l<n-1 and s[l]==s[l+1]: l+=1 r=l while {{completion}}: r+=1 if s[l]== s[r]=='B': ans+=(r-l+1)//2 if s[l]==s[r]=='A': ans+=(r-l+1)//2 if s[l]=='A' and s[r]=='B': abls.append((r-l+1)//2) if s[l]=='B' and s[r]=='A': bals.append((r-l+1)//2) l=r+1 abls.sort() bals.sort() for i in abls: if i<=c: c-=i else: d-=i-c-1 c = 0 for i in bals: if i<=d: d-=i else: c-=i-d-1 d = 0 return (c+d)<=ans t=int(input()) for _ in range(t): a,b,c,d=[int(x) for x in input().split()] s=input() res=canmake(s,a,b,c,d) if res: print('YES') else: print('NO')
r<n-1 and s[r]!=s[r+1]
[{"input": "8\n1 0 0 0\nB\n0 0 1 0\nAB\n1 1 0 1\nABAB\n1 0 1 1\nABAAB\n1 1 2 2\nBAABBABBAA\n1 1 2 3\nABABABBAABAB\n2 3 5 4\nAABAABBABAAABABBABBBABB\n1 3 3 10\nBBABABABABBBABABABABABABAABABA", "output": ["NO\nYES\nYES\nYES\nYES\nYES\nNO\nYES"]}]
control_completion_001183
control_fixed
python
Complete the code in python to solve this programming problem: Description: This is the hard version of this problem. In this version, we have queries. Note that we do not have multiple test cases in this version. You can make hacks only if both versions of the problem are solved.An array $$$b$$$ of length $$$m$$$ is good if for all $$$i$$$ the $$$i$$$-th element is greater than or equal to $$$i$$$. In other words, $$$b$$$ is good if and only if $$$b_i \geq i$$$ for all $$$i$$$ ($$$1 \leq i \leq m$$$).You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and you are asked $$$q$$$ queries. In each query, you are given two integers $$$p$$$ and $$$x$$$ ($$$1 \leq p,x \leq n$$$). You have to do $$$a_p := x$$$ (assign $$$x$$$ to $$$a_p$$$). In the updated array, find the number of pairs of indices $$$(l, r)$$$, where $$$1 \le l \le r \le n$$$, such that the array $$$[a_l, a_{l+1}, \ldots, a_r]$$$ is good.Note that all queries are independent, which means after each query, the initial array $$$a$$$ is restored. Input Specification: The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$). The third line contains an integer $$$q$$$ ($$$1 \leq q \leq 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contains two integers $$$p_j$$$ and $$$x_j$$$ ($$$1 \leq p_j, x_j \leq n$$$) – the description of the $$$j$$$-th query. Output Specification: For each query, print the number of suitable pairs of indices after making the change. Notes: NoteHere are notes for first example.In first query, after update $$$a=[2,4,1,4]$$$. Now $$$(1,1)$$$, $$$(2,2)$$$, $$$(3,3)$$$, $$$(4,4)$$$, $$$(1,2)$$$, and $$$(3,4)$$$ are suitable pairs.In second query, after update $$$a=[2,4,3,4]$$$. Now all subarrays of $$$a$$$ are good.In third query, after update $$$a=[2,1,1,4]$$$. Now $$$(1,1)$$$, $$$(2,2)$$$, $$$(3,3)$$$, $$$(4,4)$$$, and $$$(3,4)$$$ are suitable. Code: from sys import stdin, stdout from collections import defaultdict n = int(stdin.readline()) a = [int(x) for x in stdin.readline().split()] left = 0 right = 0 right_array = [] right_prefix = [0] answer = 0 second_right = 0 second_right_array = [] second_right_prefix = [0] while left < n: if right < left: right = left while right < n and a[right] >= right-left+1: right += 1 right_array.append(right) right_prefix.append(right_prefix[-1]+right_array[-1]) answer += right - left if second_right <= right: second_right = right + 1 if second_right > n: second_right = n while second_right < n and a[second_right] >= second_right-left+1: second_right += 1 second_right_array.append(second_right) second_right_prefix.append(second_right_prefix[-1]+second_right_array[-1]) left += 1 p_to_right = defaultdict(list) for i in range(n): p_to_right[right_array[i]].append(i) q = int(stdin.readline()) for _ in range(q): p, x = [int(z) for z in stdin.readline().split()] p -= 1 if x == a[p]: adjustment = 0 elif x < a[p]: if right_array[-1] <= p: adjustment = 0 else: upper = n-1 lower = -1 while upper - lower > 1: candidate = (upper + lower)//2 if right_array[candidate] > p: upper = candidate else: # TODO: Your code here if upper > p-x: adjustment = 0 else: adjustment = -(right_prefix[p-x+1] - right_prefix[upper] - p*(p-x-upper+1)) else: if len(p_to_right[p]) == 0: adjustment = 0 elif p_to_right[p][0] > p-a[p] or p_to_right[p][-1] < p-x+1: adjustment = 0 else: lower = max(p-x+1, p_to_right[p][0]) upper = min(p-a[p], p_to_right[p][-1]) adjustment = second_right_prefix[upper+1] - second_right_prefix[lower] - p*(upper-lower+1) stdout.write(str(answer+adjustment)+'\n')
from sys import stdin, stdout from collections import defaultdict n = int(stdin.readline()) a = [int(x) for x in stdin.readline().split()] left = 0 right = 0 right_array = [] right_prefix = [0] answer = 0 second_right = 0 second_right_array = [] second_right_prefix = [0] while left < n: if right < left: right = left while right < n and a[right] >= right-left+1: right += 1 right_array.append(right) right_prefix.append(right_prefix[-1]+right_array[-1]) answer += right - left if second_right <= right: second_right = right + 1 if second_right > n: second_right = n while second_right < n and a[second_right] >= second_right-left+1: second_right += 1 second_right_array.append(second_right) second_right_prefix.append(second_right_prefix[-1]+second_right_array[-1]) left += 1 p_to_right = defaultdict(list) for i in range(n): p_to_right[right_array[i]].append(i) q = int(stdin.readline()) for _ in range(q): p, x = [int(z) for z in stdin.readline().split()] p -= 1 if x == a[p]: adjustment = 0 elif x < a[p]: if right_array[-1] <= p: adjustment = 0 else: upper = n-1 lower = -1 while upper - lower > 1: candidate = (upper + lower)//2 if right_array[candidate] > p: upper = candidate else: {{completion}} if upper > p-x: adjustment = 0 else: adjustment = -(right_prefix[p-x+1] - right_prefix[upper] - p*(p-x-upper+1)) else: if len(p_to_right[p]) == 0: adjustment = 0 elif p_to_right[p][0] > p-a[p] or p_to_right[p][-1] < p-x+1: adjustment = 0 else: lower = max(p-x+1, p_to_right[p][0]) upper = min(p-a[p], p_to_right[p][-1]) adjustment = second_right_prefix[upper+1] - second_right_prefix[lower] - p*(upper-lower+1) stdout.write(str(answer+adjustment)+'\n')
lower = candidate
[{"input": "4\n2 4 1 4\n3\n2 4\n3 3\n2 1", "output": ["6\n10\n5"]}, {"input": "5\n1 1 3 2 1\n3\n1 3\n2 5\n4 5", "output": ["7\n9\n8"]}]
block_completion_007068
block
python
Complete the code in python to solve this programming problem: Description: You are given an array $$$a$$$ consisting of $$$n$$$ integers. You should divide $$$a$$$ into continuous non-empty subarrays (there are $$$2^{n-1}$$$ ways to do that).Let $$$s=a_l+a_{l+1}+\ldots+a_r$$$. The value of a subarray $$$a_l, a_{l+1}, \ldots, a_r$$$ is: $$$(r-l+1)$$$ if $$$s&gt;0$$$, $$$0$$$ if $$$s=0$$$, $$$-(r-l+1)$$$ if $$$s&lt;0$$$. What is the maximum sum of values you can get with a partition? Input Specification: The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 5 \cdot 10^5$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 5 \cdot 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$). It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$. Output Specification: For each test case print a single integer — the maximum sum of values you can get with an optimal parition. Notes: NoteTest case $$$1$$$: one optimal partition is $$$[1, 2]$$$, $$$[-3]$$$. $$$1+2&gt;0$$$ so the value of $$$[1, 2]$$$ is $$$2$$$. $$$-3&lt;0$$$, so the value of $$$[-3]$$$ is $$$-1$$$. $$$2+(-1)=1$$$.Test case $$$2$$$: the optimal partition is $$$[0, -2, 3]$$$, $$$[-4]$$$, and the sum of values is $$$3+(-1)=2$$$. Code: from collections import Counter, defaultdict, deque import bisect from sys import stdin, stdout from itertools import repeat import math MOD = 998244353 input = stdin.readline finp = [int(x) for x in stdin.buffer.read().split()] def inp(force_list=False): re = list(map(int, input().split())) if len(re) == 1 and not force_list: return re[0] return re def inst(): return input().strip() def gcd(x, y): while(y): x, y = y, x % y return x def qmod(a, b, mod=MOD): res = 1 while b: if b&1: res = (res*a)%mod b >>= 1 a = (a*a)%mod return res def inv(a): return qmod(a, MOD-2) INF = 1<<30 class Seg(object): def __init__(self, n): self._da = [-INF] * (n * 5) self._op = [-INF] * (n * 5) def update(self, p): self._op[p] = max(self._op[p*2], self._op[p*2+1]) def modify(self, pos, x, p, l, r): if l==r-1: self._da[p] = self._op[p] = x return mid = (l+r)//2 if pos < mid: self.modify(pos, x, p*2, l, mid) else: # TODO: Your code here self.update(p) def query(self, x, y, p, l, r): if x <= l and r <= y: return self._op[p] if x >= r or y<=l: return -INF mid = (l+r)//2 return max(self.query(x, y, p*2, l, mid), self.query(x, y, p*2+1, mid, r)) class Fenwick(object): def __init__(self, n): self._da = [-INF] * (n+2) self._mx = n+2 def max(self, x): res = -INF while x>0: res = max(res, self._da[x]) x = (x&(x+1))-1 return res def modify(self, p, x): while p < self._mx: self._da[p] = max(self._da[p], x) p |= p+1 def my_main(): # print(500000) # for i in range(500000): # print(1) # print(-1000000000) ii = 0 kase = finp[ii];ii+=1 pans = [] for skase in range(kase): # print("Case #%d: " % (skase+1), end='') n = finp[ii];ii+=1 da = finp[ii:ii+n];ii+=n pref = [0] for i in da: pref.append(pref[-1] + i) spos, sneg = sorted([(pref[i], -i) for i, v in enumerate(pref)]), sorted([(pref[i], i) for i, v in enumerate(pref)]) ordpos, ordneg = [0] * (n+1), [0] * (n+1) pfen, nfen = Fenwick(n), Fenwick(n) dmx = {} for i in range(n+1): ordpos[-spos[i][-1]] = i ordneg[sneg[i][-1]] = i dp = [0] * (n+1) dmx[0] = 0 pfen.modify(ordpos[0], 0) nfen.modify(n+1-ordneg[0], 0) for i in range(1, n+1): dp[i] = max(i+pfen.max(ordpos[i]), nfen.max(n+1-ordneg[i])-i, dmx.get(pref[i], -INF)) pfen.modify(ordpos[i], dp[i]-i) nfen.modify(n+1-ordneg[i], dp[i]+i) if dp[i] > dmx.get(pref[i], -INF): dmx[pref[i]] = dp[i] pans.append(str(dp[n])) print('\n'.join(pans)) my_main()
from collections import Counter, defaultdict, deque import bisect from sys import stdin, stdout from itertools import repeat import math MOD = 998244353 input = stdin.readline finp = [int(x) for x in stdin.buffer.read().split()] def inp(force_list=False): re = list(map(int, input().split())) if len(re) == 1 and not force_list: return re[0] return re def inst(): return input().strip() def gcd(x, y): while(y): x, y = y, x % y return x def qmod(a, b, mod=MOD): res = 1 while b: if b&1: res = (res*a)%mod b >>= 1 a = (a*a)%mod return res def inv(a): return qmod(a, MOD-2) INF = 1<<30 class Seg(object): def __init__(self, n): self._da = [-INF] * (n * 5) self._op = [-INF] * (n * 5) def update(self, p): self._op[p] = max(self._op[p*2], self._op[p*2+1]) def modify(self, pos, x, p, l, r): if l==r-1: self._da[p] = self._op[p] = x return mid = (l+r)//2 if pos < mid: self.modify(pos, x, p*2, l, mid) else: {{completion}} self.update(p) def query(self, x, y, p, l, r): if x <= l and r <= y: return self._op[p] if x >= r or y<=l: return -INF mid = (l+r)//2 return max(self.query(x, y, p*2, l, mid), self.query(x, y, p*2+1, mid, r)) class Fenwick(object): def __init__(self, n): self._da = [-INF] * (n+2) self._mx = n+2 def max(self, x): res = -INF while x>0: res = max(res, self._da[x]) x = (x&(x+1))-1 return res def modify(self, p, x): while p < self._mx: self._da[p] = max(self._da[p], x) p |= p+1 def my_main(): # print(500000) # for i in range(500000): # print(1) # print(-1000000000) ii = 0 kase = finp[ii];ii+=1 pans = [] for skase in range(kase): # print("Case #%d: " % (skase+1), end='') n = finp[ii];ii+=1 da = finp[ii:ii+n];ii+=n pref = [0] for i in da: pref.append(pref[-1] + i) spos, sneg = sorted([(pref[i], -i) for i, v in enumerate(pref)]), sorted([(pref[i], i) for i, v in enumerate(pref)]) ordpos, ordneg = [0] * (n+1), [0] * (n+1) pfen, nfen = Fenwick(n), Fenwick(n) dmx = {} for i in range(n+1): ordpos[-spos[i][-1]] = i ordneg[sneg[i][-1]] = i dp = [0] * (n+1) dmx[0] = 0 pfen.modify(ordpos[0], 0) nfen.modify(n+1-ordneg[0], 0) for i in range(1, n+1): dp[i] = max(i+pfen.max(ordpos[i]), nfen.max(n+1-ordneg[i])-i, dmx.get(pref[i], -INF)) pfen.modify(ordpos[i], dp[i]-i) nfen.modify(n+1-ordneg[i], dp[i]+i) if dp[i] > dmx.get(pref[i], -INF): dmx[pref[i]] = dp[i] pans.append(str(dp[n])) print('\n'.join(pans)) my_main()
self.modify(pos, x, p*2 + 1, mid, r)
[{"input": "5\n\n3\n\n1 2 -3\n\n4\n\n0 -2 3 -4\n\n5\n\n-1 -2 3 -1 -1\n\n6\n\n-1 2 -3 4 -5 6\n\n7\n\n1 -1 -1 1 -1 -1 1", "output": ["1\n2\n1\n6\n-1"]}]
block_completion_001050
block
python
Complete the code in python to solve this programming problem: Description: There are $$$n+1$$$ teleporters on a straight line, located in points $$$0$$$, $$$a_1$$$, $$$a_2$$$, $$$a_3$$$, ..., $$$a_n$$$. It's possible to teleport from point $$$x$$$ to point $$$y$$$ if there are teleporters in both of those points, and it costs $$$(x-y)^2$$$ energy.You want to install some additional teleporters so that it is possible to get from the point $$$0$$$ to the point $$$a_n$$$ (possibly through some other teleporters) spending no more than $$$m$$$ energy in total. Each teleporter you install must be located in an integer point.What is the minimum number of teleporters you have to install? Input Specification: The first line contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_1 &lt; a_2 &lt; a_3 &lt; \dots &lt; a_n \le 10^9$$$). The third line contains one integer $$$m$$$ ($$$a_n \le m \le 10^{18}$$$). Output Specification: Print one integer — the minimum number of teleporters you have to install so that it is possible to get from $$$0$$$ to $$$a_n$$$ spending at most $$$m$$$ energy. It can be shown that it's always possible under the constraints from the input format. Code: from collections import Counter n, a, m = int(input()), [*map(int, input().split())], int(input()) def energy(l, t): x,y = divmod(l, t+1) return x*x*(t+1-y)+(x+1)*(x+1)*y def getdiff(l, diff): lo, hi = 0, l while lo < hi: mid = lo + hi >> 1 if energy(l, mid) - energy(l, mid+1) < diff: hi = mid else: # TODO: Your code here return lo, energy(l, lo) def getsum(d, c=0): a1,a2 = zip(*[getdiff(i, d) for i,_ in a]); return (0 if c else sum(a1[i]*x for i, (_, x) in enumerate(a)), sum(a2[i]*x for i, (_, x) in enumerate(a))) a = [0] + a a = Counter([a[i+1]-a[i] for i in range(n)]).items() lo, hi = 1, m while lo < hi: mid = lo + hi >> 1 if getsum(mid, 1)[1] > m: hi = mid else: lo = mid + 1 lo-=1 a1, a2 = getsum(lo) print(a1-(m-a2)//lo if lo else a1)
from collections import Counter n, a, m = int(input()), [*map(int, input().split())], int(input()) def energy(l, t): x,y = divmod(l, t+1) return x*x*(t+1-y)+(x+1)*(x+1)*y def getdiff(l, diff): lo, hi = 0, l while lo < hi: mid = lo + hi >> 1 if energy(l, mid) - energy(l, mid+1) < diff: hi = mid else: {{completion}} return lo, energy(l, lo) def getsum(d, c=0): a1,a2 = zip(*[getdiff(i, d) for i,_ in a]); return (0 if c else sum(a1[i]*x for i, (_, x) in enumerate(a)), sum(a2[i]*x for i, (_, x) in enumerate(a))) a = [0] + a a = Counter([a[i+1]-a[i] for i in range(n)]).items() lo, hi = 1, m while lo < hi: mid = lo + hi >> 1 if getsum(mid, 1)[1] > m: hi = mid else: lo = mid + 1 lo-=1 a1, a2 = getsum(lo) print(a1-(m-a2)//lo if lo else a1)
lo = mid + 1
[{"input": "2\n1 5\n7", "output": ["2"]}, {"input": "2\n1 5\n6", "output": ["3"]}, {"input": "1\n5\n5", "output": ["4"]}, {"input": "1\n1000000000\n1000000043", "output": ["999999978"]}]
block_completion_003462
block
python
Complete the code in python to solve this programming problem: Description: You are given a string $$$s$$$ consisting of $$$n$$$ characters. Each character of $$$s$$$ is either 0 or 1.A substring of $$$s$$$ is a contiguous subsequence of its characters.You have to choose two substrings of $$$s$$$ (possibly intersecting, possibly the same, possibly non-intersecting — just any two substrings). After choosing them, you calculate the value of the chosen pair of substrings as follows: let $$$s_1$$$ be the first substring, $$$s_2$$$ be the second chosen substring, and $$$f(s_i)$$$ be the integer such that $$$s_i$$$ is its binary representation (for example, if $$$s_i$$$ is 11010, $$$f(s_i) = 26$$$); the value is the bitwise OR of $$$f(s_1)$$$ and $$$f(s_2)$$$. Calculate the maximum possible value you can get, and print it in binary representation without leading zeroes. Input Specification: The first line contains one integer $$$n$$$ — the number of characters in $$$s$$$. The second line contains $$$s$$$ itself, consisting of exactly $$$n$$$ characters 0 and/or 1. All non-example tests in this problem are generated randomly: every character of $$$s$$$ is chosen independently of other characters; for each character, the probability of it being 1 is exactly $$$\frac{1}{2}$$$. This problem has exactly $$$40$$$ tests. Tests from $$$1$$$ to $$$3$$$ are the examples; tests from $$$4$$$ to $$$40$$$ are generated randomly. In tests from $$$4$$$ to $$$10$$$, $$$n = 5$$$; in tests from $$$11$$$ to $$$20$$$, $$$n = 1000$$$; in tests from $$$21$$$ to $$$40$$$, $$$n = 10^6$$$. Hacks are forbidden in this problem. Output Specification: Print the maximum possible value you can get in binary representation without leading zeroes. Notes: NoteIn the first example, you can choose the substrings 11010 and 101. $$$f(s_1) = 26$$$, $$$f(s_2) = 5$$$, their bitwise OR is $$$31$$$, and the binary representation of $$$31$$$ is 11111.In the second example, you can choose the substrings 1110010 and 11100. Code: n = input() s = int(input(),2) res = 0 for i in range(100): # TODO: Your code here ans = bin(res)[2:] print(ans)
n = input() s = int(input(),2) res = 0 for i in range(100): {{completion}} ans = bin(res)[2:] print(ans)
res = max(res,(s | (s >> i)))
[{"input": "5\n11010", "output": ["11111"]}, {"input": "7\n1110010", "output": ["1111110"]}, {"input": "4\n0000", "output": ["0"]}]
block_completion_002157
block
python
Complete the code in python to solve this programming problem: Description: A class of students got bored wearing the same pair of shoes every day, so they decided to shuffle their shoes among themselves. In this problem, a pair of shoes is inseparable and is considered as a single object.There are $$$n$$$ students in the class, and you are given an array $$$s$$$ in non-decreasing order, where $$$s_i$$$ is the shoe size of the $$$i$$$-th student. A shuffling of shoes is valid only if no student gets their own shoes and if every student gets shoes of size greater than or equal to their size. You have to output a permutation $$$p$$$ of $$$\{1,2,\ldots,n\}$$$ denoting a valid shuffling of shoes, where the $$$i$$$-th student gets the shoes of the $$$p_i$$$-th student ($$$p_i \ne i$$$). And output $$$-1$$$ if a valid shuffling does not exist.A permutation is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array) and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array). Input Specification: Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1\leq n\leq10^5$$$) — the number of students. The second line of each test case contains $$$n$$$ integers $$$s_1, s_2,\ldots,s_n$$$ ($$$1\leq s_i\leq10^9$$$, and for all $$$1\le i&lt;n$$$, $$$s_i\le s_{i+1}$$$) — the shoe sizes of the students. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$. Output Specification: For each test case, print the answer in a single line using the following format. If a valid shuffling does not exist, print the number $$$-1$$$ as the answer. If a valid shuffling exists, print $$$n$$$ space-separated integers — a permutation $$$p$$$ of $$$1,2,\ldots,n$$$ denoting a valid shuffling of shoes where the $$$i$$$-th student gets the shoes of the $$$p_i$$$-th student. If there are multiple answers, then print any of them. Notes: NoteIn the first test case, any permutation $$$p$$$ of $$$1,\ldots,n$$$ where $$$p_i\ne i$$$ would represent a valid shuffling since all students have equal shoe sizes, and thus anyone can wear anyone's shoes.In the second test case, it can be shown that no valid shuffling is possible. Code: import collections import sys input = sys.stdin.readline for _ in range(int(input())): n = int(input()) data = dict(collections.Counter(map(int, input().split()))) if min(list(data.values())) > 1: last = 1 for i in data.keys(): print(last + data[i] - 1, end=' ') for j in range(last, last + data[i] - 1): # TODO: Your code here last = last + data[i] print() else: print(-1)
import collections import sys input = sys.stdin.readline for _ in range(int(input())): n = int(input()) data = dict(collections.Counter(map(int, input().split()))) if min(list(data.values())) > 1: last = 1 for i in data.keys(): print(last + data[i] - 1, end=' ') for j in range(last, last + data[i] - 1): {{completion}} last = last + data[i] print() else: print(-1)
print(j, end=' ')
[{"input": "2\n5\n1 1 1 1 1\n6\n3 6 8 13 15 21", "output": ["5 1 2 3 4 \n-1"]}]
block_completion_002398
block
python
Complete the code in python to solve this programming problem: Description: This is a hard version of the problem. The only difference between an easy and a hard version is in the number of queries.Polycarp grew a tree from $$$n$$$ vertices. We remind you that a tree of $$$n$$$ vertices is an undirected connected graph of $$$n$$$ vertices and $$$n-1$$$ edges that does not contain cycles.He calls a set of vertices passable if there is such a path in the tree that passes through each vertex of this set without passing through any edge twice. The path can visit other vertices (not from this set).In other words, a set of vertices is called passable if there is a simple path that passes through all the vertices of this set (and possibly some other).For example, for a tree below sets $$$\{3, 2, 5\}$$$, $$$\{1, 5, 4\}$$$, $$$\{1, 4\}$$$ are passable, and $$$\{1, 3, 5\}$$$, $$$\{1, 2, 3, 4, 5\}$$$ are not. Polycarp asks you to answer $$$q$$$ queries. Each query is a set of vertices. For each query, you need to determine whether the corresponding set of vertices is passable. Input Specification: The first line of input contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — number of vertices. Following $$$n - 1$$$ lines a description of the tree.. Each line contains two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — indices of vertices connected by an edge. Following line contains single integer $$$q$$$ ($$$1 \le q \le 10^5$$$) — number of queries. The following $$$2 \cdot q$$$ lines contain descriptions of sets. The first line of the description contains an integer $$$k$$$ ($$$1 \le k \le n$$$) — the size of the set. The second line of the description contains $$$k$$$ of distinct integers $$$p_1, p_2, \dots, p_k$$$ ($$$1 \le p_i \le n$$$) — indices of the vertices of the set. It is guaranteed that the sum of $$$k$$$ values for all queries does not exceed $$$2 \cdot 10^5$$$. Output Specification: Output $$$q$$$ lines, each of which contains the answer to the corresponding query. As an answer, output "YES" if the set is passable, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer). Code: input = __import__('sys').stdin.readline mapnode = lambda x: int(x)-1 n = int(input()) adj = [[] for _ in range(n)] for _ in range(n-1): u, v = map(mapnode, input().split()) adj[u].append(v) adj[v].append(u) # dfs jump = [[0] * n] depth = [0] * n stack = [(0, -1)] while len(stack) > 0: u, par = stack.pop() jump[0][u] = par for v in adj[u]: if v != par: # TODO: Your code here for _ in range(19): jump.append([jump[-1][jump[-1][u]] for u in range(n)]) # print('depth', depth) # print('jump', jump) def lca(u, v): if depth[u] < depth[v]: u, v = v, u step = depth[u] - depth[v] for i in range(19): if (step >> i) & 1 == 1: u = jump[i][u] if u == v: return u # move up together for i in range(18, -1, -1): if jump[i][u] != jump[i][v]: u, v = jump[i][u], jump[i][v] return jump[0][u] # answer queries for _ in range(int(input())): nk = int(input()) nodes = list(sorted(map(mapnode, input().split()), key=lambda u: -depth[u])) mindepth = min(depth[u] for u in nodes) # check from lowest depth tocheck = [nodes[i] for i in range(1, nk) if nodes[i] != lca(nodes[0], nodes[i])] # print(subroot+1, [x+1 for x in tocheck], [x+1 for x in nodes]) ok = len(tocheck) == 0 or all(lca(tocheck[0], u) == u for u in tocheck) and depth[lca(tocheck[0], nodes[0])] <= mindepth print('YES' if ok else 'NO')
input = __import__('sys').stdin.readline mapnode = lambda x: int(x)-1 n = int(input()) adj = [[] for _ in range(n)] for _ in range(n-1): u, v = map(mapnode, input().split()) adj[u].append(v) adj[v].append(u) # dfs jump = [[0] * n] depth = [0] * n stack = [(0, -1)] while len(stack) > 0: u, par = stack.pop() jump[0][u] = par for v in adj[u]: if v != par: {{completion}} for _ in range(19): jump.append([jump[-1][jump[-1][u]] for u in range(n)]) # print('depth', depth) # print('jump', jump) def lca(u, v): if depth[u] < depth[v]: u, v = v, u step = depth[u] - depth[v] for i in range(19): if (step >> i) & 1 == 1: u = jump[i][u] if u == v: return u # move up together for i in range(18, -1, -1): if jump[i][u] != jump[i][v]: u, v = jump[i][u], jump[i][v] return jump[0][u] # answer queries for _ in range(int(input())): nk = int(input()) nodes = list(sorted(map(mapnode, input().split()), key=lambda u: -depth[u])) mindepth = min(depth[u] for u in nodes) # check from lowest depth tocheck = [nodes[i] for i in range(1, nk) if nodes[i] != lca(nodes[0], nodes[i])] # print(subroot+1, [x+1 for x in tocheck], [x+1 for x in nodes]) ok = len(tocheck) == 0 or all(lca(tocheck[0], u) == u for u in tocheck) and depth[lca(tocheck[0], nodes[0])] <= mindepth print('YES' if ok else 'NO')
depth[v] = depth[u] + 1 stack.append((v, u))
[{"input": "5\n1 2\n2 3\n2 4\n4 5\n5\n3\n3 2 5\n5\n1 2 3 4 5\n2\n1 4\n3\n1 3 5\n3\n1 5 4", "output": ["YES\nNO\nYES\nNO\nYES"]}, {"input": "5\n1 2\n3 2\n2 4\n5 2\n4\n2\n3 1\n3\n3 4 5\n3\n2 3 5\n1\n1", "output": ["YES\nNO\nYES\nYES"]}]
block_completion_002280
block
python
Complete the code in python to solve this programming problem: Description: On an $$$8 \times 8$$$ grid, some horizontal rows have been painted red, and some vertical columns have been painted blue, in some order. The stripes are drawn sequentially, one after the other. When the stripe is drawn, it repaints all the cells through which it passes.Determine which color was used last. The red stripe was painted after the blue one, so the answer is R. Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \leq t \leq 4000$$$) — the number of test cases. The description of test cases follows. There is an empty line before each test case. Each test case consists of $$$8$$$ lines, each containing $$$8$$$ characters. Each of these characters is either 'R', 'B', or '.', denoting a red square, a blue square, and an unpainted square, respectively. It is guaranteed that the given field is obtained from a colorless one by drawing horizontal red rows and vertical blue columns. At least one stripe is painted. Output Specification: For each test case, output 'R' if a red stripe was painted last, and 'B' if a blue stripe was painted last (without quotes). Notes: NoteThe first test case is pictured in the statement.In the second test case, the first blue column is painted first, then the first and last red rows, and finally the last blue column. Since a blue stripe is painted last, the answer is B. Code: test = int(input()) for i in range(test): ans = "B" cnt =0 while cnt < 8 : t = input() if t.strip() != '': cnt +=1 if t == "RRRRRRRR": # TODO: Your code here print(ans)
test = int(input()) for i in range(test): ans = "B" cnt =0 while cnt < 8 : t = input() if t.strip() != '': cnt +=1 if t == "RRRRRRRR": {{completion}} print(ans)
ans = "R"
[{"input": "4\n\n\n\n\n....B...\n\n....B...\n\n....B...\n\nRRRRRRRR\n\n....B...\n\n....B...\n\n....B...\n\n....B...\n\n\n\n\nRRRRRRRB\n\nB......B\n\nB......B\n\nB......B\n\nB......B\n\nB......B\n\nB......B\n\nRRRRRRRB\n\n\n\n\nRRRRRRBB\n\n.B.B..BB\n\nRRRRRRBB\n\n.B.B..BB\n\n.B.B..BB\n\nRRRRRRBB\n\n.B.B..BB\n\n.B.B..BB\n\n\n\n\n........\n\n........\n\n........\n\nRRRRRRRR\n\n........\n\n........\n\n........\n\n........", "output": ["R\nB\nB\nR"]}]
block_completion_005800
block
python
Complete the code in python to solve this programming problem: Description: Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match  — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$)  — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b &lt; r \leq n$$$, $$$r+b=n$$$). Output Specification: For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. Notes: NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. Code: t = int(input()) for i in range(t): x = input().split() r = int(x[1]) b = int(x[2]) x = "" p = r%(b+1) q = r//(b+1) for i in range(p): # TODO: Your code here for i in range(b+1-p): x+= "R"*(q)+"B" print(x[:-1])
t = int(input()) for i in range(t): x = input().split() r = int(x[1]) b = int(x[2]) x = "" p = r%(b+1) q = r//(b+1) for i in range(p): {{completion}} for i in range(b+1-p): x+= "R"*(q)+"B" print(x[:-1])
x += "R"*(q+1)+"B"
[{"input": "3\n7 4 3\n6 5 1\n19 13 6", "output": ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR"]}, {"input": "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2", "output": ["RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"]}]
block_completion_008717
block
python
Complete the code in python to solve this programming problem: Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i &lt; j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i &lt; j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow. Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$. Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i &lt; j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++). Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions. Code: from collections import Counter t=int(input()) while(t!=0): n=int(input()) s = Counter(input() for x in [1]*n) cnt = 0 for x in s: for y in s: if(x!=y and (x[1]==y[1] or x[0]==y[0])): # TODO: Your code here print(cnt//2) t-=1
from collections import Counter t=int(input()) while(t!=0): n=int(input()) s = Counter(input() for x in [1]*n) cnt = 0 for x in s: for y in s: if(x!=y and (x[1]==y[1] or x[0]==y[0])): {{completion}} print(cnt//2) t-=1
cnt += s[x]*s[y]
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
block_completion_000893
block
python
Complete the code in python to solve this programming problem: Description: You are given an array $$$a$$$ of $$$n$$$ integers $$$a_1, a_2, a_3, \ldots, a_n$$$.You have to answer $$$q$$$ independent queries, each consisting of two integers $$$l$$$ and $$$r$$$. Consider the subarray $$$a[l:r]$$$ $$$=$$$ $$$[a_l, a_{l+1}, \ldots, a_r]$$$. You can apply the following operation to the subarray any number of times (possibly zero)- Choose two integers $$$L$$$, $$$R$$$ such that $$$l \le L \le R \le r$$$ and $$$R - L + 1$$$ is odd. Replace each element in the subarray from $$$L$$$ to $$$R$$$ with the XOR of the elements in the subarray $$$[L, R]$$$. The answer to the query is the minimum number of operations required to make all elements of the subarray $$$a[l:r]$$$ equal to $$$0$$$ or $$$-1$$$ if it is impossible to make all of them equal to $$$0$$$. You can find more details about XOR operation here. Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ $$$(1 \le n, q \le 2 \cdot 10^5)$$$  — the length of the array $$$a$$$ and the number of queries. The next line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ $$$(0 \le a_i \lt 2^{30})$$$  — the elements of the array $$$a$$$. The $$$i$$$-th of the next $$$q$$$ lines contains two integers $$$l_i$$$ and $$$r_i$$$ $$$(1 \le l_i \le r_i \le n)$$$  — the description of the $$$i$$$-th query. Output Specification: For each query, output a single integer  — the answer to that query. Notes: NoteIn the first query, $$$l = 3, r = 4$$$, subarray = $$$[3, 3]$$$. We can apply operation only to the subarrays of length $$$1$$$, which won't change the array; hence it is impossible to make all elements equal to $$$0$$$.In the second query, $$$l = 4, r = 6$$$, subarray = $$$[3, 1, 2]$$$. We can choose the whole subarray $$$(L = 4, R = 6)$$$ and replace all elements by their XOR $$$(3 \oplus 1 \oplus 2) = 0$$$, making the subarray $$$[0, 0, 0]$$$.In the fifth query, $$$l = 1, r = 6$$$, subarray = $$$[3, 0, 3, 3, 1, 2]$$$. We can make the operations as follows: Choose $$$L = 4, R = 6$$$, making the subarray $$$[3, 0, 3, 0, 0, 0]$$$. Choose $$$L = 1, R = 5$$$, making the subarray $$$[0, 0, 0, 0, 0, 0]$$$. Code: import sys input = sys.stdin.readline n,q = map(int,input().split()) a = [0] + list(map(int,input().split())) cml = a[::1] for i in range(1, n+1): a[i] ^= a[i-1] cml[i] += cml[i-1] qs = [list(map(int,input().split())) for i in range(q)] from collections import defaultdict d = defaultdict(list) dd = defaultdict(list) cnt = defaultdict(int) ord = [0]*(n+1) for i in range(n+1): dd[a[i]].append(i % 2) cnt[a[i]] += 1 ord[i] = cnt[a[i]] for k,v in dd.items(): dd[k] = [0] + v for i in range(len(v)+1): if i == 0: continue else: dd[k][i] += dd[k][i-1] for l,r in qs: if a[r] != a[l-1]: print(-1) else: if cml[r] - cml[l-1] == 0: print(0) elif (r-l) % 2 == 0 or a[l] == a[l-1] or a[r] == a[r-1]: print(1) else: ll = ord[l-1]-1 rr = ord[r] tot = dd[a[r]][rr] - dd[a[r]][ll] if # TODO: Your code here: print(-1) else: print(2)
import sys input = sys.stdin.readline n,q = map(int,input().split()) a = [0] + list(map(int,input().split())) cml = a[::1] for i in range(1, n+1): a[i] ^= a[i-1] cml[i] += cml[i-1] qs = [list(map(int,input().split())) for i in range(q)] from collections import defaultdict d = defaultdict(list) dd = defaultdict(list) cnt = defaultdict(int) ord = [0]*(n+1) for i in range(n+1): dd[a[i]].append(i % 2) cnt[a[i]] += 1 ord[i] = cnt[a[i]] for k,v in dd.items(): dd[k] = [0] + v for i in range(len(v)+1): if i == 0: continue else: dd[k][i] += dd[k][i-1] for l,r in qs: if a[r] != a[l-1]: print(-1) else: if cml[r] - cml[l-1] == 0: print(0) elif (r-l) % 2 == 0 or a[l] == a[l-1] or a[r] == a[r-1]: print(1) else: ll = ord[l-1]-1 rr = ord[r] tot = dd[a[r]][rr] - dd[a[r]][ll] if {{completion}}: print(-1) else: print(2)
tot == rr-ll or tot == 0
[{"input": "7 6\n3 0 3 3 1 2 3\n3 4\n4 6\n3 7\n5 6\n1 6\n2 2", "output": ["-1\n1\n1\n-1\n2\n0"]}]
control_completion_001769
control_fixed
python
Complete the code in python to solve this programming problem: Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count. Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] &lt; x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] &lt; x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked. Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO". Code: import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353 n,m=M();a=L();q=I() t=[0]*(2*m) # a is the array of initial values def build(t,n,a): for i in range(n):t[i+n]=a[i] for i in range(n-1,0,-1):t[i]=max(t[i<<1],t[(i<<1)|1]) # change value at position p to v def modify(t,n,p,v): p+=n t[p]=v while p>1: t[p>>1]=max(t[p],t[p^1]) p>>=1 # find the combined value of range [l,r) def query(t,n,l,r): resl=resr=0 l+=n;r+=n while l<r: if (l&1):# TODO: Your code here if (r&1):r-=1;resr=max(t[r],resr) l>>=1;r>>=1 return max(resl,resr) build(t,m,a) for i in range(q): xs,ys,xf,yf,k=M() if abs(xs-xf)%k or abs(ys-yf)%k:print("NO");continue p=query(t,m,min(ys,yf)-1,max(ys,yf)) z=min(xs,xf)+((n-min(xs,xf))//k)*k if z<=p:print("NO") else:print("YES")
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353 n,m=M();a=L();q=I() t=[0]*(2*m) # a is the array of initial values def build(t,n,a): for i in range(n):t[i+n]=a[i] for i in range(n-1,0,-1):t[i]=max(t[i<<1],t[(i<<1)|1]) # change value at position p to v def modify(t,n,p,v): p+=n t[p]=v while p>1: t[p>>1]=max(t[p],t[p^1]) p>>=1 # find the combined value of range [l,r) def query(t,n,l,r): resl=resr=0 l+=n;r+=n while l<r: if (l&1):{{completion}} if (r&1):r-=1;resr=max(t[r],resr) l>>=1;r>>=1 return max(resl,resr) build(t,m,a) for i in range(q): xs,ys,xf,yf,k=M() if abs(xs-xf)%k or abs(ys-yf)%k:print("NO");continue p=query(t,m,min(ys,yf)-1,max(ys,yf)) z=min(xs,xf)+((n-min(xs,xf))//k)*k if z<=p:print("NO") else:print("YES")
resl=max(resl,t[l]);l+=1
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
block_completion_002999
block
python
Complete the code in python to solve this programming problem: Description: While searching for the pizza, baby Hosssam came across two permutations $$$a$$$ and $$$b$$$ of length $$$n$$$.Recall that a permutation is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array) and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).Baby Hosssam forgot about the pizza and started playing around with the two permutations. While he was playing with them, some elements of the first permutation got mixed up with some elements of the second permutation, and to his surprise those elements also formed a permutation of size $$$n$$$.Specifically, he mixed up the permutations to form a new array $$$c$$$ in the following way. For each $$$i$$$ ($$$1\le i\le n$$$), he either made $$$c_i=a_i$$$ or $$$c_i=b_i$$$. The array $$$c$$$ is a permutation. You know permutations $$$a$$$, $$$b$$$, and values at some positions in $$$c$$$. Please count the number different permutations $$$c$$$ that are consistent with the described process and the given values. Since the answer can be large, print it modulo $$$10^9+7$$$.It is guaranteed that there exists at least one permutation $$$c$$$ that satisfies all the requirements. Input Specification: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1\le n\le 10^5$$$) — the length of the permutations. The next line contains $$$n$$$ distinct integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1\le a_i\le n$$$) — the first permutation. The next line contains $$$n$$$ distinct integers $$$b_1,b_2,\ldots,b_n$$$ ($$$1\le b_i\le n$$$) — the second permutation. The next line contains $$$n$$$ distinct integers $$$d_1,d_2,\ldots,d_n$$$ ($$$d_i$$$ is either $$$0$$$, $$$a_i$$$, or $$$b_i$$$) — the description of the known values of $$$c$$$. If $$$d_i=0$$$, then there are no requirements on the value of $$$c_i$$$. Otherwise, it is required that $$$c_i=d_i$$$. It is guaranteed that there exists at least one permutation $$$c$$$ that satisfies all the requirements. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$. Output Specification: For each test case, print the number of possible permutations $$$c$$$, modulo $$$10^9+7$$$. Notes: NoteIn the first test case, there are $$$4$$$ distinct permutation that can be made using the process: $$$[2,3,1,4,5,6,7]$$$, $$$[2,3,1,7,6,5,4]$$$, $$$[2,3,1,4,6,5,7]$$$, $$$[2,3,1,7,5,6,4]$$$.In the second test case, there is only one distinct permutation that can be made using the process: $$$[1]$$$.In the third test case, there are $$$2$$$ distinct permutation that can be made using the process: $$$[6,5,2,1,4,3]$$$, $$$[6,5,3,1,4,2]$$$.In the fourth test case, there are $$$2$$$ distinct permutation that can be made using the process: $$$[1,2,8,7,4,3,6,5]$$$, $$$[1,6,4,7,2,3,8,5]$$$.In the fifth test case, there is only one distinct permutation that can be made using the process: $$$[1,9,2,3,4,10,8,6,7,5]$$$. Code: # read interger t from input.txt and then read t lines import sys DEBUG = False def check(a, b, c): a = [0] + a b = [0] + b c = [0] + c m_ = [0] * len(a) m = [0] * len(a) for i in range(1, len(b)): m_[b[i]] = i for i in range(1, len(a)): m[i] = m_[a[i]] # print(">>>", a) # print(">>>", b) # print(">>>", m) # find cicles in permutations total_num = 1 used = [False] * len(m) # print(a, b, c) for i in range(1, len(m)): if not used[i]: j = i c_zeros = True while not used[j]: if c[j] != 0: # TODO: Your code here used[j] = True j = m[j] used[i] = True # print(i, m[i], a[i], b[i], c[i]) if c_zeros and m[i] != i: # print(">>", i) total_num = (total_num) * 2 % 1000000007 print(total_num) def main(f): t = int(f.readline()) for i in range(t): n = int(f.readline()) a = list(map(int, f.readline().split())) b = list(map(int, f.readline().split())) c = list(map(int, f.readline().split())) check(a, b, c) if DEBUG: f = open('input.txt', 'r') else: f = sys.stdin main(f) f.close()
# read interger t from input.txt and then read t lines import sys DEBUG = False def check(a, b, c): a = [0] + a b = [0] + b c = [0] + c m_ = [0] * len(a) m = [0] * len(a) for i in range(1, len(b)): m_[b[i]] = i for i in range(1, len(a)): m[i] = m_[a[i]] # print(">>>", a) # print(">>>", b) # print(">>>", m) # find cicles in permutations total_num = 1 used = [False] * len(m) # print(a, b, c) for i in range(1, len(m)): if not used[i]: j = i c_zeros = True while not used[j]: if c[j] != 0: {{completion}} used[j] = True j = m[j] used[i] = True # print(i, m[i], a[i], b[i], c[i]) if c_zeros and m[i] != i: # print(">>", i) total_num = (total_num) * 2 % 1000000007 print(total_num) def main(f): t = int(f.readline()) for i in range(t): n = int(f.readline()) a = list(map(int, f.readline().split())) b = list(map(int, f.readline().split())) c = list(map(int, f.readline().split())) check(a, b, c) if DEBUG: f = open('input.txt', 'r') else: f = sys.stdin main(f) f.close()
c_zeros = False
[{"input": "9\n7\n1 2 3 4 5 6 7\n2 3 1 7 6 5 4\n2 0 1 0 0 0 0\n1\n1\n1\n0\n6\n1 5 2 4 6 3\n6 5 3 1 4 2\n6 0 0 0 0 0\n8\n1 6 4 7 2 3 8 5\n3 2 8 1 4 5 6 7\n1 0 0 7 0 3 0 5\n10\n1 8 6 2 4 7 9 3 10 5\n1 9 2 3 4 10 8 6 7 5\n1 9 2 3 4 10 8 6 7 5\n7\n1 2 3 4 5 6 7\n2 3 1 7 6 5 4\n0 0 0 0 0 0 0\n5\n1 2 3 4 5\n1 2 3 4 5\n0 0 0 0 0\n5\n1 2 3 4 5\n1 2 3 5 4\n0 0 0 0 0\n3\n1 2 3\n3 1 2\n0 0 0", "output": ["4\n1\n2\n2\n1\n8\n1\n2\n2"]}]
block_completion_006025
block
python
Complete the code in python to solve this programming problem: Description: A triple of points $$$i$$$, $$$j$$$ and $$$k$$$ on a coordinate line is called beautiful if $$$i &lt; j &lt; k$$$ and $$$k - i \le d$$$.You are given a set of points on a coordinate line, initially empty. You have to process queries of three types: add a point; remove a point; calculate the number of beautiful triples consisting of points belonging to the set. Input Specification: The first line contains two integers $$$q$$$ and $$$d$$$ ($$$1 \le q, d \le 2 \cdot 10^5$$$) — the number of queries and the parameter for defining if a triple is beautiful, respectively. The second line contains $$$q$$$ integers $$$a_1, a_2, \dots, a_q$$$ ($$$1 \le a_i \le 2 \cdot 10^5$$$) denoting the queries. The integer $$$a_i$$$ denotes the $$$i$$$-th query in the following way: if the point $$$a_i$$$ belongs to the set, remove it; otherwise, add it; after adding or removing the point, print the number of beautiful triples. Output Specification: For each query, print one integer — the number of beautiful triples after processing the respective query. Code: import os import sys from io import BytesIO, IOBase BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") class LazySegmentTree(): def __init__(self, n, op, e, mapping, composition, id): self.n = n self.op = op self.e = e self.mapping = mapping self.composition = composition self.id = id self.log = (n - 1).bit_length() self.size = 1 << self.log self.data = [e] * (2 * self.size) self.lazy = [id] * self.size def update(self, k): self.data[k] = self.op(self.data[2 * k], self.data[2 * k + 1]) def all_apply(self, k, f): self.data[k] = self.mapping(f, self.data[k]) if k < self.size: self.lazy[k] = self.composition(f, self.lazy[k]) def push(self, k): self.all_apply(2 * k, self.lazy[k]) self.all_apply(2 * k + 1, self.lazy[k]) self.lazy[k] = self.id def build(self, arr): # assert len(arr) == self.n for i, a in enumerate(arr): self.data[self.size + i] = a for i in range(self.size - 1, 0, -1): self.update(i) def set(self, p, x): # assert 0 <= p < self.n p += self.size for i in range(self.log, 0, -1): self.push(p >> i) self.data[p] = x for i in range(1, self.log + 1): self.update(p >> i) def get(self, p): # assert 0 <= p < self.n p += self.size for i in range(self.log, 0, -1): self.push(p >> i) return self.data[p] def prod(self, l, r): # assert 0 <= l <= r <= self.n if l == r: return self.e l += self.size r += self.size for i in range(self.log, 0, -1): if ((l >> i) << i) != l: # TODO: Your code here if ((r >> i) << i) != r: self.push(r >> i) sml = smr = self.e while l < r: if l & 1: sml = self.op(sml, self.data[l]) l += 1 if r & 1: r -= 1 smr = self.op(self.data[r], smr) l >>= 1 r >>= 1 return self.op(sml, smr) def all_prod(self): return self.data[1] def apply(self, p, f): # assert 0 <= p < self.n p += self.size for i in range(self.log, 0, -1): self.push(p >> i) self.data[p] = self.mapping(f, self.data[p]) for i in range(1, self.log + 1): self.update(p >> i) def range_apply(self, l, r, f): # assert 0 <= l <= r <= self.n if l == r: return l += self.size r += self.size for i in range(self.log, 0, -1): if ((l >> i) << i) != l: self.push(l >> i) if ((r >> i) << i) != r: self.push((r - 1) >> i) l2 = l r2 = r while l < r: if l & 1: self.all_apply(l, f) l += 1 if r & 1: r -= 1 self.all_apply(r, f) l >>= 1 r >>= 1 l = l2 r = r2 for i in range(1, self.log + 1): if ((l >> i) << i) != l: self.update(l >> i) if ((r >> i) << i) != r: self.update((r - 1) >> i) n = 2*10**5+1 q, d = map(int, input().split()) arr = list(map(int, input().split())) v = [0]*n def op(x, y): return [x[0]+y[0], x[1]+y[1], x[2]+y[2], x[3]] def mapping(k, x): return [x[0], k*x[0]+x[1], k*k*x[0]+2*k*x[1]+x[2], x[3]+k] def composition(f, g): return f+g e = [0, 0, 0, 0] id = 0 st = LazySegmentTree(n, op, e, mapping, composition, id) st.build([[0, 0, 0, 0] for i in range(n)]) for x in arr: v[x] ^= 1 m = st.get(x)[3] if v[x]: st.range_apply(x+1, min(x+d+1, n), 1) st.set(x, [1, m, m*m, m]) else: st.range_apply(x+1, min(x+d+1, n), -1) st.set(x, [0, 0, 0, m]) a = st.all_prod() print((a[2]-a[1])//2)
import os import sys from io import BytesIO, IOBase BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") class LazySegmentTree(): def __init__(self, n, op, e, mapping, composition, id): self.n = n self.op = op self.e = e self.mapping = mapping self.composition = composition self.id = id self.log = (n - 1).bit_length() self.size = 1 << self.log self.data = [e] * (2 * self.size) self.lazy = [id] * self.size def update(self, k): self.data[k] = self.op(self.data[2 * k], self.data[2 * k + 1]) def all_apply(self, k, f): self.data[k] = self.mapping(f, self.data[k]) if k < self.size: self.lazy[k] = self.composition(f, self.lazy[k]) def push(self, k): self.all_apply(2 * k, self.lazy[k]) self.all_apply(2 * k + 1, self.lazy[k]) self.lazy[k] = self.id def build(self, arr): # assert len(arr) == self.n for i, a in enumerate(arr): self.data[self.size + i] = a for i in range(self.size - 1, 0, -1): self.update(i) def set(self, p, x): # assert 0 <= p < self.n p += self.size for i in range(self.log, 0, -1): self.push(p >> i) self.data[p] = x for i in range(1, self.log + 1): self.update(p >> i) def get(self, p): # assert 0 <= p < self.n p += self.size for i in range(self.log, 0, -1): self.push(p >> i) return self.data[p] def prod(self, l, r): # assert 0 <= l <= r <= self.n if l == r: return self.e l += self.size r += self.size for i in range(self.log, 0, -1): if ((l >> i) << i) != l: {{completion}} if ((r >> i) << i) != r: self.push(r >> i) sml = smr = self.e while l < r: if l & 1: sml = self.op(sml, self.data[l]) l += 1 if r & 1: r -= 1 smr = self.op(self.data[r], smr) l >>= 1 r >>= 1 return self.op(sml, smr) def all_prod(self): return self.data[1] def apply(self, p, f): # assert 0 <= p < self.n p += self.size for i in range(self.log, 0, -1): self.push(p >> i) self.data[p] = self.mapping(f, self.data[p]) for i in range(1, self.log + 1): self.update(p >> i) def range_apply(self, l, r, f): # assert 0 <= l <= r <= self.n if l == r: return l += self.size r += self.size for i in range(self.log, 0, -1): if ((l >> i) << i) != l: self.push(l >> i) if ((r >> i) << i) != r: self.push((r - 1) >> i) l2 = l r2 = r while l < r: if l & 1: self.all_apply(l, f) l += 1 if r & 1: r -= 1 self.all_apply(r, f) l >>= 1 r >>= 1 l = l2 r = r2 for i in range(1, self.log + 1): if ((l >> i) << i) != l: self.update(l >> i) if ((r >> i) << i) != r: self.update((r - 1) >> i) n = 2*10**5+1 q, d = map(int, input().split()) arr = list(map(int, input().split())) v = [0]*n def op(x, y): return [x[0]+y[0], x[1]+y[1], x[2]+y[2], x[3]] def mapping(k, x): return [x[0], k*x[0]+x[1], k*k*x[0]+2*k*x[1]+x[2], x[3]+k] def composition(f, g): return f+g e = [0, 0, 0, 0] id = 0 st = LazySegmentTree(n, op, e, mapping, composition, id) st.build([[0, 0, 0, 0] for i in range(n)]) for x in arr: v[x] ^= 1 m = st.get(x)[3] if v[x]: st.range_apply(x+1, min(x+d+1, n), 1) st.set(x, [1, m, m*m, m]) else: st.range_apply(x+1, min(x+d+1, n), -1) st.set(x, [0, 0, 0, m]) a = st.all_prod() print((a[2]-a[1])//2)
self.push(l >> i)
[{"input": "7 5\n8 5 3 2 1 5 6", "output": ["0\n0\n1\n2\n5\n1\n5"]}]
block_completion_005218
block
python
Complete the code in python to solve this programming problem: Description: Polycarp has a string $$$s$$$ consisting of lowercase Latin letters.He encodes it using the following algorithm.He goes through the letters of the string $$$s$$$ from left to right and for each letter Polycarp considers its number in the alphabet: if the letter number is single-digit number (less than $$$10$$$), then just writes it out; if the letter number is a two-digit number (greater than or equal to $$$10$$$), then it writes it out and adds the number 0 after. For example, if the string $$$s$$$ is code, then Polycarp will encode this string as follows: 'c' — is the $$$3$$$-rd letter of the alphabet. Consequently, Polycarp adds 3 to the code (the code becomes equal to 3); 'o' — is the $$$15$$$-th letter of the alphabet. Consequently, Polycarp adds 15 to the code and also 0 (the code becomes 3150); 'd' — is the $$$4$$$-th letter of the alphabet. Consequently, Polycarp adds 4 to the code (the code becomes 31504); 'e' — is the $$$5$$$-th letter of the alphabet. Therefore, Polycarp adds 5 to the code (the code becomes 315045). Thus, code of string code is 315045.You are given a string $$$t$$$ resulting from encoding the string $$$s$$$. Your task is to decode it (get the original string $$$s$$$ by $$$t$$$). Input Specification: The first line of the input contains an integer $$$q$$$ ($$$1 \le q \le 10^4$$$) — the number of test cases in the input. The descriptions of the test cases follow. The first line of description of each test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of the given code. The second line of the description of each test case contains a string $$$t$$$ of length $$$n$$$ — the given code. It is guaranteed that there exists such a string of lowercase Latin letters, as a result of encoding which the string $$$t$$$ is obtained. Output Specification: For each test case output the required string $$$s$$$ — the string that gives string $$$t$$$ as the result of encoding. It is guaranteed that such a string always exists. It can be shown that such a string is always unique. Notes: NoteThe first test case is explained above.In the second test case, the answer is aj. Indeed, the number of the letter a is equal to $$$1$$$, so 1 will be appended to the code. The number of the letter j is $$$10$$$, so 100 will be appended to the code. The resulting code is 1100.There are no zeros in the third test case, which means that the numbers of all letters are less than $$$10$$$ and are encoded as one digit. The original string is abacaba.In the fourth test case, the string $$$s$$$ is equal to ll. The letter l has the number $$$12$$$ and is encoded as 120. So ll is indeed 120120. Code: from sys import stdin from collections import deque lst = list(map(int, stdin.read().split())) _s = 0 def inp(n=1): global _s ret = lst[_s:_s + n] _s += n return ret def inp1(): return inp()[0] t = inp1() for _ in range(t): n = inp1() s = str(inp1())[::-1] alph = "0abcdefghijklmnopqrstuvwxyz" d = deque() i = 0 while i < n: if # TODO: Your code here: d.appendleft(int(s[i + 1:i + 3][::-1])) i += 3 else: d.appendleft(int(s[i])) i += 1 ret = "" for i in d: ret += alph[i] print(ret)
from sys import stdin from collections import deque lst = list(map(int, stdin.read().split())) _s = 0 def inp(n=1): global _s ret = lst[_s:_s + n] _s += n return ret def inp1(): return inp()[0] t = inp1() for _ in range(t): n = inp1() s = str(inp1())[::-1] alph = "0abcdefghijklmnopqrstuvwxyz" d = deque() i = 0 while i < n: if {{completion}}: d.appendleft(int(s[i + 1:i + 3][::-1])) i += 3 else: d.appendleft(int(s[i])) i += 1 ret = "" for i in d: ret += alph[i] print(ret)
s[i] == "0"
[{"input": "9\n\n6\n\n315045\n\n4\n\n1100\n\n7\n\n1213121\n\n6\n\n120120\n\n18\n\n315045615018035190\n\n7\n\n1111110\n\n7\n\n1111100\n\n5\n\n11111\n\n4\n\n2606", "output": ["code\naj\nabacaba\nll\ncodeforces\naaaak\naaaaj\naaaaa\nzf"]}]
control_completion_008431
control_fixed
python
Complete the code in python to solve this programming problem: Description: There are $$$n$$$ candies put from left to right on a table. The candies are numbered from left to right. The $$$i$$$-th candy has weight $$$w_i$$$. Alice and Bob eat candies. Alice can eat any number of candies from the left (she can't skip candies, she eats them in a row). Bob can eat any number of candies from the right (he can't skip candies, he eats them in a row). Of course, if Alice ate a candy, Bob can't eat it (and vice versa).They want to be fair. Their goal is to eat the same total weight of candies. What is the most number of candies they can eat in total? Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the number of candies on the table. The second line of each test case contains $$$n$$$ integers $$$w_1, w_2, \dots, w_n$$$ ($$$1 \leq w_i \leq 10^4$$$) — the weights of candies from left to right. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$. Output Specification: For each test case, print a single integer — the maximum number of candies Alice and Bob can eat in total while satisfying the condition. Notes: NoteFor the first test case, Alice will eat one candy from the left and Bob will eat one candy from the right. There is no better way for them to eat the same total amount of weight. The answer is $$$2$$$ because they eat two candies in total.For the second test case, Alice will eat the first three candies from the left (with total weight $$$7$$$) and Bob will eat the first three candies from the right (with total weight $$$7$$$). They cannot eat more candies since all the candies have been eaten, so the answer is $$$6$$$ (because they eat six candies in total).For the third test case, there is no way Alice and Bob will eat the same non-zero weight so the answer is $$$0$$$.For the fourth test case, Alice will eat candies with weights $$$[7, 3, 20]$$$ and Bob will eat candies with weights $$$[10, 8, 11, 1]$$$, they each eat $$$30$$$ weight. There is no better partition so the answer is $$$7$$$. Code: for _ in range(int(input())): n = int(input()) a = [*map(int, input().split())] x = sum(a) // 2 s, d = 0, {} for idx, i in enumerate(a): s += i if # TODO: Your code here: break d[s] = idx + 1 s, r = 0, 0 for idx, i in enumerate(a[::-1]): s += i if s in d: r = idx + 1 + d[s] print(r)
for _ in range(int(input())): n = int(input()) a = [*map(int, input().split())] x = sum(a) // 2 s, d = 0, {} for idx, i in enumerate(a): s += i if {{completion}}: break d[s] = idx + 1 s, r = 0, 0 for idx, i in enumerate(a[::-1]): s += i if s in d: r = idx + 1 + d[s] print(r)
s > x
[{"input": "4\n3\n10 20 10\n6\n2 1 4 2 4 1\n5\n1 2 4 8 16\n9\n7 3 20 5 15 1 11 8 10", "output": ["2\n6\n0\n7"]}]
control_completion_000790
control_fixed
python
Complete the code in python to solve this programming problem: Description: Tokitsukaze has a sequence $$$a$$$ of length $$$n$$$. For each operation, she selects two numbers $$$a_i$$$ and $$$a_j$$$ ($$$i \ne j$$$; $$$1 \leq i,j \leq n$$$). If $$$a_i = a_j$$$, change one of them to $$$0$$$. Otherwise change both of them to $$$\min(a_i, a_j)$$$. Tokitsukaze wants to know the minimum number of operations to change all numbers in the sequence to $$$0$$$. It can be proved that the answer always exists. Input Specification: The first line contains a single positive integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. For each test case, the first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 100$$$) — the length of the sequence $$$a$$$. The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \leq a_i \leq 100$$$) — the sequence $$$a$$$. Output Specification: For each test case, print a single integer — the minimum number of operations to change all numbers in the sequence to $$$0$$$. Notes: NoteIn the first test case, one of the possible ways to change all numbers in the sequence to $$$0$$$:In the $$$1$$$-st operation, $$$a_1 &lt; a_2$$$, after the operation, $$$a_2 = a_1 = 1$$$. Now the sequence $$$a$$$ is $$$[1,1,3]$$$.In the $$$2$$$-nd operation, $$$a_1 = a_2 = 1$$$, after the operation, $$$a_1 = 0$$$. Now the sequence $$$a$$$ is $$$[0,1,3]$$$.In the $$$3$$$-rd operation, $$$a_1 &lt; a_2$$$, after the operation, $$$a_2 = 0$$$. Now the sequence $$$a$$$ is $$$[0,0,3]$$$.In the $$$4$$$-th operation, $$$a_2 &lt; a_3$$$, after the operation, $$$a_3 = 0$$$. Now the sequence $$$a$$$ is $$$[0,0,0]$$$.So the minimum number of operations is $$$4$$$. Code: for n in [*open(0)][2::2]: *a,=map(int,n.split());b=len(a);c=a.count(0) while a: q=a.pop() if # TODO: Your code here: break print(b+(a==[])*(c==0)-c)
for n in [*open(0)][2::2]: *a,=map(int,n.split());b=len(a);c=a.count(0) while a: q=a.pop() if {{completion}}: break print(b+(a==[])*(c==0)-c)
a.count(q)>0
[{"input": "3\n3\n1 2 3\n3\n1 2 2\n3\n1 2 0", "output": ["4\n3\n2"]}]
control_completion_008021
control_fixed
python
Complete the code in python to solve this programming problem: Description: Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$)  — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. Output Specification: For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. Notes: NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. Code: import sys input = sys.stdin.readline T = int(input()) for t in range(T): N=int(input()) C=list(map(int,input().split())) ans=[0]*N k=sum(C)//N i=N-1 while i>-1 and k>0: if C[i]==N: ans[i]=1 k-=1 else: # TODO: Your code here i-=1 print(*ans)
import sys input = sys.stdin.readline T = int(input()) for t in range(T): N=int(input()) C=list(map(int,input().split())) ans=[0]*N k=sum(C)//N i=N-1 while i>-1 and k>0: if C[i]==N: ans[i]=1 k-=1 else: {{completion}} i-=1 print(*ans)
C[i-k]+=N-i
[{"input": "5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3", "output": ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"]}]
block_completion_008753
block
python
Complete the code in python to solve this programming problem: Description: You are given $$$n$$$ points on the plane, the coordinates of the $$$i$$$-th point are $$$(x_i, y_i)$$$. No two points have the same coordinates.The distance between points $$$i$$$ and $$$j$$$ is defined as $$$d(i,j) = |x_i - x_j| + |y_i - y_j|$$$.For each point, you have to choose a color, represented by an integer from $$$1$$$ to $$$n$$$. For every ordered triple of different points $$$(a,b,c)$$$, the following constraints should be met: if $$$a$$$, $$$b$$$ and $$$c$$$ have the same color, then $$$d(a,b) = d(a,c) = d(b,c)$$$; if $$$a$$$ and $$$b$$$ have the same color, and the color of $$$c$$$ is different from the color of $$$a$$$, then $$$d(a,b) &lt; d(a,c)$$$ and $$$d(a,b) &lt; d(b,c)$$$. Calculate the number of different ways to choose the colors that meet these constraints. Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the number of points. Then $$$n$$$ lines follow. The $$$i$$$-th of them contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \le x_i, y_i \le 10^8$$$). No two points have the same coordinates (i. e. if $$$i \ne j$$$, then either $$$x_i \ne x_j$$$ or $$$y_i \ne y_j$$$). Output Specification: Print one integer — the number of ways to choose the colors for the points. Since it can be large, print it modulo $$$998244353$$$. Notes: NoteIn the first test, the following ways to choose the colors are suitable: $$$[1, 1, 1]$$$; $$$[2, 2, 2]$$$; $$$[3, 3, 3]$$$; $$$[1, 2, 3]$$$; $$$[1, 3, 2]$$$; $$$[2, 1, 3]$$$; $$$[2, 3, 1]$$$; $$$[3, 1, 2]$$$; $$$[3, 2, 1]$$$. Code: input = __import__('sys').stdin.readline MOD = 998244353 fact = [1] invfact = [1] for i in range(1, 101): fact.append(fact[-1] * i % MOD) invfact.append(pow(fact[-1], MOD-2, MOD)) def C(n, k): if k < 0 or k > n: return 0 return fact[n] * invfact[k] % MOD * invfact[n-k] % MOD def P(n, k): if k < 0 or k > n: return 0 return fact[n] * invfact[n-k] % MOD n = int(input()) coords = [] for _ in range(n): x, y = map(int, input().split()) coords.append((x, y)) min_dist = [10**9] * n dist = [[-1] * n for _ in range(n)] for u in range(n): for v in range(n): dist[u][v] = abs(coords[u][0] - coords[v][0]) + abs(coords[u][1] - coords[v][1]) if u != v: min_dist[u] = min(min_dist[u], dist[u][v]) cnt = [0, 0, 0, 0, 0] vis = [False]*n for u in sorted(range(n), key=lambda x: min_dist[x]): if vis[u]: continue vis[u] = True seen = [False]*n seen[u] = True ptr = 0 found = [u] while ptr < len(found): v = found[ptr] ptr += 1 for w in range(n): if not seen[w] and dist[v][w] == min_dist[v]: seen[w] = True found.append(w) ok = all(dist[found[i]][found[j]] == min_dist[u] for i in range(len(found)) for j in range(i+1, len(found))) if len(found) == 1 or not ok: cnt[1] += 1 else: # print('found', found, ok) cnt[len(found)] += 1 for u in found: vis[u] = True # print('cnt', cnt[1:]) ans = 0 for two in range(cnt[2] + 1): for three in range(cnt[3] + 1): for four in range(cnt[4] + 1): ans += P(n, n - two - 2*three - 3*four) * C(cnt[2], two) % MOD \ * C(cnt[3], three) % MOD \ * C(cnt[4], four) % MOD if ans >= MOD: # TODO: Your code here # print(f'add P({n},{n - two - 2*three - 3*four})*C({cnt[2]},{two})*C({cnt[3]},{three})*C({cnt[4]},{four}) {ans}') print(ans)
input = __import__('sys').stdin.readline MOD = 998244353 fact = [1] invfact = [1] for i in range(1, 101): fact.append(fact[-1] * i % MOD) invfact.append(pow(fact[-1], MOD-2, MOD)) def C(n, k): if k < 0 or k > n: return 0 return fact[n] * invfact[k] % MOD * invfact[n-k] % MOD def P(n, k): if k < 0 or k > n: return 0 return fact[n] * invfact[n-k] % MOD n = int(input()) coords = [] for _ in range(n): x, y = map(int, input().split()) coords.append((x, y)) min_dist = [10**9] * n dist = [[-1] * n for _ in range(n)] for u in range(n): for v in range(n): dist[u][v] = abs(coords[u][0] - coords[v][0]) + abs(coords[u][1] - coords[v][1]) if u != v: min_dist[u] = min(min_dist[u], dist[u][v]) cnt = [0, 0, 0, 0, 0] vis = [False]*n for u in sorted(range(n), key=lambda x: min_dist[x]): if vis[u]: continue vis[u] = True seen = [False]*n seen[u] = True ptr = 0 found = [u] while ptr < len(found): v = found[ptr] ptr += 1 for w in range(n): if not seen[w] and dist[v][w] == min_dist[v]: seen[w] = True found.append(w) ok = all(dist[found[i]][found[j]] == min_dist[u] for i in range(len(found)) for j in range(i+1, len(found))) if len(found) == 1 or not ok: cnt[1] += 1 else: # print('found', found, ok) cnt[len(found)] += 1 for u in found: vis[u] = True # print('cnt', cnt[1:]) ans = 0 for two in range(cnt[2] + 1): for three in range(cnt[3] + 1): for four in range(cnt[4] + 1): ans += P(n, n - two - 2*three - 3*four) * C(cnt[2], two) % MOD \ * C(cnt[3], three) % MOD \ * C(cnt[4], four) % MOD if ans >= MOD: {{completion}} # print(f'add P({n},{n - two - 2*three - 3*four})*C({cnt[2]},{two})*C({cnt[3]},{three})*C({cnt[4]},{four}) {ans}') print(ans)
ans -= MOD
[{"input": "3\n1 0\n3 0\n2 1", "output": ["9"]}, {"input": "5\n1 2\n2 4\n3 4\n4 4\n1 3", "output": ["240"]}, {"input": "4\n1 0\n3 0\n2 1\n2 0", "output": ["24"]}]
block_completion_000546
block
python
Complete the code in python to solve this programming problem: Description: You are given an array $$$a$$$ consisting of $$$n$$$ positive integers, and an array $$$b$$$, with length $$$n$$$. Initially $$$b_i=0$$$ for each $$$1 \leq i \leq n$$$.In one move you can choose an integer $$$i$$$ ($$$1 \leq i \leq n$$$), and add $$$a_i$$$ to $$$b_i$$$ or subtract $$$a_i$$$ from $$$b_i$$$. What is the minimum number of moves needed to make $$$b$$$ increasing (that is, every element is strictly greater than every element before it)? Input Specification: The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 5000$$$). The second line contains $$$n$$$ integers, $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \leq a_i \leq 10^9$$$) — the elements of the array $$$a$$$. Output Specification: Print a single integer, the minimum number of moves to make $$$b$$$ increasing. Notes: NoteExample $$$1$$$: you can subtract $$$a_1$$$ from $$$b_1$$$, and add $$$a_3$$$, $$$a_4$$$, and $$$a_5$$$ to $$$b_3$$$, $$$b_4$$$, and $$$b_5$$$ respectively. The final array will be [$$$-1$$$, $$$0$$$, $$$3$$$, $$$4$$$, $$$5$$$] after $$$4$$$ moves.Example $$$2$$$: you can reach [$$$-3$$$, $$$-2$$$, $$$-1$$$, $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$] in $$$10$$$ moves. Code: for _ in range(1): n = int(input()) a = list(map(int, input().split())) Min = 1e18 for l in range(n): m = a[l] answer = 1 for i in range(l-1, -1, -1): answer += (m + a[i]) // a[i] m = a[i] * ((m + a[i]) // a[i]) if l + 1 < n: m = 0 for i in range(l + 2, n): # TODO: Your code here Min = min(answer, Min) print(Min)
for _ in range(1): n = int(input()) a = list(map(int, input().split())) Min = 1e18 for l in range(n): m = a[l] answer = 1 for i in range(l-1, -1, -1): answer += (m + a[i]) // a[i] m = a[i] * ((m + a[i]) // a[i]) if l + 1 < n: m = 0 for i in range(l + 2, n): {{completion}} Min = min(answer, Min) print(Min)
answer += (m + a[i]) // a[i] m = a[i] * ((m + a[i]) // a[i])
[{"input": "5\n1 2 3 4 5", "output": ["4"]}, {"input": "7\n1 2 1 2 1 2 1", "output": ["10"]}, {"input": "8\n1 8 2 7 3 6 4 5", "output": ["16"]}]
block_completion_000979
block
python
Complete the code in python to solve this programming problem: Description: Stanley has decided to buy a new desktop PC made by the company "Monoblock", and to solve captcha on their website, he needs to solve the following task.The awesomeness of an array is the minimum number of blocks of consecutive identical numbers in which the array could be split. For example, the awesomeness of an array $$$[1, 1, 1]$$$ is $$$1$$$; $$$[5, 7]$$$ is $$$2$$$, as it could be split into blocks $$$[5]$$$ and $$$[7]$$$; $$$[1, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9]$$$ is 3, as it could be split into blocks $$$[1]$$$, $$$[7, 7, 7, 7, 7, 7, 7]$$$, and $$$[9, 9, 9, 9, 9, 9, 9, 9, 9]$$$. You are given an array $$$a$$$ of length $$$n$$$. There are $$$m$$$ queries of two integers $$$i$$$, $$$x$$$. A query $$$i$$$, $$$x$$$ means that from now on the $$$i$$$-th element of the array $$$a$$$ is equal to $$$x$$$.After each query print the sum of awesomeness values among all subsegments of array $$$a$$$. In other words, after each query you need to calculate $$$$$$\sum\limits_{l = 1}^n \sum\limits_{r = l}^n g(l, r),$$$$$$ where $$$g(l, r)$$$ is the awesomeness of the array $$$b = [a_l, a_{l + 1}, \ldots, a_r]$$$. Input Specification: In the first line you are given with two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 10^5$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$. In the next $$$m$$$ lines you are given the descriptions of queries. Each line contains two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$1 \leq x \leq 10^9$$$). Output Specification: Print the answer to each query on a new line. Notes: NoteAfter the first query $$$a$$$ is equal to $$$[1, 2, 2, 4, 5]$$$, and the answer is $$$29$$$ because we can split each of the subsegments the following way: $$$[1; 1]$$$: $$$[1]$$$, 1 block; $$$[1; 2]$$$: $$$[1] + [2]$$$, 2 blocks; $$$[1; 3]$$$: $$$[1] + [2, 2]$$$, 2 blocks; $$$[1; 4]$$$: $$$[1] + [2, 2] + [4]$$$, 3 blocks; $$$[1; 5]$$$: $$$[1] + [2, 2] + [4] + [5]$$$, 4 blocks; $$$[2; 2]$$$: $$$[2]$$$, 1 block; $$$[2; 3]$$$: $$$[2, 2]$$$, 1 block; $$$[2; 4]$$$: $$$[2, 2] + [4]$$$, 2 blocks; $$$[2; 5]$$$: $$$[2, 2] + [4] + [5]$$$, 3 blocks; $$$[3; 3]$$$: $$$[2]$$$, 1 block; $$$[3; 4]$$$: $$$[2] + [4]$$$, 2 blocks; $$$[3; 5]$$$: $$$[2] + [4] + [5]$$$, 3 blocks; $$$[4; 4]$$$: $$$[4]$$$, 1 block; $$$[4; 5]$$$: $$$[4] + [5]$$$, 2 blocks; $$$[5; 5]$$$: $$$[5]$$$, 1 block; which is $$$1 + 2 + 2 + 3 + 4 + 1 + 1 + 2 + 3 + 1 + 2 + 3 + 1 + 2 + 1 = 29$$$ in total. Code: from sys import stdin input = stdin.readline inp = lambda : list(map(int,input().split())) def update(i , t): global ans if(i + 1 < n and a[i] == a[i + 1]): ans += t * (i + 1) else: ans += t * (n - i) * (i + 1) return ans def answer(): global ans ans = 0 for i in range(n): update(i , 1) for q in range(m): i , x = inp() i -= 1 if# TODO: Your code here:update(i - 1 , -1) update(i , -1) a[i] = x if(i >= 0):update(i - 1 , 1) update(i , 1) print(ans) for T in range(1): n , m = inp() a = inp() answer()
from sys import stdin input = stdin.readline inp = lambda : list(map(int,input().split())) def update(i , t): global ans if(i + 1 < n and a[i] == a[i + 1]): ans += t * (i + 1) else: ans += t * (n - i) * (i + 1) return ans def answer(): global ans ans = 0 for i in range(n): update(i , 1) for q in range(m): i , x = inp() i -= 1 if{{completion}}:update(i - 1 , -1) update(i , -1) a[i] = x if(i >= 0):update(i - 1 , 1) update(i , 1) print(ans) for T in range(1): n , m = inp() a = inp() answer()
(i >= 0)
[{"input": "5 5\n1 2 3 4 5\n3 2\n4 2\n3 1\n2 1\n2 2", "output": ["29\n23\n35\n25\n35"]}]
control_completion_000077
control_fixed
python
Complete the code in python to solve this programming problem: Description: There is a grid with $$$n$$$ rows and $$$m$$$ columns, and three types of cells: An empty cell, denoted with '.'. A stone, denoted with '*'. An obstacle, denoted with the lowercase Latin letter 'o'. All stones fall down until they meet the floor (the bottom row), an obstacle, or other stone which is already immovable. (In other words, all the stones just fall down as long as they can fall.)Simulate the process. What does the resulting grid look like? Input Specification: The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 50$$$) — the number of rows and the number of columns in the grid, respectively. Then $$$n$$$ lines follow, each containing $$$m$$$ characters. Each of these characters is either '.', '*', or 'o' — an empty cell, a stone, or an obstacle, respectively. Output Specification: For each test case, output a grid with $$$n$$$ rows and $$$m$$$ columns, showing the result of the process. You don't need to output a new line after each test, it is in the samples just for clarity. Code: t = int(input()) for i in range (t): n, m = map(int,input().split()) arr = [[0]*m]*n for j in range(n): arr[j] = list(input()) # for h in range(m): # print(arr[j][h]) for k in range(m): for l in range(n-1, -1, -1): if arr[l][k]=='.': # print("yes") for f in range(l-1,-1,-1): if arr[f][k]=='o': break elif # TODO: Your code here: # print("yes") arr[f][k]='.' arr[l][k]='*' break for g in range(n): for h in range(m-1): print(arr[g][h],end="") print(arr[g][m-1],end="\n")
t = int(input()) for i in range (t): n, m = map(int,input().split()) arr = [[0]*m]*n for j in range(n): arr[j] = list(input()) # for h in range(m): # print(arr[j][h]) for k in range(m): for l in range(n-1, -1, -1): if arr[l][k]=='.': # print("yes") for f in range(l-1,-1,-1): if arr[f][k]=='o': break elif {{completion}}: # print("yes") arr[f][k]='.' arr[l][k]='*' break for g in range(n): for h in range(m-1): print(arr[g][h],end="") print(arr[g][m-1],end="\n")
arr[f][k]=='*'
[{"input": "3\n6 10\n.*.*....*.\n.*.......*\n...o....o.\n.*.*....*.\n..........\n.o......o*\n2 9\n...***ooo\n.*o.*o.*o\n5 5\n*****\n*....\n*****\n....*\n*****", "output": ["..........\n...*....*.\n.*.o....o.\n.*........\n.*......**\n.o.*....o*\n\n....**ooo\n.*o**o.*o\n\n.....\n*...*\n*****\n*****\n*****"]}]
control_completion_000838
control_fixed
python
Complete the code in python to solve this programming problem: Description: Monocarp plays "Rage of Empires II: Definitive Edition" — a strategic computer game. Right now he's planning to attack his opponent in the game, but Monocarp's forces cannot enter the opponent's territory since the opponent has built a wall.The wall consists of $$$n$$$ sections, aligned in a row. The $$$i$$$-th section initially has durability $$$a_i$$$. If durability of some section becomes $$$0$$$ or less, this section is considered broken.To attack the opponent, Monocarp needs to break at least two sections of the wall (any two sections: possibly adjacent, possibly not). To do this, he plans to use an onager — a special siege weapon. The onager can be used to shoot any section of the wall; the shot deals $$$2$$$ damage to the target section and $$$1$$$ damage to adjacent sections. In other words, if the onager shoots at the section $$$x$$$, then the durability of the section $$$x$$$ decreases by $$$2$$$, and the durability of the sections $$$x - 1$$$ and $$$x + 1$$$ (if they exist) decreases by $$$1$$$ each. Monocarp can shoot at any sections any number of times, he can even shoot at broken sections.Monocarp wants to calculate the minimum number of onager shots needed to break at least two sections. Help him! Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the number of sections. The second line contains the sequence of integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$), where $$$a_i$$$ is the initial durability of the $$$i$$$-th section. Output Specification: Print one integer — the minimum number of onager shots needed to break at least two sections of the wall. Notes: NoteIn the first example, it is possible to break the $$$2$$$-nd and the $$$4$$$-th section in $$$10$$$ shots, for example, by shooting the third section $$$10$$$ times. After that, the durabilities become $$$[20, 0, 10, 0, 20]$$$. Another way of doing it is firing $$$5$$$ shots at the $$$2$$$-nd section, and another $$$5$$$ shots at the $$$4$$$-th section. After that, the durabilities become $$$[15, 0, 20, 0, 15]$$$.In the second example, it is enough to shoot the $$$2$$$-nd section once. Then the $$$1$$$-st and the $$$3$$$-rd section will be broken.In the third example, it is enough to shoot the $$$2$$$-nd section twice (then the durabilities become $$$[5, 2, 4, 8, 5, 8]$$$), and then shoot the $$$3$$$-rd section twice (then the durabilities become $$$[5, 0, 0, 6, 5, 8]$$$). So, four shots are enough to break the $$$2$$$-nd and the $$$3$$$-rd section. Code: N=int(input()) A=[int(x) for x in input().split()] B=sorted(A) ans=-(-B[0]//2)-(-B[1]//2) for i in range(N-2): # TODO: Your code here for i in range(N-1): score=max(-(-(A[i]+A[i+1])//3),-(-A[i]//2),-(-A[i+1]//2)) ans=min(score,ans) print(ans)
N=int(input()) A=[int(x) for x in input().split()] B=sorted(A) ans=-(-B[0]//2)-(-B[1]//2) for i in range(N-2): {{completion}} for i in range(N-1): score=max(-(-(A[i]+A[i+1])//3),-(-A[i]//2),-(-A[i+1]//2)) ans=min(score,ans) print(ans)
ans=min(ans,-(-(A[i]+A[i+2])//2))
[{"input": "5\n20 10 30 10 20", "output": ["10"]}, {"input": "3\n1 8 1", "output": ["1"]}, {"input": "6\n7 6 6 8 5 8", "output": ["4"]}, {"input": "6\n14 3 8 10 15 4", "output": ["4"]}, {"input": "4\n1 100 100 1", "output": ["2"]}, {"input": "3\n40 10 10", "output": ["7"]}]
block_completion_007904
block
python
Complete the code in python to solve this programming problem: Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock. Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query. Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase. Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$. Code: arr=[int(i) for i in input().split()] ans=[] prices=[int(i) for i in input().split()] prices.sort(reverse=True) for i in range(1,arr[0]): prices[i]=prices[i]+prices[i-1] for i in range(arr[1]): xy=[int(i) for i in input().split()] if# TODO: Your code here: ans.append(prices[xy[0]-1]) else: ans.append(prices[xy[0]-1]-prices[xy[0]-xy[1]-1]) for ele in ans: print(ele)
arr=[int(i) for i in input().split()] ans=[] prices=[int(i) for i in input().split()] prices.sort(reverse=True) for i in range(1,arr[0]): prices[i]=prices[i]+prices[i-1] for i in range(arr[1]): xy=[int(i) for i in input().split()] if{{completion}}: ans.append(prices[xy[0]-1]) else: ans.append(prices[xy[0]-1]-prices[xy[0]-xy[1]-1]) for ele in ans: print(ele)
(xy[0]==xy[1])
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
control_completion_000508
control_fixed
python
Complete the code in python to solve this programming problem: Description: You are given $$$n$$$ segments on the coordinate axis. The $$$i$$$-th segment is $$$[l_i, r_i]$$$. Let's denote the set of all integer points belonging to the $$$i$$$-th segment as $$$S_i$$$.Let $$$A \cup B$$$ be the union of two sets $$$A$$$ and $$$B$$$, $$$A \cap B$$$ be the intersection of two sets $$$A$$$ and $$$B$$$, and $$$A \oplus B$$$ be the symmetric difference of $$$A$$$ and $$$B$$$ (a set which contains all elements of $$$A$$$ and all elements of $$$B$$$, except for the ones that belong to both sets).Let $$$[\mathbin{op}_1, \mathbin{op}_2, \dots, \mathbin{op}_{n-1}]$$$ be an array where each element is either $$$\cup$$$, $$$\oplus$$$, or $$$\cap$$$. Over all $$$3^{n-1}$$$ ways to choose this array, calculate the sum of the following values:$$$$$$|(((S_1\ \mathbin{op}_1\ S_2)\ \mathbin{op}_2\ S_3)\ \mathbin{op}_3\ S_4)\ \dots\ \mathbin{op}_{n-1}\ S_n|$$$$$$In this expression, $$$|S|$$$ denotes the size of the set $$$S$$$. Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 3 \cdot 10^5$$$). Then, $$$n$$$ lines follow. The $$$i$$$-th of them contains two integers $$$l_i$$$ and $$$r_i$$$ ($$$0 \le l_i \le r_i \le 3 \cdot 10^5$$$). Output Specification: Print one integer — the sum of $$$|(((S_1\ \mathbin{op}_1\ S_2)\ \mathbin{op}_2\ S_3)\ \mathbin{op}_3\ S_4)\ \dots\ \mathbin{op}_{n-1}\ S_n|$$$ over all possible ways to choose $$$[\mathbin{op}_1, \mathbin{op}_2, \dots, \mathbin{op}_{n-1}]$$$. Since the answer can be huge, print it modulo $$$998244353$$$. Code: import sys import heapq from collections import Counter # sys.setrecursionlimit(10000) def input_general(): return sys.stdin.readline().rstrip('\r\n') def input_num(): return int(sys.stdin.readline().rstrip("\r\n")) def input_multi(x=int): return map(x, sys.stdin.readline().rstrip("\r\n").split()) def input_list(x=int): return list(input_multi(x)) def main(): def mod_pow(p, a, e): base = a answer = 1 while e: if e & 1: # TODO: Your code here base = (base * base) % p e >>= 1 return answer n = input_num() hp = [] pos = [[] for _ in range(300001)] for i in range(n): l, r = input_multi() pos[l].append((i, r + 1)) P = 998244353 two_inv = (P + 1) // 2 loc = [-1] * 300001 for i in range(300001): for (idx, r) in pos[i]: heapq.heappush(hp, (-idx, r)) while hp and hp[0][1] <= i: heapq.heappop(hp) if hp: loc[i] = -hp[0][0] ctr = Counter(loc) max_loc = max(ctr.keys()) curr = mod_pow(P, 2, n - 1) answer = (curr * (ctr[0] + ctr[1])) % P for i in range(2, max_loc + 1): curr = (curr * two_inv * 3) % P answer = (answer + curr * ctr[i]) % P print(answer) if __name__ == "__main__": # cases = input_num() # # for _ in range(cases): main()
import sys import heapq from collections import Counter # sys.setrecursionlimit(10000) def input_general(): return sys.stdin.readline().rstrip('\r\n') def input_num(): return int(sys.stdin.readline().rstrip("\r\n")) def input_multi(x=int): return map(x, sys.stdin.readline().rstrip("\r\n").split()) def input_list(x=int): return list(input_multi(x)) def main(): def mod_pow(p, a, e): base = a answer = 1 while e: if e & 1: {{completion}} base = (base * base) % p e >>= 1 return answer n = input_num() hp = [] pos = [[] for _ in range(300001)] for i in range(n): l, r = input_multi() pos[l].append((i, r + 1)) P = 998244353 two_inv = (P + 1) // 2 loc = [-1] * 300001 for i in range(300001): for (idx, r) in pos[i]: heapq.heappush(hp, (-idx, r)) while hp and hp[0][1] <= i: heapq.heappop(hp) if hp: loc[i] = -hp[0][0] ctr = Counter(loc) max_loc = max(ctr.keys()) curr = mod_pow(P, 2, n - 1) answer = (curr * (ctr[0] + ctr[1])) % P for i in range(2, max_loc + 1): curr = (curr * two_inv * 3) % P answer = (answer + curr * ctr[i]) % P print(answer) if __name__ == "__main__": # cases = input_num() # # for _ in range(cases): main()
answer = (answer * base) % p
[{"input": "4\n3 5\n4 8\n2 2\n1 9", "output": ["162"]}, {"input": "4\n1 9\n3 5\n4 8\n2 2", "output": ["102"]}]
block_completion_002199
block
python
Complete the code in python to solve this programming problem: Description: Little Leon lives in the forest. He has recently noticed that some trees near his favourite path are withering, while the other ones are overhydrated so he decided to learn how to control the level of the soil moisture to save the trees.There are $$$n$$$ trees growing near the path, the current levels of moisture of each tree are denoted by the array $$$a_1, a_2, \dots, a_n$$$. Leon has learned three abilities which will help him to dry and water the soil. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$1, 2, \dots, i$$$ by $$$1$$$. Choose a position $$$i$$$ and decrease the level of moisture of the trees $$$i, i + 1, \dots, n$$$ by $$$1$$$. Increase the level of moisture of all trees by $$$1$$$. Leon wants to know the minimum number of actions he needs to perform to make the moisture of each tree equal to $$$0$$$. Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$)  — the number of test cases. The description of $$$t$$$ test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 200\,000$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \leq a_i \leq 10^9$$$) — the initial levels of trees moisture. It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$200\,000$$$. Output Specification: For each test case output a single integer — the minimum number of actions. It can be shown that the answer exists. Notes: NoteIn the first test case it's enough to apply the operation of adding $$$1$$$ to the whole array $$$2$$$ times. In the second test case you can apply the operation of decreasing $$$4$$$ times on the prefix of length $$$3$$$ and get an array $$$6, 0, 3$$$. After that apply the operation of decreasing $$$6$$$ times on the prefix of length $$$1$$$ and $$$3$$$ times on the suffix of length $$$1$$$. In total, the number of actions will be $$$4 + 6 + 3 = 13$$$. It can be shown that it's impossible to perform less actions to get the required array, so the answer is $$$13$$$. Code: import sys T = int(sys.stdin.readline()) for t in range(T): n = int(sys.stdin.readline()) a = [int(x) for x in sys.stdin.readline().strip().split(' ')] l, r = 0, 0 for i in range(len(a) - 1): x, y = a[i], a[i+1] if x > y: l += x - y elif # TODO: Your code here: r += y - x print(abs(a[-1]-r)+l+r) # 4 # 3 # -2 -2 -2 # 3 # 10 4 7 # 4 # 4 -4 4 -4 # 5 # 1 -2 3 -4 5
import sys T = int(sys.stdin.readline()) for t in range(T): n = int(sys.stdin.readline()) a = [int(x) for x in sys.stdin.readline().strip().split(' ')] l, r = 0, 0 for i in range(len(a) - 1): x, y = a[i], a[i+1] if x > y: l += x - y elif {{completion}}: r += y - x print(abs(a[-1]-r)+l+r) # 4 # 3 # -2 -2 -2 # 3 # 10 4 7 # 4 # 4 -4 4 -4 # 5 # 1 -2 3 -4 5
x < y
[{"input": "4\n3\n-2 -2 -2\n3\n10 4 7\n4\n4 -4 4 -4\n5\n1 -2 3 -4 5", "output": ["2\n13\n36\n33"]}]
control_completion_004124
control_fixed
python
Complete the code in python to solve this programming problem: Description: Suppose you have an integer $$$v$$$. In one operation, you can: either set $$$v = (v + 1) \bmod 32768$$$ or set $$$v = (2 \cdot v) \bmod 32768$$$. You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. What is the minimum number of operations you need to make each $$$a_i$$$ equal to $$$0$$$? Input Specification: The first line contains the single integer $$$n$$$ ($$$1 \le n \le 32768$$$) — the number of integers. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i &lt; 32768$$$). Output Specification: Print $$$n$$$ integers. The $$$i$$$-th integer should be equal to the minimum number of operations required to make $$$a_i$$$ equal to $$$0$$$. Notes: NoteLet's consider each $$$a_i$$$: $$$a_1 = 19$$$. You can, firstly, increase it by one to get $$$20$$$ and then multiply it by two $$$13$$$ times. You'll get $$$0$$$ in $$$1 + 13 = 14$$$ steps. $$$a_2 = 32764$$$. You can increase it by one $$$4$$$ times: $$$32764 \rightarrow 32765 \rightarrow 32766 \rightarrow 32767 \rightarrow 0$$$. $$$a_3 = 10240$$$. You can multiply it by two $$$4$$$ times: $$$10240 \rightarrow 20480 \rightarrow 8192 \rightarrow 16384 \rightarrow 0$$$. $$$a_4 = 49$$$. You can multiply it by two $$$15$$$ times. Code: n,s=open(0) for x in s.split():# TODO: Your code here
n,s=open(0) for x in s.split():{{completion}}
print(min(-int(x)%2**i-i+15for i in range(16)))
[{"input": "4\n19 32764 10240 49", "output": ["14 4 4 15"]}]
block_completion_003352
block
python
Complete the code in python to solve this programming problem: Description: Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from 'a' to 'k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i &lt; j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position.In other words, count the number of pairs $$$(i, j)$$$ ($$$i &lt; j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$.The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow. Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lowercase Latin letters from 'a' to 'k'. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$. Output Specification: For each test case, print a single integer — the number of pairs $$$(i, j)$$$ ($$$i &lt; j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$${s_{i}}_{p} \neq {s_{j}}_{p}$$$. Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++). Notes: NoteFor the first test case the pairs that differ in exactly one position are: ("ab", "cb"), ("ab", "db"), ("ab", "aa"), ("cb", "db") and ("cb", "cc").For the second test case the pairs that differ in exactly one position are: ("aa", "ac"), ("aa", "ca"), ("cc", "ac"), ("cc", "ca"), ("ac", "aa") and ("ca", "aa").For the third test case, the are no pairs satisfying the conditions. Code: for i in range(int(input())): data = [[0 for l in range(11)] for k in range(11)] for j in range(int(input())): first, second = input() data[ord(first)-ord('a')][ord(second)-ord('a')] += 1 answer = 0 for j in range(11): for k in range(11): for l in range(11): if # TODO: Your code here: answer += data[j][k]*data[l][k] if k != l: answer += data[j][k]*data[j][l] print(answer//2)
for i in range(int(input())): data = [[0 for l in range(11)] for k in range(11)] for j in range(int(input())): first, second = input() data[ord(first)-ord('a')][ord(second)-ord('a')] += 1 answer = 0 for j in range(11): for k in range(11): for l in range(11): if {{completion}}: answer += data[j][k]*data[l][k] if k != l: answer += data[j][k]*data[j][l] print(answer//2)
j != l
[{"input": "4\n6\nab\ncb\ndb\naa\ncc\nef\n7\naa\nbb\ncc\nac\nca\nbb\naa\n4\nkk\nkk\nab\nab\n5\njf\njf\njk\njk\njk", "output": ["5\n6\n0\n6"]}]
control_completion_000867
control_fixed
python
Complete the code in python to solve this programming problem: Description: The Narrator has an integer array $$$a$$$ of length $$$n$$$, but he will only tell you the size $$$n$$$ and $$$q$$$ statements, each of them being three integers $$$i, j, x$$$, which means that $$$a_i \mid a_j = x$$$, where $$$|$$$ denotes the bitwise OR operation.Find the lexicographically smallest array $$$a$$$ that satisfies all the statements.An array $$$a$$$ is lexicographically smaller than an array $$$b$$$ of the same length if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the array $$$a$$$ has a smaller element than the corresponding element in $$$b$$$. Input Specification: In the first line you are given with two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 10^5$$$, $$$0 \le q \le 2 \cdot 10^5$$$). In the next $$$q$$$ lines you are given with three integers $$$i$$$, $$$j$$$, and $$$x$$$ ($$$1 \le i, j \le n$$$, $$$0 \le x &lt; 2^{30}$$$) — the statements. It is guaranteed that all $$$q$$$ statements hold for at least one array. Output Specification: On a single line print $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i &lt; 2^{30}$$$) — array $$$a$$$. Notes: NoteIn the first sample, these are all the arrays satisfying the statements: $$$[0, 3, 2, 2]$$$, $$$[2, 1, 0, 0]$$$, $$$[2, 1, 0, 2]$$$, $$$[2, 1, 2, 0]$$$, $$$[2, 1, 2, 2]$$$, $$$[2, 3, 0, 0]$$$, $$$[2, 3, 0, 2]$$$, $$$[2, 3, 2, 0]$$$, $$$[2, 3, 2, 2]$$$. Code: n,q = map(int,input().split()) graph = [set() for _ in range(n)] start = [0xffffffff]*n for _ in range(q): i,j,x = map(int,input().split()) i -= 1; j -= 1 graph[i].add(j) graph[j].add(i) start[i] &= x start[j] &= x for i in range(n): if i in graph[i]: continue val = start[i] for # TODO: Your code here: val &= start[j] start[i] ^= val print(*start)
n,q = map(int,input().split()) graph = [set() for _ in range(n)] start = [0xffffffff]*n for _ in range(q): i,j,x = map(int,input().split()) i -= 1; j -= 1 graph[i].add(j) graph[j].add(i) start[i] &= x start[j] &= x for i in range(n): if i in graph[i]: continue val = start[i] for {{completion}}: val &= start[j] start[i] ^= val print(*start)
j in graph[i]
[{"input": "4 3\n1 2 3\n1 3 2\n4 1 2", "output": ["0 3 2 2"]}, {"input": "1 0", "output": ["0"]}, {"input": "2 1\n1 1 1073741823", "output": ["1073741823 0"]}]
control_completion_000017
control_fixed
python
Complete the code in python to solve this programming problem: Description: Your friend Ivan asked you to help him rearrange his desktop. The desktop can be represented as a rectangle matrix of size $$$n \times m$$$ consisting of characters '.' (empty cell of the desktop) and '*' (an icon).The desktop is called good if all its icons are occupying some prefix of full columns and, possibly, the prefix of the next column (and there are no icons outside this figure). In other words, some amount of first columns will be filled with icons and, possibly, some amount of first cells of the next (after the last full column) column will be also filled with icons (and all the icons on the desktop belong to this figure). This is pretty much the same as the real life icons arrangement.In one move, you can take one icon and move it to any empty cell in the desktop.Ivan loves to add some icons to his desktop and remove them from it, so he is asking you to answer $$$q$$$ queries: what is the minimum number of moves required to make the desktop good after adding/removing one icon?Note that queries are permanent and change the state of the desktop. Input Specification: The first line of the input contains three integers $$$n$$$, $$$m$$$ and $$$q$$$ ($$$1 \le n, m \le 1000; 1 \le q \le 2 \cdot 10^5$$$) — the number of rows in the desktop, the number of columns in the desktop and the number of queries, respectively. The next $$$n$$$ lines contain the description of the desktop. The $$$i$$$-th of them contains $$$m$$$ characters '.' and '*' — the description of the $$$i$$$-th row of the desktop. The next $$$q$$$ lines describe queries. The $$$i$$$-th of them contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$1 \le x_i \le n; 1 \le y_i \le m$$$) — the position of the cell which changes its state (if this cell contained the icon before, then this icon is removed, otherwise an icon appears in this cell). Output Specification: Print $$$q$$$ integers. The $$$i$$$-th of them should be the minimum number of moves required to make the desktop good after applying the first $$$i$$$ queries. Code: h, w, q = map(int, input().split()) p = [False] * (h * w) c = cc = 0 def query(y, x): global c, cc i = x*h+y p[i] = not p[i] if p[i]: if i < c: cc += 1 c += 1 if p[c-1]: cc += 1 else: if # TODO: Your code here: cc -= 1 c -= 1 if p[c]: cc -= 1 return c - cc for y in range(h): inp = input() for x in range(w): if inp[x] == '*': query(y, x) for _ in range(q): x, y = map(int, input().split()) print(query(x-1, y-1))
h, w, q = map(int, input().split()) p = [False] * (h * w) c = cc = 0 def query(y, x): global c, cc i = x*h+y p[i] = not p[i] if p[i]: if i < c: cc += 1 c += 1 if p[c-1]: cc += 1 else: if {{completion}}: cc -= 1 c -= 1 if p[c]: cc -= 1 return c - cc for y in range(h): inp = input() for x in range(w): if inp[x] == '*': query(y, x) for _ in range(q): x, y = map(int, input().split()) print(query(x-1, y-1))
i < c
[{"input": "4 4 8\n..**\n.*..\n*...\n...*\n1 3\n2 3\n3 1\n2 3\n3 4\n4 3\n2 3\n2 2", "output": ["3\n4\n4\n3\n4\n5\n5\n5"]}, {"input": "2 5 5\n*...*\n*****\n1 3\n2 2\n1 3\n1 5\n2 3", "output": ["2\n3\n3\n3\n2"]}]
control_completion_007735
control_fixed
python
Complete the code in python to solve this programming problem: Description: On a beach there are $$$n$$$ huts in a perfect line, hut $$$1$$$ being at the left and hut $$$i+1$$$ being $$$100$$$ meters to the right of hut $$$i$$$, for all $$$1 \le i \le n - 1$$$. In hut $$$i$$$ there are $$$p_i$$$ people.There are $$$m$$$ ice cream sellers, also aligned in a perfect line with all the huts. The $$$i$$$-th ice cream seller has their shop $$$x_i$$$ meters to the right of the first hut. All ice cream shops are at distinct locations, but they may be at the same location as a hut.You want to open a new ice cream shop and you wonder what the best location for your shop is. You can place your ice cream shop anywhere on the beach (not necessarily at an integer distance from the first hut) as long as it is aligned with the huts and the other ice cream shops, even if there is already another ice cream shop or a hut at that location. You know that people would come to your shop only if it is strictly closer to their hut than any other ice cream shop.If every person living in the huts wants to buy exactly one ice cream, what is the maximum number of ice creams that you can sell if you place the shop optimally? Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 200\,000$$$, $$$1 \le m \le 200\,000$$$) — the number of huts and the number of ice cream sellers. The second line contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le 10^9$$$) — the number of people in each hut. The third line contains $$$m$$$ integers $$$x_1, x_2, \ldots, x_m$$$ ($$$0 \le x_i \le 10^9$$$, $$$x_i \ne x_j$$$ for $$$i \ne j$$$) — the location of each ice cream shop. Output Specification: Print the maximum number of ice creams that can be sold by choosing optimally the location of the new shop. Notes: NoteIn the first sample, you can place the shop (coloured orange in the picture below) $$$150$$$ meters to the right of the first hut (for example) so that it is the closest shop to the first two huts, which have $$$2$$$ and $$$5$$$ people, for a total of $$$7$$$ sold ice creams. In the second sample, you can place the shop $$$170$$$ meters to the right of the first hut (for example) so that it is the closest shop to the last two huts, which have $$$7$$$ and $$$8$$$ people, for a total of $$$15$$$ sold ice creams. Code: from itertools import chain from sys import stdin (n, m), population, shops = [[int(x) for x in line.split()] for line in stdin.readlines()] shops.sort() shops = chain([float('-inf')], (v / 100 for v in shops), [float('inf')]) shop_left, shop_right = next(shops), next(shops) hut_left_idx = max_score = score = 0 for hut_right_idx, hut_right_score in enumerate(population): score += hut_right_score # print(f'{score=}') while shop_right <= hut_right_idx: shop_left, shop_right = shop_right, next(shops) # print(f'{hut_right_idx=} {shop_left=} {shop_right=}') shop_delta = shop_right - shop_left while shop_left >= hut_left_idx or 2 * (hut_right_idx - hut_left_idx) >= shop_delta: # TODO: Your code here # print(f'{score=} {hut_left_idx=} {hut_right_idx=} {shop_delta=}') if score > max_score: max_score = score print(max_score)
from itertools import chain from sys import stdin (n, m), population, shops = [[int(x) for x in line.split()] for line in stdin.readlines()] shops.sort() shops = chain([float('-inf')], (v / 100 for v in shops), [float('inf')]) shop_left, shop_right = next(shops), next(shops) hut_left_idx = max_score = score = 0 for hut_right_idx, hut_right_score in enumerate(population): score += hut_right_score # print(f'{score=}') while shop_right <= hut_right_idx: shop_left, shop_right = shop_right, next(shops) # print(f'{hut_right_idx=} {shop_left=} {shop_right=}') shop_delta = shop_right - shop_left while shop_left >= hut_left_idx or 2 * (hut_right_idx - hut_left_idx) >= shop_delta: {{completion}} # print(f'{score=} {hut_left_idx=} {hut_right_idx=} {shop_delta=}') if score > max_score: max_score = score print(max_score)
score -= population[hut_left_idx] hut_left_idx += 1
[{"input": "3 1\n2 5 6\n169", "output": ["7"]}, {"input": "4 2\n1 2 7 8\n35 157", "output": ["15"]}, {"input": "4 1\n272203905 348354708 848256926 939404176\n20", "output": ["2136015810"]}, {"input": "3 2\n1 1 1\n300 99", "output": ["2"]}]
block_completion_001150
block
python
Complete the code in python to solve this programming problem: Description: Recently in Divanovo, a huge river locks system was built. There are now $$$n$$$ locks, the $$$i$$$-th of them has the volume of $$$v_i$$$ liters, so that it can contain any amount of water between $$$0$$$ and $$$v_i$$$ liters. Each lock has a pipe attached to it. When the pipe is open, $$$1$$$ liter of water enters the lock every second.The locks system is built in a way to immediately transfer all water exceeding the volume of the lock $$$i$$$ to the lock $$$i + 1$$$. If the lock $$$i + 1$$$ is also full, water will be transferred further. Water exceeding the volume of the last lock pours out to the river. The picture illustrates $$$5$$$ locks with two open pipes at locks $$$1$$$ and $$$3$$$. Because locks $$$1$$$, $$$3$$$, and $$$4$$$ are already filled, effectively the water goes to locks $$$2$$$ and $$$5$$$. Note that the volume of the $$$i$$$-th lock may be greater than the volume of the $$$i + 1$$$-th lock.To make all locks work, you need to completely fill each one of them. The mayor of Divanovo is interested in $$$q$$$ independent queries. For each query, suppose that initially all locks are empty and all pipes are closed. Then, some pipes are opened simultaneously. For the $$$j$$$-th query the mayor asks you to calculate the minimum number of pipes to open so that all locks are filled no later than after $$$t_j$$$ seconds.Please help the mayor to solve this tricky problem and answer his queries. Input Specification: The first lines contains one integer $$$n$$$ ($$$1 \le n \le 200\,000$$$) — the number of locks. The second lines contains $$$n$$$ integers $$$v_1, v_2, \dots, v_n$$$ ($$$1 \le v_i \le 10^9$$$)) — volumes of the locks. The third line contains one integer $$$q$$$ ($$$1 \le q \le 200\,000$$$) — the number of queries. Each of the next $$$q$$$ lines contains one integer $$$t_j$$$ ($$$1 \le t_j \le 10^9$$$) — the number of seconds you have to fill all the locks in the query $$$j$$$. Output Specification: Print $$$q$$$ integers. The $$$j$$$-th of them should be equal to the minimum number of pipes to turn on so that after $$$t_j$$$ seconds all of the locks are filled. If it is impossible to fill all of the locks in given time, print $$$-1$$$. Notes: NoteThere are $$$6$$$ queries in the first example test. In the queries $$$1, 3, 4$$$ the answer is $$$-1$$$. We need to wait $$$4$$$ seconds to fill the first lock even if we open all the pipes. In the sixth query we can open pipes in locks $$$1$$$, $$$3$$$, and $$$4$$$. After $$$4$$$ seconds the locks $$$1$$$ and $$$4$$$ are full. In the following $$$1$$$ second $$$1$$$ liter of water is transferred to the locks $$$2$$$ and $$$5$$$. The lock $$$3$$$ is filled by its own pipe. Similarly, in the second query one can open pipes in locks $$$1$$$, $$$3$$$, and $$$4$$$.In the fifth query one can open pipes $$$1, 2, 3, 4$$$. Code: number = int(input()) V = [int(i) for i in input().split()] time,total = 0,0 for i in range(number): total += V[i] time = max(time,(total+i)//(i+1)) pass for q in range(int(input())): t = int(input()) if# TODO: Your code here: print(-1) else: print((total+t-1)//t) pass
number = int(input()) V = [int(i) for i in input().split()] time,total = 0,0 for i in range(number): total += V[i] time = max(time,(total+i)//(i+1)) pass for q in range(int(input())): t = int(input()) if{{completion}}: print(-1) else: print((total+t-1)//t) pass
(t<time)
[{"input": "5\n4 1 5 4 1\n6\n1\n6\n2\n3\n4\n5", "output": ["-1\n3\n-1\n-1\n4\n3"]}, {"input": "5\n4 4 4 4 4\n6\n1\n3\n6\n5\n2\n4", "output": ["-1\n-1\n4\n4\n-1\n5"]}]
control_completion_004185
control_fixed
python
Complete the code in python to solve this programming problem: Description: You are given an array $$$a$$$ that contains $$$n$$$ integers. You can choose any proper subsegment $$$a_l, a_{l + 1}, \ldots, a_r$$$ of this array, meaning you can choose any two integers $$$1 \le l \le r \le n$$$, where $$$r - l + 1 &lt; n$$$. We define the beauty of a given subsegment as the value of the following expression:$$$$$$\max(a_{1}, a_{2}, \ldots, a_{l-1}, a_{r+1}, a_{r+2}, \ldots, a_{n}) - \min(a_{1}, a_{2}, \ldots, a_{l-1}, a_{r+1}, a_{r+2}, \ldots, a_{n}) + \max(a_{l}, \ldots, a_{r}) - \min(a_{l}, \ldots, a_{r}).$$$$$$Please find the maximum beauty among all proper subsegments. Input Specification: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Then follow the descriptions of each test case. The first line of each test case contains a single integer $$$n$$$ $$$(4 \leq n \leq 10^5)$$$ — the length of the array. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_{i} \leq 10^9$$$) — the elements of the given array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$. Output Specification: For each testcase print a single integer — the maximum beauty of a proper subsegment. Notes: NoteIn the first test case, the optimal segment is $$$l = 7$$$, $$$r = 8$$$. The beauty of this segment equals to $$$(6 - 1) + (5 - 1) = 9$$$.In the second test case, the optimal segment is $$$l = 2$$$, $$$r = 4$$$. The beauty of this segment equals $$$(100 - 2) + (200 - 1) = 297$$$. Code: for # TODO: Your code here:p,q,*_,r,s=sorted(map(int,sdr.split()));print(r+s-p-q)
for {{completion}}:p,q,*_,r,s=sorted(map(int,sdr.split()));print(r+s-p-q)
sdr in[*open(0)][2::2]
[{"input": "4\n8\n1 2 2 3 1 5 6 1\n5\n1 2 3 100 200\n4\n3 3 3 3\n6\n7 8 3 1 1 8", "output": ["9\n297\n0\n14"]}]
control_completion_005298
control_fixed
python
Complete the code in python to solve this programming problem: Description: Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$)  — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. Output Specification: For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. Notes: NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. Code: t = int(input()) for _ in range(t): n = int(input()) c = list(map(int, input().split())) a, e, se, s = [], [0]*n, 0, sum(c) for # TODO: Your code here: se -= e[i-1] n1 = s//i t = 0 if c[i-1] < i + se else 1 a.append(t) s -= (n1 + (i-1)*t) e[i-n1-1] += 1 se += 1 print(*reversed(a))
t = int(input()) for _ in range(t): n = int(input()) c = list(map(int, input().split())) a, e, se, s = [], [0]*n, 0, sum(c) for {{completion}}: se -= e[i-1] n1 = s//i t = 0 if c[i-1] < i + se else 1 a.append(t) s -= (n1 + (i-1)*t) e[i-n1-1] += 1 se += 1 print(*reversed(a))
i in range(n, 0, -1)
[{"input": "5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3", "output": ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"]}]
control_completion_008599
control_fixed
python
Complete the code in python to solve this programming problem: Description: Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match  — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$)  — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b &lt; r \leq n$$$, $$$r+b=n$$$). Output Specification: For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. Notes: NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. Code: for _ in range(int(input())): # TODO: Your code here
for _ in range(int(input())): {{completion}}
n,r,b=map(int,input().split()) t=((r-1)//(b+1)+1) k=t*(b+1)-r res=('R'*t+'B')*(b+1-k)+('R'*(t-1)+'B')*k print(res[:-1]) #
[{"input": "3\n7 4 3\n6 5 1\n19 13 6", "output": ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR"]}, {"input": "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2", "output": ["RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"]}]
block_completion_008710
block
python
Complete the code in python to solve this programming problem: Description: You are given an integer $$$x$$$ and an array of integers $$$a_1, a_2, \ldots, a_n$$$. You have to determine if the number $$$a_1! + a_2! + \ldots + a_n!$$$ is divisible by $$$x!$$$.Here $$$k!$$$ is a factorial of $$$k$$$ — the product of all positive integers less than or equal to $$$k$$$. For example, $$$3! = 1 \cdot 2 \cdot 3 = 6$$$, and $$$5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120$$$. Input Specification: The first line contains two integers $$$n$$$ and $$$x$$$ ($$$1 \le n \le 500\,000$$$, $$$1 \le x \le 500\,000$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le x$$$) — elements of given array. Output Specification: In the only line print "Yes" (without quotes) if $$$a_1! + a_2! + \ldots + a_n!$$$ is divisible by $$$x!$$$, and "No" (without quotes) otherwise. Notes: NoteIn the first example $$$3! + 2! + 2! + 2! + 3! + 3! = 6 + 2 + 2 + 2 + 6 + 6 = 24$$$. Number $$$24$$$ is divisible by $$$4! = 24$$$.In the second example $$$3! + 2! + 2! + 2! + 2! + 2! + 1! + 1! = 18$$$, is divisible by $$$3! = 6$$$.In the third example $$$7! + 7! + 7! + 7! + 7! + 7! + 7! = 7 \cdot 7!$$$. It is easy to prove that this number is not divisible by $$$8!$$$. Code: b=[0]*500001 l=list(map(int,input("").split())) n=l[0] m=l[1] a=list(map(int,input("").split())) e=1 for i in a: b[i]+=1 for i in range(1,l[1]): if b[i]%(i+1)==0: b[i+1]+=(b[i]//(i+1)) else: print("No") e=0 break if e==1: if b[m]!=0 : print("Yes") else: # TODO: Your code here
b=[0]*500001 l=list(map(int,input("").split())) n=l[0] m=l[1] a=list(map(int,input("").split())) e=1 for i in a: b[i]+=1 for i in range(1,l[1]): if b[i]%(i+1)==0: b[i+1]+=(b[i]//(i+1)) else: print("No") e=0 break if e==1: if b[m]!=0 : print("Yes") else: {{completion}}
print("No")
[{"input": "6 4\n3 2 2 2 3 3", "output": ["Yes"]}, {"input": "8 3\n3 2 2 2 2 2 1 1", "output": ["Yes"]}, {"input": "7 8\n7 7 7 7 7 7 7", "output": ["No"]}, {"input": "10 5\n4 3 2 1 4 3 2 4 3 4", "output": ["No"]}, {"input": "2 500000\n499999 499999", "output": ["No"]}]
block_completion_006097
block
python
Complete the code in python to solve this programming problem: Description: Pak Chanek is given an array $$$a$$$ of $$$n$$$ integers. For each $$$i$$$ ($$$1 \leq i \leq n$$$), Pak Chanek will write the one-element set $$$\{a_i\}$$$ on a whiteboard.After that, in one operation, Pak Chanek may do the following: Choose two different sets $$$S$$$ and $$$T$$$ on the whiteboard such that $$$S \cap T = \varnothing$$$ ($$$S$$$ and $$$T$$$ do not have any common elements). Erase $$$S$$$ and $$$T$$$ from the whiteboard and write $$$S \cup T$$$ (the union of $$$S$$$ and $$$T$$$) onto the whiteboard. After performing zero or more operations, Pak Chanek will construct a multiset $$$M$$$ containing the sizes of all sets written on the whiteboard. In other words, each element in $$$M$$$ corresponds to the size of a set after the operations.How many distinct$$$^\dagger$$$ multisets $$$M$$$ can be created by this process? Since the answer may be large, output it modulo $$$998\,244\,353$$$.$$$^\dagger$$$ Multisets $$$B$$$ and $$$C$$$ are different if and only if there exists a value $$$k$$$ such that the number of elements with value $$$k$$$ in $$$B$$$ is different than the number of elements with value $$$k$$$ in $$$C$$$. Input Specification: The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2000$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq n$$$). Output Specification: Output the number of distinct multisets $$$M$$$ modulo $$$998\,244\,353$$$. Notes: NoteIn the first example, the possible multisets $$$M$$$ are $$$\{1,1,1,1,1,1\}$$$, $$$\{1,1,1,1,2\}$$$, $$$\{1,1,1,3\}$$$, $$$\{1,1,2,2\}$$$, $$$\{1,1,4\}$$$, $$$\{1,2,3\}$$$, and $$$\{2,2,2\}$$$.As an example, let's consider a possible sequence of operations. In the beginning, the sets are $$$\{1\}$$$, $$$\{1\}$$$, $$$\{2\}$$$, $$$\{1\}$$$, $$$\{4\}$$$, and $$$\{3\}$$$. Do an operation on sets $$$\{1\}$$$ and $$$\{3\}$$$. Now, the sets are $$$\{1\}$$$, $$$\{1\}$$$, $$$\{2\}$$$, $$$\{4\}$$$, and $$$\{1,3\}$$$. Do an operation on sets $$$\{2\}$$$ and $$$\{4\}$$$. Now, the sets are $$$\{1\}$$$, $$$\{1\}$$$, $$$\{1,3\}$$$, and $$$\{2,4\}$$$. Do an operation on sets $$$\{1,3\}$$$ and $$$\{2,4\}$$$. Now, the sets are $$$\{1\}$$$, $$$\{1\}$$$, and $$$\{1,2,3,4\}$$$. The multiset $$$M$$$ that is constructed is $$$\{1,1,4\}$$$. Code: from sys import stdin, stdout from collections import defaultdict N = 998244353 n = int(stdin.readline()) a = [int(x)-1 for x in stdin.readline().split()] count = [0]*n row_values = [0]*n for i in range(n): row_values[count[a[i]]] += 1 count[a[i]] += 1 row_values_pref = [row_values[0]] for i in range(1,n): row_values_pref.append(row_values_pref[-1]+row_values[i]) dp = [[0]*(n+1) for bar in range(n+1)] for i in range(1,row_values[0]+1): dp[i][i] = 1 for i in range(1,n): for total in range(n-1,0,-1): current = 0 for last in range(total//i, 0, -1): current += dp[total][last] current %= N dp[total][last] = 0 if # TODO: Your code here: dp[total+last][last] += current dp[total+last][last] %= N answer = 0 for i in range(1,n+1): answer += dp[n][i] answer %= N stdout.write(str(answer)+'\n')
from sys import stdin, stdout from collections import defaultdict N = 998244353 n = int(stdin.readline()) a = [int(x)-1 for x in stdin.readline().split()] count = [0]*n row_values = [0]*n for i in range(n): row_values[count[a[i]]] += 1 count[a[i]] += 1 row_values_pref = [row_values[0]] for i in range(1,n): row_values_pref.append(row_values_pref[-1]+row_values[i]) dp = [[0]*(n+1) for bar in range(n+1)] for i in range(1,row_values[0]+1): dp[i][i] = 1 for i in range(1,n): for total in range(n-1,0,-1): current = 0 for last in range(total//i, 0, -1): current += dp[total][last] current %= N dp[total][last] = 0 if {{completion}}: dp[total+last][last] += current dp[total+last][last] %= N answer = 0 for i in range(1,n+1): answer += dp[n][i] answer %= N stdout.write(str(answer)+'\n')
last <= min(row_values_pref[i]-total, n-total)
[{"input": "6\n1 1 2 1 4 3", "output": ["7"]}, {"input": "7\n3 5 4 3 7 4 5", "output": ["11"]}]
control_completion_004651
control_fixed
python
Complete the code in python to solve this programming problem: Description: Once upon a time Mike and Mike decided to come up with an outstanding problem for some stage of ROI (rare olympiad in informatics). One of them came up with a problem prototype but another stole the idea and proposed that problem for another stage of the same olympiad. Since then the first Mike has been waiting for an opportunity to propose the original idea for some other contest... Mike waited until this moment!You are given an array $$$a$$$ of $$$n$$$ integers. You are also given $$$q$$$ queries of two types: Replace $$$i$$$-th element in the array with integer $$$x$$$. Replace each element in the array with integer $$$x$$$. After performing each query you have to calculate the sum of all elements in the array. Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of queries, respectively. The second line contains $$$n$$$ integers $$$a_1, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — elements of the array $$$a$$$. Each of the following $$$q$$$ lines contains a description of the corresponding query. Description begins with integer $$$t$$$ ($$$t \in \{1, 2\}$$$) which denotes a type of the query: If $$$t = 1$$$, then two integers $$$i$$$ and $$$x$$$ are following ($$$1 \le i \le n$$$, $$$1 \le x \le 10^9$$$) — position of replaced element and it's new value. If $$$t = 2$$$, then integer $$$x$$$ is following ($$$1 \le x \le 10^9$$$) — new value of each element in the array. Output Specification: Print $$$q$$$ integers, each on a separate line. In the $$$i$$$-th line print the sum of all elements in the array after performing the first $$$i$$$ queries. Notes: NoteConsider array from the example and the result of performing each query: Initial array is $$$[1, 2, 3, 4, 5]$$$. After performing the first query, array equals to $$$[5, 2, 3, 4, 5]$$$. The sum of all elements is $$$19$$$. After performing the second query, array equals to $$$[10, 10, 10, 10, 10]$$$. The sum of all elements is $$$50$$$. After performing the third query, array equals to $$$[10, 10, 10, 10, 11$$$]. The sum of all elements is $$$51$$$. After performing the fourth query, array equals to $$$[10, 10, 10, 1, 11]$$$. The sum of all elements is $$$42$$$. After performing the fifth query, array equals to $$$[1, 1, 1, 1, 1]$$$. The sum of all elements is $$$5$$$. Code: f = open(0) def R(): return map(int, next(f).split()) n, q = R() d = {} i = v = r = 0 for x in R(): r += x i += 1 d[i] = x while q: q -= 1 t, *x = R() if t & 1: i, x = x r += x - d.get(i, v) d[i] = x else: # TODO: Your code here print(r)
f = open(0) def R(): return map(int, next(f).split()) n, q = R() d = {} i = v = r = 0 for x in R(): r += x i += 1 d[i] = x while q: q -= 1 t, *x = R() if t & 1: i, x = x r += x - d.get(i, v) d[i] = x else: {{completion}} print(r)
d = {} v, = x r = v * n
[{"input": "5 5\n1 2 3 4 5\n1 1 5\n2 10\n1 5 11\n1 4 1\n2 1", "output": ["19\n50\n51\n42\n5"]}]
block_completion_005606
block
python
Complete the code in python to solve this programming problem: Description: Pupils Alice and Ibragim are best friends. It's Ibragim's birthday soon, so Alice decided to gift him a new puzzle. The puzzle can be represented as a matrix with $$$2$$$ rows and $$$n$$$ columns, every element of which is either $$$0$$$ or $$$1$$$. In one move you can swap two values in neighboring cells.More formally, let's number rows $$$1$$$ to $$$2$$$ from top to bottom, and columns $$$1$$$ to $$$n$$$ from left to right. Also, let's denote a cell in row $$$x$$$ and column $$$y$$$ as $$$(x, y)$$$. We consider cells $$$(x_1, y_1)$$$ and $$$(x_2, y_2)$$$ neighboring if $$$|x_1 - x_2| + |y_1 - y_2| = 1$$$.Alice doesn't like the way in which the cells are currently arranged, so she came up with her own arrangement, with which she wants to gift the puzzle to Ibragim. Since you are her smartest friend, she asked you to help her find the minimal possible number of operations in which she can get the desired arrangement. Find this number, or determine that it's not possible to get the new arrangement. Input Specification: The first line contains an integer $$$n$$$ ($$$1 \leq n \leq 200\,000$$$) — the number of columns in the puzzle. Following two lines describe the current arrangement on the puzzle. Each line contains $$$n$$$ integers, every one of which is either $$$0$$$ or $$$1$$$. The last two lines describe Alice's desired arrangement in the same format. Output Specification: If it is possible to get the desired arrangement, print the minimal possible number of steps, otherwise print $$$-1$$$. Notes: NoteIn the first example the following sequence of swaps will suffice: $$$(2, 1), (1, 1)$$$, $$$(1, 2), (1, 3)$$$, $$$(2, 2), (2, 3)$$$, $$$(1, 4), (1, 5)$$$, $$$(2, 5), (2, 4)$$$. It can be shown that $$$5$$$ is the minimal possible answer in this case.In the second example no matter what swaps you do, you won't get the desired arrangement, so the answer is $$$-1$$$. Code: # 1 0 0 1 0 0 # 0 1 0 0 0 1 # 1 1 1 2 2 2 # 0 1 1 1 1 2 # swap two same number is useless # each time we swap i and i+1, only prefix of i is changed, either increased by 1 or decreased by 1 # same argument as of inversion, each time inversion can only increased by 1 or decreased by 1 # so in one step we can always move a prefix closer to the expected by 1 # Let s = sum(diff(prefix[a] - prefix[a'])) , we can always achive this in minimum time # prefix + (0, 1) | (1,0) # we would like to make every prefix of a is same as a' (with a' is fixed) # when we swap pair (0,1) in two row, say we are at column j, then the prefix of one row decreased by 1 and the other is increased by 1 form the column j # 1 1 1 0 0 0 # 0 0 0 1 1 1 # i # Now let's construct array diff of two row # Reprahsed of the problem # In one step, we can either # 1. increase / decrease any diff by 1 # 2. at a certain column, increase diff of either by 1 and decrease the other by 1 till the last column # Analysis # Go from the start since we have increamnt suffix operation # If both element is same sign ,then add their abs since suffix operation not reduce their diff but also increase the total number of operatons, if on suffix operatons help the rest move closer to the target then we can just apply from the i+1 position to get the same result and take 1 less move # If there are of different sign # k = min(abs(row1[i]]), abs(row2[i])) normaly we would take k to make one ddiff move to zeros, but now with suffix opertions, we also move the other closer to 0 by k, so we are have free k more free operations, if this suffix make the rest worst then just apply k reverse suffix operation on i+1 to cancel the efffect so this algorithm always move to a better answer n = int(input()) a, b, x, y = [list(map(int, input().split())) for _ in range(4)] s0 = s1 = ans = 0 for m, n, p, q in zip(a, b, x, y): s0 += m s0 -= p s1 += n s1 -= q while s0 > 0 and s1 < 0: # TODO: Your code here while s0 < 0 and s1 > 0: ans += 1 s0 += 1 s1 -= 1 ans += abs(s0) + abs(s1) print(-1 if s1 or s0 else ans) # finally but prefix must be 0
# 1 0 0 1 0 0 # 0 1 0 0 0 1 # 1 1 1 2 2 2 # 0 1 1 1 1 2 # swap two same number is useless # each time we swap i and i+1, only prefix of i is changed, either increased by 1 or decreased by 1 # same argument as of inversion, each time inversion can only increased by 1 or decreased by 1 # so in one step we can always move a prefix closer to the expected by 1 # Let s = sum(diff(prefix[a] - prefix[a'])) , we can always achive this in minimum time # prefix + (0, 1) | (1,0) # we would like to make every prefix of a is same as a' (with a' is fixed) # when we swap pair (0,1) in two row, say we are at column j, then the prefix of one row decreased by 1 and the other is increased by 1 form the column j # 1 1 1 0 0 0 # 0 0 0 1 1 1 # i # Now let's construct array diff of two row # Reprahsed of the problem # In one step, we can either # 1. increase / decrease any diff by 1 # 2. at a certain column, increase diff of either by 1 and decrease the other by 1 till the last column # Analysis # Go from the start since we have increamnt suffix operation # If both element is same sign ,then add their abs since suffix operation not reduce their diff but also increase the total number of operatons, if on suffix operatons help the rest move closer to the target then we can just apply from the i+1 position to get the same result and take 1 less move # If there are of different sign # k = min(abs(row1[i]]), abs(row2[i])) normaly we would take k to make one ddiff move to zeros, but now with suffix opertions, we also move the other closer to 0 by k, so we are have free k more free operations, if this suffix make the rest worst then just apply k reverse suffix operation on i+1 to cancel the efffect so this algorithm always move to a better answer n = int(input()) a, b, x, y = [list(map(int, input().split())) for _ in range(4)] s0 = s1 = ans = 0 for m, n, p, q in zip(a, b, x, y): s0 += m s0 -= p s1 += n s1 -= q while s0 > 0 and s1 < 0: {{completion}} while s0 < 0 and s1 > 0: ans += 1 s0 += 1 s1 -= 1 ans += abs(s0) + abs(s1) print(-1 if s1 or s0 else ans) # finally but prefix must be 0
ans += 1 s0 -= 1 s1 += 1
[{"input": "5\n0 1 0 1 0\n1 1 0 0 1\n1 0 1 0 1\n0 0 1 1 0", "output": ["5"]}, {"input": "3\n1 0 0\n0 0 0\n0 0 0\n0 0 0", "output": ["-1"]}]
block_completion_004255
block
python
Complete the code in python to solve this programming problem: Description: Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$)  — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. Output Specification: For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. Notes: NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. Code: t = int(input()) for _ in range(t): n = int(input()) c = list(map(int, input().split())) a, e, se, s = [], [0]*n, 0, sum(c) for i in range(n, 0, -1): # TODO: Your code here print(*reversed(a))
t = int(input()) for _ in range(t): n = int(input()) c = list(map(int, input().split())) a, e, se, s = [], [0]*n, 0, sum(c) for i in range(n, 0, -1): {{completion}} print(*reversed(a))
se -= e[i-1] n1 = s//i t = 0 if c[i-1] < i + se else 1 a.append(t) s -= (n1 + (i-1)*t) e[i-n1-1] += 1 se += 1
[{"input": "5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3", "output": ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"]}]
block_completion_008749
block
python
Complete the code in python to solve this programming problem: Description: You are walking with your dog, and now you are at the promenade. The promenade can be represented as an infinite line. Initially, you are in the point $$$0$$$ with your dog. You decided to give some freedom to your dog, so you untied her and let her run for a while. Also, you watched what your dog is doing, so you have some writings about how she ran. During the $$$i$$$-th minute, the dog position changed from her previous position by the value $$$a_i$$$ (it means, that the dog ran for $$$a_i$$$ meters during the $$$i$$$-th minute). If $$$a_i$$$ is positive, the dog ran $$$a_i$$$ meters to the right, otherwise (if $$$a_i$$$ is negative) she ran $$$a_i$$$ meters to the left.During some minutes, you were chatting with your friend, so you don't have writings about your dog movement during these minutes. These values $$$a_i$$$ equal zero.You want your dog to return to you after the end of the walk, so the destination point of the dog after $$$n$$$ minutes should be $$$0$$$.Now you are wondering: what is the maximum possible number of different integer points of the line your dog could visit on her way, if you replace every $$$0$$$ with some integer from $$$-k$$$ to $$$k$$$ (and your dog should return to $$$0$$$ after the walk)? The dog visits an integer point if she runs through that point or reaches in it at the end of any minute. Point $$$0$$$ is always visited by the dog, since she is initially there.If the dog cannot return to the point $$$0$$$ after $$$n$$$ minutes regardless of the integers you place, print -1. Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3000; 1 \le k \le 10^9$$$) — the number of minutes and the maximum possible speed of your dog during the minutes without records. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$), where $$$a_i$$$ is the number of meters your dog ran during the $$$i$$$-th minutes (to the left if $$$a_i$$$ is negative, to the right otherwise). If $$$a_i = 0$$$ then this value is unknown and can be replaced with any integer from the range $$$[-k; k]$$$. Output Specification: If the dog cannot return to the point $$$0$$$ after $$$n$$$ minutes regardless of the set of integers you place, print -1. Otherwise, print one integer — the maximum number of different integer points your dog could visit if you fill all the unknown values optimally and the dog will return to the point $$$0$$$ at the end of the walk. Code: n, k = map(int, input().split()) A = list(map(int, input().split())) ans = 0 for i in range(n): C = [0]*n for j in range(n-1, -1, -1): if A[j] == 0: C[j] = 1 if j+1 < n: C[j] += C[j+1] B = A.copy() s = sum(B) flag = True for j in range(n): if B[j] == 0: if j+1 < n: x = C[j+1] else: # TODO: Your code here B[j] = min(k, x*k-s) if B[j] < -k: flag = False s += B[j] if flag: pos = 0 mn = 0 mx = 0 for j in range(n): pos += B[j] mn = min(mn, pos) mx = max(mx, pos) if pos == 0: ans = max(ans, mx-mn+1) A = A[1:]+A[0:1] if ans != 0: print(ans) else: print(-1)
n, k = map(int, input().split()) A = list(map(int, input().split())) ans = 0 for i in range(n): C = [0]*n for j in range(n-1, -1, -1): if A[j] == 0: C[j] = 1 if j+1 < n: C[j] += C[j+1] B = A.copy() s = sum(B) flag = True for j in range(n): if B[j] == 0: if j+1 < n: x = C[j+1] else: {{completion}} B[j] = min(k, x*k-s) if B[j] < -k: flag = False s += B[j] if flag: pos = 0 mn = 0 mx = 0 for j in range(n): pos += B[j] mn = min(mn, pos) mx = max(mx, pos) if pos == 0: ans = max(ans, mx-mn+1) A = A[1:]+A[0:1] if ans != 0: print(ans) else: print(-1)
x = 0
[{"input": "3 2\n5 0 -4", "output": ["6"]}, {"input": "6 4\n1 -2 0 3 -4 5", "output": ["7"]}, {"input": "3 1000000000\n0 0 0", "output": ["1000000001"]}, {"input": "5 9\n-7 -3 8 12 0", "output": ["-1"]}, {"input": "5 3\n-1 0 3 3 0", "output": ["7"]}, {"input": "5 4\n0 2 0 3 0", "output": ["9"]}]
block_completion_000201
block
python
Complete the code in python to solve this programming problem: Description: Leslie and Leon entered a labyrinth. The labyrinth consists of $$$n$$$ halls and $$$m$$$ one-way passages between them. The halls are numbered from $$$1$$$ to $$$n$$$.Leslie and Leon start their journey in the hall $$$s$$$. Right away, they quarrel and decide to explore the labyrinth separately. However, they want to meet again at the end of their journey.To help Leslie and Leon, your task is to find two different paths from the given hall $$$s$$$ to some other hall $$$t$$$, such that these two paths do not share halls other than the staring hall $$$s$$$ and the ending hall $$$t$$$. The hall $$$t$$$ has not been determined yet, so you can choose any of the labyrinth's halls as $$$t$$$ except $$$s$$$.Leslie's and Leon's paths do not have to be the shortest ones, but their paths must be simple, visiting any hall at most once. Also, they cannot visit any common halls except $$$s$$$ and $$$t$$$ during their journey, even at different times. Input Specification: The first line contains three integers $$$n$$$, $$$m$$$, and $$$s$$$, where $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) is the number of vertices, $$$m$$$ ($$$0 \le m \le 2 \cdot 10^5$$$) is the number of edges in the labyrinth, and $$$s$$$ ($$$1 \le s \le n$$$) is the starting hall. Then $$$m$$$ lines with descriptions of passages follow. Each description contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$; $$$u_i \neq v_i$$$), denoting a passage from the hall $$$u_i$$$ to the hall $$$v_i$$$. The passages are one-way. Each tuple $$$(u_i, v_i)$$$ is present in the input at most once. The labyrinth can contain cycles and is not necessarily connected in any way. Output Specification: If it is possible to find the desired two paths, output "Possible", otherwise output "Impossible". If the answer exists, output two path descriptions. Each description occupies two lines. The first line of the description contains an integer $$$h$$$ ($$$2 \le h \le n$$$) — the number of halls in a path, and the second line contains distinct integers $$$w_1, w_2, \dots, w_h$$$ ($$$w_1 = s$$$; $$$1 \le w_j \le n$$$; $$$w_h = t$$$) — the halls in the path in the order of passing. Both paths must end at the same vertex $$$t$$$. The paths must be different, and all intermediate halls in these paths must be distinct. Code: def DFS(start): nodes=set() stack=[start] while stack: parent=stack.pop() if(not visited[parent]): nodes.add(parent) visited[parent]=True for child in graph[parent]: if # TODO: Your code here: stack.append(child) else: if child not in nodes and child!=s: return child else: if parent not in nodes and parent != s: return parent return -1 def DFS_get_path(start): stack=[start] parent_list[start]=-1 while stack: parent=stack.pop() if parent==end: visited[end]=False return True if(not visited[parent]): visited[parent]=True for child in graph[parent]: if (not visited[child]): stack.append(child) parent_list[child]=parent return False def get_path(node): path=[] while node!=-1: path.append(node) node=parent_list[node] path.reverse() return path n,m,s=map(int,input().split()) s-=1 graph=[[] for _ in range(n)] for _ in range(m): a,b=map(int,input().split()) a-=1 b-=1 graph[a].append(b) visited=[False]*n visited[s]=True for child in graph[s]: end=DFS(child) if end!=-1: visited = [False] * n parent_list=[-1]*n visited[s]=True ans=[] for child in graph[s]: if DFS_get_path(child): ans.append([s]+get_path(end)) if len(ans)==2: break print("Possible") for i in ans: print(len(i)) print(*[j+1 for j in i]) break else: print("Impossible") # 3 3 1 # 1 2 # 2 1 # 1 3
def DFS(start): nodes=set() stack=[start] while stack: parent=stack.pop() if(not visited[parent]): nodes.add(parent) visited[parent]=True for child in graph[parent]: if {{completion}}: stack.append(child) else: if child not in nodes and child!=s: return child else: if parent not in nodes and parent != s: return parent return -1 def DFS_get_path(start): stack=[start] parent_list[start]=-1 while stack: parent=stack.pop() if parent==end: visited[end]=False return True if(not visited[parent]): visited[parent]=True for child in graph[parent]: if (not visited[child]): stack.append(child) parent_list[child]=parent return False def get_path(node): path=[] while node!=-1: path.append(node) node=parent_list[node] path.reverse() return path n,m,s=map(int,input().split()) s-=1 graph=[[] for _ in range(n)] for _ in range(m): a,b=map(int,input().split()) a-=1 b-=1 graph[a].append(b) visited=[False]*n visited[s]=True for child in graph[s]: end=DFS(child) if end!=-1: visited = [False] * n parent_list=[-1]*n visited[s]=True ans=[] for child in graph[s]: if DFS_get_path(child): ans.append([s]+get_path(end)) if len(ans)==2: break print("Possible") for i in ans: print(len(i)) print(*[j+1 for j in i]) break else: print("Impossible") # 3 3 1 # 1 2 # 2 1 # 1 3
(not visited[child])
[{"input": "5 5 1\n1 2\n2 3\n1 4\n4 3\n3 5", "output": ["Possible\n3\n1 2 3\n3\n1 4 3"]}, {"input": "5 5 1\n1 2\n2 3\n3 4\n2 5\n5 4", "output": ["Impossible"]}, {"input": "3 3 2\n1 2\n2 3\n3 1", "output": ["Impossible"]}]
control_completion_003114
control_fixed
python
Complete the code in python to solve this programming problem: Description: A basketball competition is held where the number of players in a team does not have a maximum or minimum limit (not necessarily $$$5$$$ players in one team for each match). There are $$$N$$$ candidate players in the competition that will be trained by Pak Chanek, the best basketball coach on earth. The $$$i$$$-th candidate player has a power of $$$P_i$$$.Pak Chanek will form zero or more teams from the $$$N$$$ candidate players on the condition that each candidate player may only join in at most one team. Each of Pak Chanek's teams will be sent to compete once with an enemy team that has a power of $$$D$$$. In each match, the team sent is said to defeat the enemy team if the sum of powers from the formed players is strictly greater than $$$D$$$.One of Pak Chanek's skills is that when a team that has been formed plays in a match, he can change the power of each player in the team to be equal to the biggest player power from the team.Determine the maximum number of wins that can be achieved by Pak Chanek. Input Specification: The first line contains two integers $$$N$$$ and $$$D$$$ ($$$1 \le N \le 10^5$$$, $$$1 \le D \le 10^9$$$) — the number of candidate players and the power of the enemy team. The second line contains $$$N$$$ integers $$$P_1, P_2, \ldots, P_N$$$ ($$$1 \le P_i \le 10^9$$$) — the powers of all candidate players. Output Specification: A line containing an integer representing the maximum number of wins that can be achieved by Pak Chanek. Notes: NoteThe $$$1$$$-st team formed is a team containing players $$$4$$$ and $$$6$$$. The power of each player in the team becomes $$$100$$$. So the total power of the team is $$$100 + 100 = 200 &gt; 180$$$.The $$$2$$$-nd team formed is a team containing players $$$1$$$, $$$2$$$, and $$$5$$$. The power of each player in the team becomes $$$90$$$. So the total power of the team is $$$90 + 90 + 90 = 270 &gt; 180$$$. Code: def solve(): n, d = [int(i) for i in input().split(' ')] power = [int(i) for i in input().split(' ')] power.sort() used = 0 w = 0 for i in range(len(power)-1, -1, -1): min_players = -(d // -power[i]) p = power[i] * min_players if(p > d): used += min_players elif# TODO: Your code here: used += min_players + 1 if(used > n): break w += 1 print(w) solve()
def solve(): n, d = [int(i) for i in input().split(' ')] power = [int(i) for i in input().split(' ')] power.sort() used = 0 w = 0 for i in range(len(power)-1, -1, -1): min_players = -(d // -power[i]) p = power[i] * min_players if(p > d): used += min_players elif{{completion}}: used += min_players + 1 if(used > n): break w += 1 print(w) solve()
(p == d)
[{"input": "6 180\n90 80 70 60 50 100", "output": ["2"]}]
control_completion_003664
control_fixed
python
Complete the code in python to solve this programming problem: Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock. Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query. Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase. Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$. Code: n,q=map(int,input().split()) a=[0] for x in sorted(map(int,input().split()))[::-1]:a+=a[-1]+x, for _ in[0]*q:# TODO: Your code here
n,q=map(int,input().split()) a=[0] for x in sorted(map(int,input().split()))[::-1]:a+=a[-1]+x, for _ in[0]*q:{{completion}}
x,y=map(int,input().split());print(a[x]-a[x-y])
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
block_completion_000513
block
python
Complete the code in python to solve this programming problem: Description: A ticket is a string consisting of six digits. A ticket is considered lucky if the sum of the first three digits is equal to the sum of the last three digits. Given a ticket, output if it is lucky or not. Note that a ticket can have leading zeroes. Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of testcases. The description of each test consists of one line containing one string consisting of six digits. Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. Output "YES" if the given ticket is lucky, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer). Notes: NoteIn the first test case, the sum of the first three digits is $$$2 + 1 + 3 = 6$$$ and the sum of the last three digits is $$$1 + 3 + 2 = 6$$$, they are equal so the answer is "YES".In the second test case, the sum of the first three digits is $$$9 + 7 + 3 = 19$$$ and the sum of the last three digits is $$$8 + 9 + 4 = 21$$$, they are not equal so the answer is "NO".In the third test case, the sum of the first three digits is $$$0 + 4 + 5 = 9$$$ and the sum of the last three digits is $$$2 + 0 + 7 = 9$$$, they are equal so the answer is "YES". Code: t = int(input()) for i in range(1, t + 1): summa = 0 a = int(input()) a6 = a % 10 a5 = (a // 10) % 10 a4 = (a // 100) % 10 a3 = (a // 1000) % 10 a2 = (a // 10000) % 10 a1 = (a // 100000) % 10 if # TODO: Your code here: print('YES') else: print("NO")
t = int(input()) for i in range(1, t + 1): summa = 0 a = int(input()) a6 = a % 10 a5 = (a // 10) % 10 a4 = (a // 100) % 10 a3 = (a // 1000) % 10 a2 = (a // 10000) % 10 a1 = (a // 100000) % 10 if {{completion}}: print('YES') else: print("NO")
a1 + a2 + a3 == a4 + a5 + a6
[{"input": "5\n213132\n973894\n045207\n000000\n055776", "output": ["YES\nNO\nYES\nYES\nNO"]}]
control_completion_007488
control_fixed
python
Complete the code in python to solve this programming problem: Description: There is a grid with $$$n$$$ rows and $$$m$$$ columns, and three types of cells: An empty cell, denoted with '.'. A stone, denoted with '*'. An obstacle, denoted with the lowercase Latin letter 'o'. All stones fall down until they meet the floor (the bottom row), an obstacle, or other stone which is already immovable. (In other words, all the stones just fall down as long as they can fall.)Simulate the process. What does the resulting grid look like? Input Specification: The input consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. The description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n, m \leq 50$$$) — the number of rows and the number of columns in the grid, respectively. Then $$$n$$$ lines follow, each containing $$$m$$$ characters. Each of these characters is either '.', '*', or 'o' — an empty cell, a stone, or an obstacle, respectively. Output Specification: For each test case, output a grid with $$$n$$$ rows and $$$m$$$ columns, showing the result of the process. You don't need to output a new line after each test, it is in the samples just for clarity. Code: for _ in range(int(input())): n, _ = map(int, input().split()) a = map("".join, zip(*(input() for _ in range(n)))) a = ("o".join("".join(sorted(y, reverse=True)) for y in x.split("o")) for x in a) for x in zip(*a): # TODO: Your code here
for _ in range(int(input())): n, _ = map(int, input().split()) a = map("".join, zip(*(input() for _ in range(n)))) a = ("o".join("".join(sorted(y, reverse=True)) for y in x.split("o")) for x in a) for x in zip(*a): {{completion}}
print("".join(x))
[{"input": "3\n6 10\n.*.*....*.\n.*.......*\n...o....o.\n.*.*....*.\n..........\n.o......o*\n2 9\n...***ooo\n.*o.*o.*o\n5 5\n*****\n*....\n*****\n....*\n*****", "output": ["..........\n...*....*.\n.*.o....o.\n.*........\n.*......**\n.o.*....o*\n\n....**ooo\n.*o**o.*o\n\n.....\n*...*\n*****\n*****\n*****"]}]
block_completion_000845
block
python
Complete the code in python to solve this programming problem: Description: There is a grid, consisting of $$$n$$$ rows and $$$m$$$ columns. The rows are numbered from $$$1$$$ to $$$n$$$ from bottom to top. The columns are numbered from $$$1$$$ to $$$m$$$ from left to right. The $$$i$$$-th column has the bottom $$$a_i$$$ cells blocked (the cells in rows $$$1, 2, \dots, a_i$$$), the remaining $$$n - a_i$$$ cells are unblocked.A robot is travelling across this grid. You can send it commands — move up, right, down or left. If a robot attempts to move into a blocked cell or outside the grid, it explodes.However, the robot is broken — it executes each received command $$$k$$$ times. So if you tell it to move up, for example, it will move up $$$k$$$ times ($$$k$$$ cells). You can't send it commands while the robot executes the current one.You are asked $$$q$$$ queries about the robot. Each query has a start cell, a finish cell and a value $$$k$$$. Can you send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times?The robot must stop in the finish cell. If it visits the finish cell while still executing commands, it doesn't count. Input Specification: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 10^9$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of rows and columns of the grid. The second line contains $$$m$$$ integers $$$a_1, a_2, \dots, a_m$$$ ($$$0 \le a_i \le n$$$) — the number of blocked cells on the bottom of the $$$i$$$-th column. The third line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Each of the next $$$q$$$ lines contain five integers $$$x_s, y_s, x_f, y_f$$$ and $$$k$$$ ($$$a[y_s] &lt; x_s \le n$$$; $$$1 \le y_s \le m$$$; $$$a[y_f] &lt; x_f \le n$$$; $$$1 \le y_f \le m$$$; $$$1 \le k \le 10^9$$$) — the row and the column of the start cell, the row and the column of the finish cell and the number of times each your command is executed. The start and the finish cell of each query are unblocked. Output Specification: For each query, print "YES" if you can send the robot an arbitrary number of commands (possibly, zero) so that it reaches the finish cell from the start cell, given that it executes each command $$$k$$$ times. Otherwise, print "NO". Code: from sys import stdin, stdout from math import floor, ceil, log input, print = stdin.readline, stdout.write def main(): n, m = map(int, input().split()) arr = [0] + [int(x) for x in input().split()] st = construct(arr, m) for _ in range(int(input())): x1, y1, x2, y2, k = map(int, input().split()) if # TODO: Your code here: print("NO\n") continue if (x1 <= arr[y1] or x2 <= arr[y2]): print("NO\n") continue max_x = ((n-x1)//k)*k + x1 if (max_x <= getMax(st, m, min(y1, y2), max(y1, y2))): print("NO\n") continue print("YES\n") def construct(arr, n): x = ceil(log(n, 2)) max_size = 2 * pow(2, x) - 1 st = [0]*max_size construct2(arr, 0, n-1, st, 0) return st def construct2(arr, ss, se, st, si): if (ss == se): st[si] = arr[ss] return arr[ss] mid = getMid(ss, se) st[si] = max(construct2(arr, ss, mid, st, si * 2 + 1), construct2(arr, mid + 1, se, st, si * 2 + 2)) return st[si] def getMid(s, e): return s + (e - s) // 2 def getMax(st, n, l, r): return MaxUtil(st, 0, n - 1, l, r, 0) def MaxUtil(st, ss, se, l, r, node): if (l <= ss and r >= se): return st[node] if (se < l or ss > r): return -1 mid = getMid(ss, se) return max(MaxUtil(st, ss, mid, l, r, 2 * node + 1), MaxUtil(st, mid + 1, se, l, r, 2 * node + 2)) main()
from sys import stdin, stdout from math import floor, ceil, log input, print = stdin.readline, stdout.write def main(): n, m = map(int, input().split()) arr = [0] + [int(x) for x in input().split()] st = construct(arr, m) for _ in range(int(input())): x1, y1, x2, y2, k = map(int, input().split()) if {{completion}}: print("NO\n") continue if (x1 <= arr[y1] or x2 <= arr[y2]): print("NO\n") continue max_x = ((n-x1)//k)*k + x1 if (max_x <= getMax(st, m, min(y1, y2), max(y1, y2))): print("NO\n") continue print("YES\n") def construct(arr, n): x = ceil(log(n, 2)) max_size = 2 * pow(2, x) - 1 st = [0]*max_size construct2(arr, 0, n-1, st, 0) return st def construct2(arr, ss, se, st, si): if (ss == se): st[si] = arr[ss] return arr[ss] mid = getMid(ss, se) st[si] = max(construct2(arr, ss, mid, st, si * 2 + 1), construct2(arr, mid + 1, se, st, si * 2 + 2)) return st[si] def getMid(s, e): return s + (e - s) // 2 def getMax(st, n, l, r): return MaxUtil(st, 0, n - 1, l, r, 0) def MaxUtil(st, ss, se, l, r, node): if (l <= ss and r >= se): return st[node] if (se < l or ss > r): return -1 mid = getMid(ss, se) return max(MaxUtil(st, ss, mid, l, r, 2 * node + 1), MaxUtil(st, mid + 1, se, l, r, 2 * node + 2)) main()
((y2 - y1) % k != 0 or (x2 - x1) % k != 0)
[{"input": "11 10\n9 0 0 10 3 4 8 11 10 8\n6\n1 2 1 3 1\n1 2 1 3 2\n4 3 4 5 2\n5 3 11 5 3\n5 3 11 5 2\n11 9 9 10 1", "output": ["YES\nNO\nNO\nNO\nYES\nYES"]}]
control_completion_002942
control_fixed
python
Complete the code in python to solve this programming problem: Description: This is the easy version of the problem. The only difference between the versions is the constraints on $$$n$$$, $$$k$$$, $$$a_i$$$, and the sum of $$$n$$$ over all test cases. You can make hacks only if both versions of the problem are solved.Note the unusual memory limit.You are given an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$, and an integer $$$k$$$.The cost of an array of integers $$$p_1, p_2, \ldots, p_n$$$ of length $$$n$$$ is $$$$$$\max\limits_{1 \le i \le n}\left(\left \lfloor \frac{a_i}{p_i} \right \rfloor \right) - \min\limits_{1 \le i \le n}\left(\left \lfloor \frac{a_i}{p_i} \right \rfloor \right).$$$$$$Here, $$$\lfloor \frac{x}{y} \rfloor$$$ denotes the integer part of the division of $$$x$$$ by $$$y$$$. Find the minimum cost of an array $$$p$$$ such that $$$1 \le p_i \le k$$$ for all $$$1 \le i \le n$$$. Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n, k \le 3000$$$). The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_1 \le a_2 \le \ldots \le a_n \le 3000$$$). It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3000$$$. Output Specification: For each test case, print a single integer — the minimum possible cost of an array $$$p$$$ satisfying the condition above. Notes: NoteIn the first test case, the optimal array is $$$p = [1, 1, 1, 2, 2]$$$. The resulting array of values of $$$\lfloor \frac{a_i}{p_i} \rfloor$$$ is $$$[4, 5, 6, 4, 5]$$$. The cost of $$$p$$$ is $$$\max\limits_{1 \le i \le n}(\lfloor \frac{a_i}{p_i} \rfloor) - \min\limits_{1 \le i \le n}(\lfloor \frac{a_i}{p_i} \rfloor) = 6 - 4 = 2$$$. We can show that there is no array (satisfying the condition from the statement) with a smaller cost.In the second test case, one of the optimal arrays is $$$p = [12, 12, 12, 12, 12]$$$, which results in all $$$\lfloor \frac{a_i}{p_i} \rfloor$$$ being $$$0$$$.In the third test case, the only possible array is $$$p = [1, 1, 1]$$$. Code: import sys tokens = ''.join(sys.stdin.readlines()).split()[::-1] def next(): return tokens.pop() def nextInt(): return int(next()) def nextFloat(): return float(next()) def getIntArray(n): return [nextInt() for _ in range(n)] def getFloatArray(n): return [nextFloat() for _ in range(n)] def getStringArray(n): return [next() for _ in range(n)] testcase = True def solve(testcase = 1): N, K = map(int, getIntArray(2)) A = getIntArray(N) s = [set() for i in range(3005)] for i in range(N): for k in range(1, K + 1): s[A[i] // k].add(i) ans = 999999999999999999999999999999999999999999999999999999999999999999999 r = 0 freq = {} for l in range(len(s)): while len(freq) < N and r < len(s): for # TODO: Your code here: if v not in freq: freq[v] = 0 freq[v] += 1 r += 1 if len(freq) < N: break ans = min(ans, r - l - 1) for v in s[l]: if freq[v] == 1: del freq[v] else: freq[v] -= 1 print(ans) if testcase is None: testcaseCount = 1 while tokens: solve(testcaseCount) testcaseCount += 1 else: testcaseCount = nextInt() if testcase else 1 for tc in range(testcaseCount): solve(tc + 1) assert not tokens
import sys tokens = ''.join(sys.stdin.readlines()).split()[::-1] def next(): return tokens.pop() def nextInt(): return int(next()) def nextFloat(): return float(next()) def getIntArray(n): return [nextInt() for _ in range(n)] def getFloatArray(n): return [nextFloat() for _ in range(n)] def getStringArray(n): return [next() for _ in range(n)] testcase = True def solve(testcase = 1): N, K = map(int, getIntArray(2)) A = getIntArray(N) s = [set() for i in range(3005)] for i in range(N): for k in range(1, K + 1): s[A[i] // k].add(i) ans = 999999999999999999999999999999999999999999999999999999999999999999999 r = 0 freq = {} for l in range(len(s)): while len(freq) < N and r < len(s): for {{completion}}: if v not in freq: freq[v] = 0 freq[v] += 1 r += 1 if len(freq) < N: break ans = min(ans, r - l - 1) for v in s[l]: if freq[v] == 1: del freq[v] else: freq[v] -= 1 print(ans) if testcase is None: testcaseCount = 1 while tokens: solve(testcaseCount) testcaseCount += 1 else: testcaseCount = nextInt() if testcase else 1 for tc in range(testcaseCount): solve(tc + 1) assert not tokens
v in s[r]
[{"input": "7\n\n5 2\n\n4 5 6 8 11\n\n5 12\n\n4 5 6 8 11\n\n3 1\n\n2 9 15\n\n7 3\n\n2 3 5 5 6 9 10\n\n6 56\n\n54 286 527 1436 2450 2681\n\n3 95\n\n16 340 2241\n\n2 2\n\n1 3", "output": ["2\n0\n13\n1\n4\n7\n0"]}]
control_completion_003595
control_fixed
python
Complete the code in python to solve this programming problem: Description: Let's call a string $$$s$$$ perfectly balanced if for all possible triplets $$$(t,u,v)$$$ such that $$$t$$$ is a non-empty substring of $$$s$$$ and $$$u$$$ and $$$v$$$ are characters present in $$$s$$$, the difference between the frequencies of $$$u$$$ and $$$v$$$ in $$$t$$$ is not more than $$$1$$$.For example, the strings "aba" and "abc" are perfectly balanced but "abb" is not because for the triplet ("bb",'a','b'), the condition is not satisfied.You are given a string $$$s$$$ consisting of lowercase English letters only. Your task is to determine whether $$$s$$$ is perfectly balanced or not.A string $$$b$$$ is called a substring of another string $$$a$$$ if $$$b$$$ can be obtained by deleting some characters (possibly $$$0$$$) from the start and some characters (possibly $$$0$$$) from the end of $$$a$$$. Input Specification: The first line of input contains a single integer $$$t$$$ ($$$1\leq t\leq 2\cdot 10^4$$$) denoting the number of testcases. Each of the next $$$t$$$ lines contain a single string $$$s$$$ ($$$1\leq |s|\leq 2\cdot 10^5$$$), consisting of lowercase English letters. It is guaranteed that the sum of $$$|s|$$$ over all testcases does not exceed $$$2\cdot 10^5$$$. Output Specification: For each test case, print "YES" if $$$s$$$ is a perfectly balanced string, and "NO" otherwise. You may print each letter in any case (for example, "YES", "Yes", "yes", "yEs" will all be recognized as positive answer). Notes: NoteLet $$$f_t(c)$$$ represent the frequency of character $$$c$$$ in string $$$t$$$.For the first testcase we have $$$t$$$$$$f_t(a)$$$$$$f_t(b)$$$$$$a$$$$$$1$$$$$$0$$$$$$ab$$$$$$1$$$$$$1$$$$$$aba$$$$$$2$$$$$$1$$$$$$b$$$$$$0$$$$$$1$$$$$$ba$$$$$$1$$$$$$1$$$ It can be seen that for any substring $$$t$$$ of $$$s$$$, the difference between $$$f_t(a)$$$ and $$$f_t(b)$$$ is not more than $$$1$$$. Hence the string $$$s$$$ is perfectly balanced.For the second testcase we have $$$t$$$$$$f_t(a)$$$$$$f_t(b)$$$$$$a$$$$$$1$$$$$$0$$$$$$ab$$$$$$1$$$$$$1$$$$$$abb$$$$$$1$$$$$$2$$$$$$b$$$$$$0$$$$$$1$$$$$$bb$$$$$$0$$$$$$2$$$ It can be seen that for the substring $$$t=bb$$$, the difference between $$$f_t(a)$$$ and $$$f_t(b)$$$ is $$$2$$$ which is greater than $$$1$$$. Hence the string $$$s$$$ is not perfectly balanced.For the third testcase we have $$$t$$$$$$f_t(a)$$$$$$f_t(b)$$$$$$f_t(c)$$$$$$a$$$$$$1$$$$$$0$$$$$$0$$$$$$ab$$$$$$1$$$$$$1$$$$$$0$$$$$$abc$$$$$$1$$$$$$1$$$$$$1$$$$$$b$$$$$$0$$$$$$1$$$$$$0$$$$$$bc$$$$$$0$$$$$$1$$$$$$1$$$$$$c$$$$$$0$$$$$$0$$$$$$1$$$It can be seen that for any substring $$$t$$$ of $$$s$$$ and any two characters $$$u,v\in\{a,b,c\}$$$, the difference between $$$f_t(u)$$$ and $$$f_t(v)$$$ is not more than $$$1$$$. Hence the string $$$s$$$ is perfectly balanced. Code: for _ in range(int(input())): # TODO: Your code here
for _ in range(int(input())): {{completion}}
s = input() c = len(set(s)) print("Yes" if all(len(set(s[i::c])) == 1 for i in range(c)) else "No")
[{"input": "5\naba\nabb\nabc\naaaaa\nabcba", "output": ["YES\nNO\nYES\nYES\nNO"]}]
block_completion_004801
block
python
Complete the code in python to solve this programming problem: Description: After watching a certain anime before going to sleep, Mark dreams of standing in an old classroom with a blackboard that has a sequence of $$$n$$$ positive integers $$$a_1, a_2,\dots,a_n$$$ on it.Then, professor Koro comes in. He can perform the following operation: select an integer $$$x$$$ that appears at least $$$2$$$ times on the board, erase those $$$2$$$ appearances, and write $$$x+1$$$ on the board. Professor Koro then asks Mark the question, "what is the maximum possible number that could appear on the board after some operations?"Mark quickly solves this question, but he is still slower than professor Koro. Thus, professor Koro decides to give Mark additional challenges. He will update the initial sequence of integers $$$q$$$ times. Each time, he will choose positive integers $$$k$$$ and $$$l$$$, then change $$$a_k$$$ to $$$l$$$. After each update, he will ask Mark the same question again.Help Mark answer these questions faster than Professor Koro!Note that the updates are persistent. Changes made to the sequence $$$a$$$ will apply when processing future updates. Input Specification: The first line of the input contains two integers $$$n$$$ and $$$q$$$ ($$$2\leq n\leq 2\cdot 10^5$$$, $$$1\leq q\leq 2\cdot 10^5$$$) — the length of the sequence $$$a$$$ and the number of updates, respectively. The second line contains $$$n$$$ integers $$$a_1,a_2,\dots,a_n$$$ ($$$1\leq a_i\leq 2\cdot 10^5$$$) Then, $$$q$$$ lines follow, each consisting of two integers $$$k$$$ and $$$l$$$ ($$$1\leq k\leq n$$$, $$$1\leq l\leq 2\cdot 10^5$$$), telling to update $$$a_k$$$ to $$$l$$$. Output Specification: Print $$$q$$$ lines. The $$$i$$$-th line should consist of a single integer — the answer after the $$$i$$$-th update. Notes: NoteIn the first example test, the program must proceed through $$$4$$$ updates.The sequence after the first update is $$$[2,3,2,4,5]$$$. One sequence of operations that achieves the number $$$6$$$ the following. Initially, the blackboard has numbers $$$[2,3,2,4,5]$$$. Erase two copies of $$$2$$$ and write $$$3$$$, yielding $$$[3,4,5,\color{red}{3}]$$$. Erase two copies of $$$3$$$ and write $$$4$$$, yielding $$$[4,5,\color{red}{4}]$$$. Erase two copies of $$$4$$$ and write $$$5$$$, yielding $$$[5,\color{red}{5}]$$$. Erase two copies of $$$5$$$ and write $$$6$$$, yielding $$$[\color{red}{6}]$$$. Then, in the second update, the array is changed to $$$[2,3,2,4,3]$$$. This time, Mark cannot achieve $$$6$$$. However, one sequence that Mark can use to achieve $$$5$$$ is shown below. Initially, the blackboard has $$$[2,3,2,4,3]$$$. Erase two copies of $$$2$$$ and write $$$3$$$, yielding $$$[3,4,3,\color{red}{3}]$$$. Erase two copies of $$$3$$$ and write $$$4$$$, yielding $$$[3,4,\color{red}{4}]$$$. Erase two copies of $$$4$$$ and write $$$5$$$, yielding $$$[3,\color{red}{5}]$$$. In the third update, the array is changed to $$$[2,3,2,1,3]$$$. One way to achieve $$$4$$$ is shown below. Initially, the blackboard has $$$[2,3,2,1,3]$$$. Erase two copies of $$$3$$$ and write $$$4$$$, yielding $$$[2,2,1,\color{red}{4}]$$$. Code: import sys import os from io import BytesIO, IOBase # region fastio BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") if sys.version_info[0] < 3: sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size)) else: sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) file = sys.stdin if os.environ.get('USER') == "loic": file = open("data.in") line = lambda: file.readline().split() ui = lambda: int(line()[0]) ti = lambda: map(int,line()) li = lambda: list(ti()) ####################################################################### class BitSet: ADDRESS_BITS_PER_WORD = 6 WORD_SZ = 1 << ADDRESS_BITS_PER_WORD MASK = -0x1 MASK_MAX = 0x7fffffffffffffff MASK_MIN = ~MASK_MAX def __init__(self, sz): self.sz = sz self.words = [0] * (self._wordIndex(sz - 1) + 1) self.last = -1 def _wordIndex(self, bitIndex): if bitIndex >= self.sz: raise ValueError("out of bound index", bitIndex) return bitIndex >> BitSet.ADDRESS_BITS_PER_WORD def _shift_one_left(self, shift): if shift == BitSet.WORD_SZ - 1: return BitSet.MASK_MIN return 1 << (shift % BitSet.WORD_SZ) def _shift_mask_right(self, shift): if shift == 0: return BitSet.MASK return BitSet.MASK_MAX >> (shift - 1) def _shift_mask_left(self, shift): if shift == 0: return BitSet.MASK return ~(BitSet.MASK_MAX >> (BitSet.WORD_SZ - shift - 1)) def flip(self, bitIndex): wordIndex = self._wordIndex(bitIndex) self.words[wordIndex] ^= self._shift_one_left(bitIndex % BitSet.WORD_SZ) def flip_range(self, l, r, pos): startWordIndex = self._wordIndex(l) endWordIndex = self._wordIndex(r) firstWordMask = self._shift_mask_left(l % BitSet.WORD_SZ) lastWordMask = self._shift_mask_right(BitSet.WORD_SZ - 1 - r % BitSet.WORD_SZ) if startWordIndex == endWordIndex: self.words[startWordIndex] ^= (firstWordMask & lastWordMask) else: self.words[startWordIndex] ^= firstWordMask for i in range(startWordIndex + 1, endWordIndex): self.words[i] ^= BitSet.MASK self.words[endWordIndex] ^= lastWordMask if pos: self.last = max(self.last, r) elif r == self.last: self.last = self.previousSetBit(r-1) def __setitem__(self, bitIndex, value): wordIndex = self._wordIndex(bitIndex) if value: self.words[wordIndex] |= self._shift_one_left(bitIndex % BitSet.WORD_SZ) else: self.words[wordIndex] &= ~self._shift_one_left(bitIndex % BitSet.WORD_SZ) def __getitem__(self, bitIndex): wordIndex = self._wordIndex(bitIndex) return (self.words[wordIndex] >> (bitIndex % BitSet.WORD_SZ)) & 1 ''' return len(bitset) if there is no "1" after fromIndex ''' def nextSetBit(self, fromIndex): wordIndex = self._wordIndex(fromIndex) word = self.words[wordIndex] & self._shift_mask_left(fromIndex % BitSet.WORD_SZ) while True: if word != 0: return wordIndex * BitSet.WORD_SZ + (word & -word).bit_length() - 1 wordIndex += 1 if wordIndex > len(self.words) - 1: return self.sz word = self.words[wordIndex] ''' return len(bitset) if there is no "0" after fromIndex ''' def nextClearBit(self, fromIndex): wordIndex = self._wordIndex(fromIndex) word = ~self.words[wordIndex] & self._shift_mask_left(fromIndex % BitSet.WORD_SZ) while True: if word != 0: return wordIndex * BitSet.WORD_SZ + (word & -word).bit_length() - 1 wordIndex += 1 if wordIndex > len(self.words) - 1: return self.sz word = ~self.words[wordIndex] ''' return -1 if there is no "1" before fromIndex ''' def previousSetBit(self, fromIndex): wordIndex = self._wordIndex(fromIndex) rem = fromIndex % BitSet.WORD_SZ word = self.words[wordIndex] & self._shift_mask_right(BitSet.WORD_SZ - 1 - rem) while True: if word != 0: return wordIndex * BitSet.WORD_SZ - 1 + (word.bit_length() if word > 0 else BitSet.WORD_SZ) wordIndex -= 1 if wordIndex < 0: return -1 word = self.words[wordIndex] ''' return -1 if there is no "0" before fromIndex ''' def previousClearBit(self,fromIndex): wordIndex = self._wordIndex(fromIndex) rem = fromIndex % BitSet.WORD_SZ word = ~self.words[wordIndex] & self._shift_mask_right(BitSet.WORD_SZ - 1 - rem) while True: if word != 0: return wordIndex * BitSet.WORD_SZ - 1 + (word.bit_length() if word > 0 else BitSet.WORD_SZ) wordIndex -= 1 if wordIndex < 0: return -1 word = ~self.words[wordIndex] def __str__(self): res = [] st = 0 while True: i = self.nextSetBit(st) if i != self.sz: res += [0] * (i - st) j = self.nextClearBit(i) if j != self.sz: res += [1] * (j-i) st = j else: # TODO: Your code here else: res += [0] * (self.sz - st) break return "".join(str(v) for v in res) def __repr__(self): return "Bitset(%s)" % str(self) def __iter__(self): for i in self[:]: yield i def __len__(self): return self.sz def add(bs,val): bs.flip_range(val, bs.nextClearBit(val), 1) def rem(bs,val): bs.flip_range(val, bs.nextSetBit(val), 0) def solve(): res = [] bs = BitSet(Z) for val in A: add(bs,val) for _ in range(Q): idx, val = ti() idx -= 1 rem(bs,A[idx]) A[idx] = val add(bs,val) res.append(bs.last) return "\n".join(str(v) for v in res) Z = 200030 for test in range(1,1+1): N,Q = ti() A = li() print(solve()) file.close()
import sys import os from io import BytesIO, IOBase # region fastio BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") if sys.version_info[0] < 3: sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size)) else: sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) file = sys.stdin if os.environ.get('USER') == "loic": file = open("data.in") line = lambda: file.readline().split() ui = lambda: int(line()[0]) ti = lambda: map(int,line()) li = lambda: list(ti()) ####################################################################### class BitSet: ADDRESS_BITS_PER_WORD = 6 WORD_SZ = 1 << ADDRESS_BITS_PER_WORD MASK = -0x1 MASK_MAX = 0x7fffffffffffffff MASK_MIN = ~MASK_MAX def __init__(self, sz): self.sz = sz self.words = [0] * (self._wordIndex(sz - 1) + 1) self.last = -1 def _wordIndex(self, bitIndex): if bitIndex >= self.sz: raise ValueError("out of bound index", bitIndex) return bitIndex >> BitSet.ADDRESS_BITS_PER_WORD def _shift_one_left(self, shift): if shift == BitSet.WORD_SZ - 1: return BitSet.MASK_MIN return 1 << (shift % BitSet.WORD_SZ) def _shift_mask_right(self, shift): if shift == 0: return BitSet.MASK return BitSet.MASK_MAX >> (shift - 1) def _shift_mask_left(self, shift): if shift == 0: return BitSet.MASK return ~(BitSet.MASK_MAX >> (BitSet.WORD_SZ - shift - 1)) def flip(self, bitIndex): wordIndex = self._wordIndex(bitIndex) self.words[wordIndex] ^= self._shift_one_left(bitIndex % BitSet.WORD_SZ) def flip_range(self, l, r, pos): startWordIndex = self._wordIndex(l) endWordIndex = self._wordIndex(r) firstWordMask = self._shift_mask_left(l % BitSet.WORD_SZ) lastWordMask = self._shift_mask_right(BitSet.WORD_SZ - 1 - r % BitSet.WORD_SZ) if startWordIndex == endWordIndex: self.words[startWordIndex] ^= (firstWordMask & lastWordMask) else: self.words[startWordIndex] ^= firstWordMask for i in range(startWordIndex + 1, endWordIndex): self.words[i] ^= BitSet.MASK self.words[endWordIndex] ^= lastWordMask if pos: self.last = max(self.last, r) elif r == self.last: self.last = self.previousSetBit(r-1) def __setitem__(self, bitIndex, value): wordIndex = self._wordIndex(bitIndex) if value: self.words[wordIndex] |= self._shift_one_left(bitIndex % BitSet.WORD_SZ) else: self.words[wordIndex] &= ~self._shift_one_left(bitIndex % BitSet.WORD_SZ) def __getitem__(self, bitIndex): wordIndex = self._wordIndex(bitIndex) return (self.words[wordIndex] >> (bitIndex % BitSet.WORD_SZ)) & 1 ''' return len(bitset) if there is no "1" after fromIndex ''' def nextSetBit(self, fromIndex): wordIndex = self._wordIndex(fromIndex) word = self.words[wordIndex] & self._shift_mask_left(fromIndex % BitSet.WORD_SZ) while True: if word != 0: return wordIndex * BitSet.WORD_SZ + (word & -word).bit_length() - 1 wordIndex += 1 if wordIndex > len(self.words) - 1: return self.sz word = self.words[wordIndex] ''' return len(bitset) if there is no "0" after fromIndex ''' def nextClearBit(self, fromIndex): wordIndex = self._wordIndex(fromIndex) word = ~self.words[wordIndex] & self._shift_mask_left(fromIndex % BitSet.WORD_SZ) while True: if word != 0: return wordIndex * BitSet.WORD_SZ + (word & -word).bit_length() - 1 wordIndex += 1 if wordIndex > len(self.words) - 1: return self.sz word = ~self.words[wordIndex] ''' return -1 if there is no "1" before fromIndex ''' def previousSetBit(self, fromIndex): wordIndex = self._wordIndex(fromIndex) rem = fromIndex % BitSet.WORD_SZ word = self.words[wordIndex] & self._shift_mask_right(BitSet.WORD_SZ - 1 - rem) while True: if word != 0: return wordIndex * BitSet.WORD_SZ - 1 + (word.bit_length() if word > 0 else BitSet.WORD_SZ) wordIndex -= 1 if wordIndex < 0: return -1 word = self.words[wordIndex] ''' return -1 if there is no "0" before fromIndex ''' def previousClearBit(self,fromIndex): wordIndex = self._wordIndex(fromIndex) rem = fromIndex % BitSet.WORD_SZ word = ~self.words[wordIndex] & self._shift_mask_right(BitSet.WORD_SZ - 1 - rem) while True: if word != 0: return wordIndex * BitSet.WORD_SZ - 1 + (word.bit_length() if word > 0 else BitSet.WORD_SZ) wordIndex -= 1 if wordIndex < 0: return -1 word = ~self.words[wordIndex] def __str__(self): res = [] st = 0 while True: i = self.nextSetBit(st) if i != self.sz: res += [0] * (i - st) j = self.nextClearBit(i) if j != self.sz: res += [1] * (j-i) st = j else: {{completion}} else: res += [0] * (self.sz - st) break return "".join(str(v) for v in res) def __repr__(self): return "Bitset(%s)" % str(self) def __iter__(self): for i in self[:]: yield i def __len__(self): return self.sz def add(bs,val): bs.flip_range(val, bs.nextClearBit(val), 1) def rem(bs,val): bs.flip_range(val, bs.nextSetBit(val), 0) def solve(): res = [] bs = BitSet(Z) for val in A: add(bs,val) for _ in range(Q): idx, val = ti() idx -= 1 rem(bs,A[idx]) A[idx] = val add(bs,val) res.append(bs.last) return "\n".join(str(v) for v in res) Z = 200030 for test in range(1,1+1): N,Q = ti() A = li() print(solve()) file.close()
res += [1] * (self.sz - i) break
[{"input": "5 4\n2 2 2 4 5\n2 3\n5 3\n4 1\n1 4", "output": ["6\n5\n4\n5"]}, {"input": "2 1\n200000 1\n2 200000", "output": ["200001"]}]
block_completion_005933
block
python
Complete the code in python to solve this programming problem: Description: The derby between Milan and Inter is happening soon, and you have been chosen as the assistant referee for the match, also known as linesman. Your task is to move along the touch-line, namely the side of the field, always looking very carefully at the match to check for offside positions and other offences.Football is an extremely serious matter in Italy, and thus it is fundamental that you keep very close track of the ball for as much time as possible. This means that you want to maximise the number of kicks which you monitor closely. You are able to monitor closely a kick if, when it happens, you are in the position along the touch-line with minimum distance from the place where the kick happens.Fortunately, expert analysts have been able to accurately predict all the kicks which will occur during the game. That is, you have been given two lists of integers, $$$t_1, \ldots, t_n$$$ and $$$a_1, \ldots, a_n$$$, indicating that $$$t_i$$$ seconds after the beginning of the match the ball will be kicked and you can monitor closely such kick if you are at the position $$$a_i$$$ along the touch-line. At the beginning of the game you start at position $$$0$$$ and the maximum speed at which you can walk along the touch-line is $$$v$$$ units per second (i.e., you can change your position by at most $$$v$$$ each second). What is the maximum number of kicks that you can monitor closely? Input Specification: The first line contains two integers $$$n$$$ and $$$v$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le v \le 10^6$$$) — the number of kicks that will take place and your maximum speed. The second line contains $$$n$$$ integers $$$t_1, \ldots, t_n$$$ ($$$1 \le t_i \le 10^9$$$) — the times of the kicks in the match. The sequence of times is guaranteed to be strictly increasing, i.e., $$$t_1 &lt; t_2 &lt; \cdots &lt; t_n$$$. The third line contains $$$n$$$ integers $$$a_1, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the positions along the touch-line where you have to be to monitor closely each kick. Output Specification: Print the maximum number of kicks that you can monitor closely. Notes: NoteIn the first sample, it is possible to move to the right at maximum speed for the first $$$3.5$$$ seconds and stay at position $$$7$$$ until the first kick happens, and then immediately move right also at maximum speed to watch the second kick at position $$$17$$$. There is no way to monitor closely the third kick after the second kick, so at most $$$2$$$ kicks can be seen. Code: from bisect import bisect_right,bisect_left n,v = map(int,input().split()) t = [*map(int,input().split())] a = [*map(int,input().split())] res = [] for i in range(n): xi,yi = t[i]*v+a[i],t[i]*v-a[i] if# TODO: Your code here: res.append((xi,yi)) res.sort() dp = [float("inf")]*(n+3) dp[0] = 0 dp[n+2] = 0 for i in range(len(res)): pos = bisect_right(dp,res[i][1],0,n+2) dp[pos] = res[i][1] for i in range(n,-1,-1): if(dp[i]!=float("inf")): print(i) break
from bisect import bisect_right,bisect_left n,v = map(int,input().split()) t = [*map(int,input().split())] a = [*map(int,input().split())] res = [] for i in range(n): xi,yi = t[i]*v+a[i],t[i]*v-a[i] if{{completion}}: res.append((xi,yi)) res.sort() dp = [float("inf")]*(n+3) dp[0] = 0 dp[n+2] = 0 for i in range(len(res)): pos = bisect_right(dp,res[i][1],0,n+2) dp[pos] = res[i][1] for i in range(n,-1,-1): if(dp[i]!=float("inf")): print(i) break
(xi>=0 and yi>=0)
[{"input": "3 2\n5 10 15\n7 17 29", "output": ["2"]}, {"input": "5 1\n5 7 8 11 13\n3 3 -2 -2 4", "output": ["3"]}, {"input": "1 2\n3\n7", "output": ["0"]}]
control_completion_001083
control_fixed
python
Complete the code in python to solve this programming problem: Description: Recently in Divanovo, a huge river locks system was built. There are now $$$n$$$ locks, the $$$i$$$-th of them has the volume of $$$v_i$$$ liters, so that it can contain any amount of water between $$$0$$$ and $$$v_i$$$ liters. Each lock has a pipe attached to it. When the pipe is open, $$$1$$$ liter of water enters the lock every second.The locks system is built in a way to immediately transfer all water exceeding the volume of the lock $$$i$$$ to the lock $$$i + 1$$$. If the lock $$$i + 1$$$ is also full, water will be transferred further. Water exceeding the volume of the last lock pours out to the river. The picture illustrates $$$5$$$ locks with two open pipes at locks $$$1$$$ and $$$3$$$. Because locks $$$1$$$, $$$3$$$, and $$$4$$$ are already filled, effectively the water goes to locks $$$2$$$ and $$$5$$$. Note that the volume of the $$$i$$$-th lock may be greater than the volume of the $$$i + 1$$$-th lock.To make all locks work, you need to completely fill each one of them. The mayor of Divanovo is interested in $$$q$$$ independent queries. For each query, suppose that initially all locks are empty and all pipes are closed. Then, some pipes are opened simultaneously. For the $$$j$$$-th query the mayor asks you to calculate the minimum number of pipes to open so that all locks are filled no later than after $$$t_j$$$ seconds.Please help the mayor to solve this tricky problem and answer his queries. Input Specification: The first lines contains one integer $$$n$$$ ($$$1 \le n \le 200\,000$$$) — the number of locks. The second lines contains $$$n$$$ integers $$$v_1, v_2, \dots, v_n$$$ ($$$1 \le v_i \le 10^9$$$)) — volumes of the locks. The third line contains one integer $$$q$$$ ($$$1 \le q \le 200\,000$$$) — the number of queries. Each of the next $$$q$$$ lines contains one integer $$$t_j$$$ ($$$1 \le t_j \le 10^9$$$) — the number of seconds you have to fill all the locks in the query $$$j$$$. Output Specification: Print $$$q$$$ integers. The $$$j$$$-th of them should be equal to the minimum number of pipes to turn on so that after $$$t_j$$$ seconds all of the locks are filled. If it is impossible to fill all of the locks in given time, print $$$-1$$$. Notes: NoteThere are $$$6$$$ queries in the first example test. In the queries $$$1, 3, 4$$$ the answer is $$$-1$$$. We need to wait $$$4$$$ seconds to fill the first lock even if we open all the pipes. In the sixth query we can open pipes in locks $$$1$$$, $$$3$$$, and $$$4$$$. After $$$4$$$ seconds the locks $$$1$$$ and $$$4$$$ are full. In the following $$$1$$$ second $$$1$$$ liter of water is transferred to the locks $$$2$$$ and $$$5$$$. The lock $$$3$$$ is filled by its own pipe. Similarly, in the second query one can open pipes in locks $$$1$$$, $$$3$$$, and $$$4$$$.In the fifth query one can open pipes $$$1, 2, 3, 4$$$. Code: import itertools m=0 n = int(input()) v = list(itertools.accumulate(map(int, input().split()))) for i in range(n): m=max((v[i]-1)//(i+1)+1,m) for # TODO: Your code here: t = int(input()) print((v[-1] - 1) // t + 1 if t >= m else -1)
import itertools m=0 n = int(input()) v = list(itertools.accumulate(map(int, input().split()))) for i in range(n): m=max((v[i]-1)//(i+1)+1,m) for {{completion}}: t = int(input()) print((v[-1] - 1) // t + 1 if t >= m else -1)
_ in range(int(input()))
[{"input": "5\n4 1 5 4 1\n6\n1\n6\n2\n3\n4\n5", "output": ["-1\n3\n-1\n-1\n4\n3"]}, {"input": "5\n4 4 4 4 4\n6\n1\n3\n6\n5\n2\n4", "output": ["-1\n-1\n4\n4\n-1\n5"]}]
control_completion_004184
control_fixed
python
Complete the code in python to solve this programming problem: Description: Monocarp plays "Rage of Empires II: Definitive Edition" — a strategic computer game. Right now he's planning to attack his opponent in the game, but Monocarp's forces cannot enter the opponent's territory since the opponent has built a wall.The wall consists of $$$n$$$ sections, aligned in a row. The $$$i$$$-th section initially has durability $$$a_i$$$. If durability of some section becomes $$$0$$$ or less, this section is considered broken.To attack the opponent, Monocarp needs to break at least two sections of the wall (any two sections: possibly adjacent, possibly not). To do this, he plans to use an onager — a special siege weapon. The onager can be used to shoot any section of the wall; the shot deals $$$2$$$ damage to the target section and $$$1$$$ damage to adjacent sections. In other words, if the onager shoots at the section $$$x$$$, then the durability of the section $$$x$$$ decreases by $$$2$$$, and the durability of the sections $$$x - 1$$$ and $$$x + 1$$$ (if they exist) decreases by $$$1$$$ each. Monocarp can shoot at any sections any number of times, he can even shoot at broken sections.Monocarp wants to calculate the minimum number of onager shots needed to break at least two sections. Help him! Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$) — the number of sections. The second line contains the sequence of integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$), where $$$a_i$$$ is the initial durability of the $$$i$$$-th section. Output Specification: Print one integer — the minimum number of onager shots needed to break at least two sections of the wall. Notes: NoteIn the first example, it is possible to break the $$$2$$$-nd and the $$$4$$$-th section in $$$10$$$ shots, for example, by shooting the third section $$$10$$$ times. After that, the durabilities become $$$[20, 0, 10, 0, 20]$$$. Another way of doing it is firing $$$5$$$ shots at the $$$2$$$-nd section, and another $$$5$$$ shots at the $$$4$$$-th section. After that, the durabilities become $$$[15, 0, 20, 0, 15]$$$.In the second example, it is enough to shoot the $$$2$$$-nd section once. Then the $$$1$$$-st and the $$$3$$$-rd section will be broken.In the third example, it is enough to shoot the $$$2$$$-nd section twice (then the durabilities become $$$[5, 2, 4, 8, 5, 8]$$$), and then shoot the $$$3$$$-rd section twice (then the durabilities become $$$[5, 0, 0, 6, 5, 8]$$$). So, four shots are enough to break the $$$2$$$-nd and the $$$3$$$-rd section. Code: N=int(input()) A=[int(x) for x in input().split()] B=sorted(A) ans=-(-B[0]//2)-(-B[1]//2) for # TODO: Your code here: ans=min(ans,-(-(A[i]+A[i+2])//2)) for i in range(N-1): score=max(-(-(A[i]+A[i+1])//3),-(-A[i]//2),-(-A[i+1]//2)) ans=min(score,ans) print(ans)
N=int(input()) A=[int(x) for x in input().split()] B=sorted(A) ans=-(-B[0]//2)-(-B[1]//2) for {{completion}}: ans=min(ans,-(-(A[i]+A[i+2])//2)) for i in range(N-1): score=max(-(-(A[i]+A[i+1])//3),-(-A[i]//2),-(-A[i+1]//2)) ans=min(score,ans) print(ans)
i in range(N-2)
[{"input": "5\n20 10 30 10 20", "output": ["10"]}, {"input": "3\n1 8 1", "output": ["1"]}, {"input": "6\n7 6 6 8 5 8", "output": ["4"]}, {"input": "6\n14 3 8 10 15 4", "output": ["4"]}, {"input": "4\n1 100 100 1", "output": ["2"]}, {"input": "3\n40 10 10", "output": ["7"]}]
control_completion_007774
control_fixed
python
Complete the code in python to solve this programming problem: Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once. Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$. Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$. Code: n, k = [int(i) for i in input().split()] a = [int(i) for i in input().split()] dd = [0]*(len(a)+5) add = 0 diff = 0 moves = 0 for key, i in reversed([*enumerate(a)]): add += diff i += add diff += dd[-1] dd.pop() if # TODO: Your code here: K = min(k, key+1) dd[-K] -= (i+K-1)//K diff += (i+K-1)//K moves += (i+K-1)//K add -= K*((i+K-1)//K) print(moves)
n, k = [int(i) for i in input().split()] a = [int(i) for i in input().split()] dd = [0]*(len(a)+5) add = 0 diff = 0 moves = 0 for key, i in reversed([*enumerate(a)]): add += diff i += add diff += dd[-1] dd.pop() if {{completion}}: K = min(k, key+1) dd[-K] -= (i+K-1)//K diff += (i+K-1)//K moves += (i+K-1)//K add -= K*((i+K-1)//K) print(moves)
i > 0
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
control_completion_003388
control_fixed
python
Complete the code in python to solve this programming problem: Description: A class of students got bored wearing the same pair of shoes every day, so they decided to shuffle their shoes among themselves. In this problem, a pair of shoes is inseparable and is considered as a single object.There are $$$n$$$ students in the class, and you are given an array $$$s$$$ in non-decreasing order, where $$$s_i$$$ is the shoe size of the $$$i$$$-th student. A shuffling of shoes is valid only if no student gets their own shoes and if every student gets shoes of size greater than or equal to their size. You have to output a permutation $$$p$$$ of $$$\{1,2,\ldots,n\}$$$ denoting a valid shuffling of shoes, where the $$$i$$$-th student gets the shoes of the $$$p_i$$$-th student ($$$p_i \ne i$$$). And output $$$-1$$$ if a valid shuffling does not exist.A permutation is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array) and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array). Input Specification: Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Description of the test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1\leq n\leq10^5$$$) — the number of students. The second line of each test case contains $$$n$$$ integers $$$s_1, s_2,\ldots,s_n$$$ ($$$1\leq s_i\leq10^9$$$, and for all $$$1\le i&lt;n$$$, $$$s_i\le s_{i+1}$$$) — the shoe sizes of the students. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$. Output Specification: For each test case, print the answer in a single line using the following format. If a valid shuffling does not exist, print the number $$$-1$$$ as the answer. If a valid shuffling exists, print $$$n$$$ space-separated integers — a permutation $$$p$$$ of $$$1,2,\ldots,n$$$ denoting a valid shuffling of shoes where the $$$i$$$-th student gets the shoes of the $$$p_i$$$-th student. If there are multiple answers, then print any of them. Notes: NoteIn the first test case, any permutation $$$p$$$ of $$$1,\ldots,n$$$ where $$$p_i\ne i$$$ would represent a valid shuffling since all students have equal shoe sizes, and thus anyone can wear anyone's shoes.In the second test case, it can be shown that no valid shuffling is possible. Code: from collections import Counter for _ in range(int(input())): n=int(input()); l=list(map(int, input().split()));c = Counter(l); if any(x == 1 for x in c.values()): # TODO: Your code here print(*[i if i!=0 and l[i]==l[i-1] else i+c[l[i]] for i in range(n)])
from collections import Counter for _ in range(int(input())): n=int(input()); l=list(map(int, input().split()));c = Counter(l); if any(x == 1 for x in c.values()): {{completion}} print(*[i if i!=0 and l[i]==l[i-1] else i+c[l[i]] for i in range(n)])
print(-1); continue
[{"input": "2\n5\n1 1 1 1 1\n6\n3 6 8 13 15 21", "output": ["5 1 2 3 4 \n-1"]}]
block_completion_002395
block
python
Complete the code in python to solve this programming problem: Description: You like the card board game "Set". Each card contains $$$k$$$ features, each of which is equal to a value from the set $$$\{0, 1, 2\}$$$. The deck contains all possible variants of cards, that is, there are $$$3^k$$$ different cards in total.A feature for three cards is called good if it is the same for these cards or pairwise distinct. Three cards are called a set if all $$$k$$$ features are good for them.For example, the cards $$$(0, 0, 0)$$$, $$$(0, 2, 1)$$$, and $$$(0, 1, 2)$$$ form a set, but the cards $$$(0, 2, 2)$$$, $$$(2, 1, 2)$$$, and $$$(1, 2, 0)$$$ do not, as, for example, the last feature is not good.A group of five cards is called a meta-set, if there is strictly more than one set among them. How many meta-sets there are among given $$$n$$$ distinct cards? Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 10^3$$$, $$$1 \le k \le 20$$$) — the number of cards on a table and the number of card features. The description of the cards follows in the next $$$n$$$ lines. Each line describing a card contains $$$k$$$ integers $$$c_{i, 1}, c_{i, 2}, \ldots, c_{i, k}$$$ ($$$0 \le c_{i, j} \le 2$$$) — card features. It is guaranteed that all cards are distinct. Output Specification: Output one integer — the number of meta-sets. Notes: NoteLet's draw the cards indicating the first four features. The first feature will indicate the number of objects on a card: $$$1$$$, $$$2$$$, $$$3$$$. The second one is the color: red, green, purple. The third is the shape: oval, diamond, squiggle. The fourth is filling: open, striped, solid.You can see the first three tests below. For the first two tests, the meta-sets are highlighted.In the first test, the only meta-set is the five cards $$$(0000,\ 0001,\ 0002,\ 0010,\ 0020)$$$. The sets in it are the triples $$$(0000,\ 0001,\ 0002)$$$ and $$$(0000,\ 0010,\ 0020)$$$. Also, a set is the triple $$$(0100,\ 1000,\ 2200)$$$ which does not belong to any meta-set. In the second test, the following groups of five cards are meta-sets: $$$(0000,\ 0001,\ 0002,\ 0010,\ 0020)$$$, $$$(0000,\ 0001,\ 0002,\ 0100,\ 0200)$$$, $$$(0000,\ 0010,\ 0020,\ 0100,\ 0200)$$$. In there third test, there are $$$54$$$ meta-sets. Code: import sys; R = sys.stdin.readline n,k = map(int,R().split()) deck = [tuple(map(int,R().split())) for _ in range(n)] dic = {} for i in range(n): dic[deck[i]] = i res = [0]*n for p in range(n-2): for q in range(p+1,n-1): last = [0]*k for j in range(k): last[j] = deck[p][j] if deck[p][j]==deck[q][j] else 3-deck[p][j]-deck[q][j] last = tuple(last) if last in dic and dic[last]>q: # TODO: Your code here print(sum((s*(s-1))//2 for s in res))
import sys; R = sys.stdin.readline n,k = map(int,R().split()) deck = [tuple(map(int,R().split())) for _ in range(n)] dic = {} for i in range(n): dic[deck[i]] = i res = [0]*n for p in range(n-2): for q in range(p+1,n-1): last = [0]*k for j in range(k): last[j] = deck[p][j] if deck[p][j]==deck[q][j] else 3-deck[p][j]-deck[q][j] last = tuple(last) if last in dic and dic[last]>q: {{completion}} print(sum((s*(s-1))//2 for s in res))
res[p] += 1; res[q] += 1; res[dic[last]] += 1
[{"input": "8 4\n0 0 0 0\n0 0 0 1\n0 0 0 2\n0 0 1 0\n0 0 2 0\n0 1 0 0\n1 0 0 0\n2 2 0 0", "output": ["1"]}, {"input": "7 4\n0 0 0 0\n0 0 0 1\n0 0 0 2\n0 0 1 0\n0 0 2 0\n0 1 0 0\n0 2 0 0", "output": ["3"]}, {"input": "9 2\n0 0\n0 1\n0 2\n1 0\n1 1\n1 2\n2 0\n2 1\n2 2", "output": ["54"]}, {"input": "20 4\n0 2 0 0\n0 2 2 2\n0 2 2 1\n0 2 0 1\n1 2 2 0\n1 2 1 0\n1 2 2 1\n1 2 0 1\n1 1 2 2\n1 1 0 2\n1 1 2 1\n1 1 1 1\n2 1 2 0\n2 1 1 2\n2 1 2 1\n2 1 1 1\n0 1 1 2\n0 0 1 0\n2 2 0 0\n2 0 0 2", "output": ["0"]}]
block_completion_005319
block
python
Complete the code in python to solve this programming problem: Description: You are given a rooted tree of $$$2^n - 1$$$ vertices. Every vertex of this tree has either $$$0$$$ children, or $$$2$$$ children. All leaves of this tree have the same distance from the root, and for every non-leaf vertex, one of its children is the left one, and the other child is the right one. Formally, you are given a perfect binary tree.The vertices of the tree are numbered in the following order: the root has index $$$1$$$; if a vertex has index $$$x$$$, then its left child has index $$$2x$$$, and its right child has index $$$2x+1$$$. Every vertex of the tree has a letter written on it, either A or B. Let's define the character on the vertex $$$x$$$ as $$$s_x$$$.Let the preorder string of some vertex $$$x$$$ be defined in the following way: if the vertex $$$x$$$ is a leaf, then the preorder string of $$$x$$$ be consisting of only one character $$$s_x$$$; otherwise, the preorder string of $$$x$$$ is $$$s_x + f(l_x) + f(r_x)$$$, where $$$+$$$ operator defines concatenation of strings, $$$f(l_x)$$$ is the preorder string of the left child of $$$x$$$, and $$$f(r_x)$$$ is the preorder string of the right child of $$$x$$$. The preorder string of the tree is the preorder string of its root.Now, for the problem itself...You have to calculate the number of different strings that can be obtained as the preorder string of the given tree, if you are allowed to perform the following operation any number of times before constructing the preorder string of the tree: choose any non-leaf vertex $$$x$$$, and swap its children (so, the left child becomes the right one, and vice versa). Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 18$$$). The second line contains a sequence of $$$2^n-1$$$ characters $$$s_1, s_2, \dots, s_{2^n-1}$$$. Each character is either A or B. The characters are not separated by spaces or anything else. Output Specification: Print one integer — the number of different strings that can be obtained as the preorder string of the given tree, if you can apply any number of operations described in the statement. Since it can be very large, print it modulo $$$998244353$$$. Code: MOD = 998244353 n, s = int(input()), input() def calc(u: int) -> tuple: if u >= (1 << n): # TODO: Your code here t1, t2 = calc(u * 2), calc(u * 2 + 1) return (t1[0] + t2[0] + (t1[1] != t2[1]), hash((min(t1[1], t2[1]), max(t1[1], t2[1]), s[u - 1]))) print(pow(2, calc(1)[0], MOD))
MOD = 998244353 n, s = int(input()), input() def calc(u: int) -> tuple: if u >= (1 << n): {{completion}} t1, t2 = calc(u * 2), calc(u * 2 + 1) return (t1[0] + t2[0] + (t1[1] != t2[1]), hash((min(t1[1], t2[1]), max(t1[1], t2[1]), s[u - 1]))) print(pow(2, calc(1)[0], MOD))
return (0, 0)
[{"input": "4\nBAAAAAAAABBABAB", "output": ["16"]}, {"input": "2\nBAA", "output": ["1"]}, {"input": "2\nABA", "output": ["2"]}, {"input": "2\nAAB", "output": ["2"]}, {"input": "2\nAAA", "output": ["1"]}]
block_completion_001705
block
python
Complete the code in python to solve this programming problem: Description: You are given two arrays: an array $$$a$$$ consisting of $$$n$$$ zeros and an array $$$b$$$ consisting of $$$n$$$ integers.You can apply the following operation to the array $$$a$$$ an arbitrary number of times: choose some subsegment of $$$a$$$ of length $$$k$$$ and add the arithmetic progression $$$1, 2, \ldots, k$$$ to this subsegment — i. e. add $$$1$$$ to the first element of the subsegment, $$$2$$$ to the second element, and so on. The chosen subsegment should be inside the borders of the array $$$a$$$ (i.e., if the left border of the chosen subsegment is $$$l$$$, then the condition $$$1 \le l \le l + k - 1 \le n$$$ should be satisfied). Note that the progression added is always $$$1, 2, \ldots, k$$$ but not the $$$k, k - 1, \ldots, 1$$$ or anything else (i.e., the leftmost element of the subsegment always increases by $$$1$$$, the second element always increases by $$$2$$$ and so on).Your task is to find the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Note that the condition $$$a_i \ge b_i$$$ should be satisfied for all elements at once. Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 3 \cdot 10^5$$$) — the number of elements in both arrays and the length of the subsegment, respectively. The second line of the input contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10^{12}$$$), where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$. Output Specification: Print one integer — the minimum possible number of operations required to satisfy the condition $$$a_i \ge b_i$$$ for each $$$i$$$ from $$$1$$$ to $$$n$$$. Notes: NoteConsider the first example. In this test, we don't really have any choice, so we need to add at least five progressions to make the first element equals $$$5$$$. The array $$$a$$$ becomes $$$[5, 10, 15]$$$.Consider the second example. In this test, let's add one progression on the segment $$$[1; 3]$$$ and two progressions on the segment $$$[4; 6]$$$. Then, the array $$$a$$$ becomes $$$[1, 2, 3, 2, 4, 6]$$$. Code: """ take element as "the tail" will use more less operations, use variables s,cnt and closed, to avoid the inner iteration(update neighbor k elements every time), complexity is O(n). """ row=lambda:map(int,input().split()) n,k=row() a=list(row()) closed=[0]*n s=cnt=res=0 for i in range(n-1,-1,-1): # print(i,a[i],'s,cnt:',s,cnt,'closed:',closed) s-=cnt cnt-=closed[i] a[i]-=s if a[i]<=0: continue th=min(i+1,k) need=(a[i]+th-1)//th#equals ceil() s+=need*th cnt+=need res+=need if # TODO: Your code here: closed[i-th]+=need print(res)
""" take element as "the tail" will use more less operations, use variables s,cnt and closed, to avoid the inner iteration(update neighbor k elements every time), complexity is O(n). """ row=lambda:map(int,input().split()) n,k=row() a=list(row()) closed=[0]*n s=cnt=res=0 for i in range(n-1,-1,-1): # print(i,a[i],'s,cnt:',s,cnt,'closed:',closed) s-=cnt cnt-=closed[i] a[i]-=s if a[i]<=0: continue th=min(i+1,k) need=(a[i]+th-1)//th#equals ceil() s+=need*th cnt+=need res+=need if {{completion}}: closed[i-th]+=need print(res)
i>=th
[{"input": "3 3\n5 4 6", "output": ["5"]}, {"input": "6 3\n1 2 3 2 2 3", "output": ["3"]}, {"input": "6 3\n1 2 4 1 2 3", "output": ["3"]}, {"input": "7 3\n50 17 81 25 42 39 96", "output": ["92"]}]
control_completion_003390
control_fixed
python
Complete the code in python to solve this programming problem: Description: You are walking with your dog, and now you are at the promenade. The promenade can be represented as an infinite line. Initially, you are in the point $$$0$$$ with your dog. You decided to give some freedom to your dog, so you untied her and let her run for a while. Also, you watched what your dog is doing, so you have some writings about how she ran. During the $$$i$$$-th minute, the dog position changed from her previous position by the value $$$a_i$$$ (it means, that the dog ran for $$$a_i$$$ meters during the $$$i$$$-th minute). If $$$a_i$$$ is positive, the dog ran $$$a_i$$$ meters to the right, otherwise (if $$$a_i$$$ is negative) she ran $$$a_i$$$ meters to the left.During some minutes, you were chatting with your friend, so you don't have writings about your dog movement during these minutes. These values $$$a_i$$$ equal zero.You want your dog to return to you after the end of the walk, so the destination point of the dog after $$$n$$$ minutes should be $$$0$$$.Now you are wondering: what is the maximum possible number of different integer points of the line your dog could visit on her way, if you replace every $$$0$$$ with some integer from $$$-k$$$ to $$$k$$$ (and your dog should return to $$$0$$$ after the walk)? The dog visits an integer point if she runs through that point or reaches in it at the end of any minute. Point $$$0$$$ is always visited by the dog, since she is initially there.If the dog cannot return to the point $$$0$$$ after $$$n$$$ minutes regardless of the integers you place, print -1. Input Specification: The first line of the input contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3000; 1 \le k \le 10^9$$$) — the number of minutes and the maximum possible speed of your dog during the minutes without records. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$), where $$$a_i$$$ is the number of meters your dog ran during the $$$i$$$-th minutes (to the left if $$$a_i$$$ is negative, to the right otherwise). If $$$a_i = 0$$$ then this value is unknown and can be replaced with any integer from the range $$$[-k; k]$$$. Output Specification: If the dog cannot return to the point $$$0$$$ after $$$n$$$ minutes regardless of the set of integers you place, print -1. Otherwise, print one integer — the maximum number of different integer points your dog could visit if you fill all the unknown values optimally and the dog will return to the point $$$0$$$ at the end of the walk. Code: n, k = map(int, input().split()) A = list(map(int, input().split())) ans = 0 for i in range(n): C = [0]*n for j in range(n-1, -1, -1): if A[j] == 0: C[j] = 1 if j+1 < n: C[j] += C[j+1] B = A.copy() s = sum(B) flag = True for j in range(n): if B[j] == 0: if # TODO: Your code here: x = C[j+1] else: x = 0 B[j] = min(k, x*k-s) if B[j] < -k: flag = False s += B[j] if flag: pos = 0 mn = 0 mx = 0 for j in range(n): pos += B[j] mn = min(mn, pos) mx = max(mx, pos) if pos == 0: ans = max(ans, mx-mn+1) A = A[1:]+A[0:1] if ans != 0: print(ans) else: print(-1)
n, k = map(int, input().split()) A = list(map(int, input().split())) ans = 0 for i in range(n): C = [0]*n for j in range(n-1, -1, -1): if A[j] == 0: C[j] = 1 if j+1 < n: C[j] += C[j+1] B = A.copy() s = sum(B) flag = True for j in range(n): if B[j] == 0: if {{completion}}: x = C[j+1] else: x = 0 B[j] = min(k, x*k-s) if B[j] < -k: flag = False s += B[j] if flag: pos = 0 mn = 0 mx = 0 for j in range(n): pos += B[j] mn = min(mn, pos) mx = max(mx, pos) if pos == 0: ans = max(ans, mx-mn+1) A = A[1:]+A[0:1] if ans != 0: print(ans) else: print(-1)
j+1 < n
[{"input": "3 2\n5 0 -4", "output": ["6"]}, {"input": "6 4\n1 -2 0 3 -4 5", "output": ["7"]}, {"input": "3 1000000000\n0 0 0", "output": ["1000000001"]}, {"input": "5 9\n-7 -3 8 12 0", "output": ["-1"]}, {"input": "5 3\n-1 0 3 3 0", "output": ["7"]}, {"input": "5 4\n0 2 0 3 0", "output": ["9"]}]
control_completion_000199
control_fixed
python
Complete the code in python to solve this programming problem: Description: Your friend Ivan asked you to help him rearrange his desktop. The desktop can be represented as a rectangle matrix of size $$$n \times m$$$ consisting of characters '.' (empty cell of the desktop) and '*' (an icon).The desktop is called good if all its icons are occupying some prefix of full columns and, possibly, the prefix of the next column (and there are no icons outside this figure). In other words, some amount of first columns will be filled with icons and, possibly, some amount of first cells of the next (after the last full column) column will be also filled with icons (and all the icons on the desktop belong to this figure). This is pretty much the same as the real life icons arrangement.In one move, you can take one icon and move it to any empty cell in the desktop.Ivan loves to add some icons to his desktop and remove them from it, so he is asking you to answer $$$q$$$ queries: what is the minimum number of moves required to make the desktop good after adding/removing one icon?Note that queries are permanent and change the state of the desktop. Input Specification: The first line of the input contains three integers $$$n$$$, $$$m$$$ and $$$q$$$ ($$$1 \le n, m \le 1000; 1 \le q \le 2 \cdot 10^5$$$) — the number of rows in the desktop, the number of columns in the desktop and the number of queries, respectively. The next $$$n$$$ lines contain the description of the desktop. The $$$i$$$-th of them contains $$$m$$$ characters '.' and '*' — the description of the $$$i$$$-th row of the desktop. The next $$$q$$$ lines describe queries. The $$$i$$$-th of them contains two integers $$$x_i$$$ and $$$y_i$$$ ($$$1 \le x_i \le n; 1 \le y_i \le m$$$) — the position of the cell which changes its state (if this cell contained the icon before, then this icon is removed, otherwise an icon appears in this cell). Output Specification: Print $$$q$$$ integers. The $$$i$$$-th of them should be the minimum number of moves required to make the desktop good after applying the first $$$i$$$ queries. Code: n,m,q = map(int, input().split()) s = [input() for _ in range(n)] s = [s[j][i] for i in range(m) for j in range(n)] qrr = [list(map(int, input().split())) for _ in range(q)] qrr = [n * (q[1] - 1) + q[0] - 1 for q in qrr] count = s.count('*') correct = s[:count].count('*') for q in qrr: count += 1 if s[q] == '.' else -1 if s[q] == '.': correct += 1 if q < count else 0 correct += 1 if s[count-1] == '*' else 0 else: # TODO: Your code here print(count - correct) s[q] = '.' if s[q] == '*' else '*'
n,m,q = map(int, input().split()) s = [input() for _ in range(n)] s = [s[j][i] for i in range(m) for j in range(n)] qrr = [list(map(int, input().split())) for _ in range(q)] qrr = [n * (q[1] - 1) + q[0] - 1 for q in qrr] count = s.count('*') correct = s[:count].count('*') for q in qrr: count += 1 if s[q] == '.' else -1 if s[q] == '.': correct += 1 if q < count else 0 correct += 1 if s[count-1] == '*' else 0 else: {{completion}} print(count - correct) s[q] = '.' if s[q] == '*' else '*'
correct -= 1 if q < count else 0 correct -= 1 if s[count] == '*' else 0
[{"input": "4 4 8\n..**\n.*..\n*...\n...*\n1 3\n2 3\n3 1\n2 3\n3 4\n4 3\n2 3\n2 2", "output": ["3\n4\n4\n3\n4\n5\n5\n5"]}, {"input": "2 5 5\n*...*\n*****\n1 3\n2 2\n1 3\n1 5\n2 3", "output": ["2\n3\n3\n3\n2"]}]
block_completion_007866
block
python
Complete the code in python to solve this programming problem: Description: You are given two non-empty strings $$$s$$$ and $$$t$$$, consisting of Latin letters.In one move, you can choose an occurrence of the string $$$t$$$ in the string $$$s$$$ and replace it with dots.Your task is to remove all occurrences of the string $$$t$$$ in the string $$$s$$$ in the minimum number of moves, and also calculate how many different sequences of moves of the minimum length exist.Two sequences of moves are considered different if the sets of indices at which the removed occurrences of the string $$$t$$$ in $$$s$$$ begin differ. For example, the sets $$$\{1, 2, 3\}$$$ and $$$\{1, 2, 4\}$$$ are considered different, the sets $$$\{2, 4, 6\}$$$ and $$$\{2, 6\}$$$ — too, but sets $$$\{3, 5\}$$$ and $$$\{5, 3\}$$$ — not.For example, let the string $$$s =$$$ "abababacababa" and the string $$$t =$$$ "aba". We can remove all occurrences of the string $$$t$$$ in $$$2$$$ moves by cutting out the occurrences of the string $$$t$$$ at the $$$3$$$th and $$$9$$$th positions. In this case, the string $$$s$$$ is an example of the form "ab...bac...ba". It is also possible to cut occurrences of the string $$$t$$$ at the $$$3$$$th and $$$11$$$th positions. There are two different sequences of minimum length moves.Since the answer can be large, output it modulo $$$10^9 + 7$$$. Input Specification: The first line of the input contains a single integer $$$q$$$ ($$$1 \le q \le 50$$$) — the number of test cases. The descriptions of the sets follow. The first line of each set contains a non-empty string $$$s$$$ ($$$1 \le |s| \le 500$$$) consisting of lowercase Latin letters. The second line of each set contains a non-empty string $$$t$$$ ($$$1 \le |t| \le 500$$$) consisting of lowercase Latin letters. It is guaranteed that the sum of string lengths $$$s$$$ over all test cases does not exceed $$$500$$$. Similarly, it is guaranteed that the sum of string lengths $$$t$$$ over all test cases does not exceed $$$500$$$. Output Specification: For each test case print two integers — the minimum number of moves and the number of different optimal sequences, modulo $$$10^9 + 7$$$. Notes: NoteThe first test case is explained in the statement.In the second case, it is enough to cut any of the four occurrences.In the third case, string $$$s$$$ is the concatenation of two strings $$$t =$$$ "xyz", so there is a unique optimal sequence of $$$2$$$ moves.In the fourth and sixth cases, the string $$$s$$$ initially contains no occurrences of the string $$$t$$$.In the fifth case, the string $$$s$$$ contains exactly one occurrence of the string $$$t$$$. Code: # from __future__ import annotations import csv import datetime import string import sys import time from collections import defaultdict from contextlib import contextmanager from typing import List, Optional def solve() -> None: s = next_token() t = next_token() ls = len(s) lt = len(t) is_start = [s[i:i + lt] == t for i in range(ls)] d: List[List[Optional[List[int]]]] = [[None for _ in range(j + 1)] for j in range(ls)] for ln in range(1, ls + 1): for j in range(ln - 1, ls): i = j - ln + 1 for k in range(i, j + 1): if k + lt - 1 <= j and is_start[k]: l = (d[k - 1][i] if k - 1 >= i else None) or [0, 1] if l[0] == 0: r = (d[j][k + lt] if j >= k + lt else None) or [0, 1] tt = d[j][i] if tt is None or tt[0] > l[0] + r[0] + 1: # TODO: Your code here elif tt[0] == l[0] + r[0] + 1: tt[1] = tt[1] + r[1] d[j][i] = tt if d[j][i]: d[j][i][1] %= 1000000007 print(*(d[ls - 1][0] or [0, 1])) def global_init() -> None: pass RUN_N_TESTS_IN_PROD = True PRINT_CASE_NUMBER = False ASSERT_IN_PROD = False LOG_TO_FILE = False READ_FROM_CONSOLE_IN_DEBUG = False WRITE_TO_CONSOLE_IN_DEBUG = True TEST_TIMER = False IS_DEBUG = "DEBUG_MODE" in sys.argv __output_file = None __input_file = None __input_last_line = None def run() -> None: global __input_file, __input_last_line, __output_file __output_file = sys.stdout if not IS_DEBUG or WRITE_TO_CONSOLE_IN_DEBUG else open("../output.txt", "w") try: __input_file = sys.stdin if not IS_DEBUG or READ_FROM_CONSOLE_IN_DEBUG else open("../input.txt") try: with timer("total"): global_init() t = next_int() if RUN_N_TESTS_IN_PROD or IS_DEBUG else 1 for i in range(t): if PRINT_CASE_NUMBER: fprint(f"Case #{i + 1}: ") if TEST_TIMER: with timer(f"test #{i + 1}"): solve() else: solve() if IS_DEBUG: __output_file.flush() finally: __input_last_line = None __input_file.close() __input_file = None finally: __output_file.flush() __output_file.close() def fprint(*objects, **kwargs): print(*objects, end="", file=__output_file, **kwargs) def fprintln(*objects, **kwargs): print(*objects, file=__output_file, **kwargs) def next_line() -> str: global __input_last_line __input_last_line = None return __input_file.readline() def next_token() -> str: global __input_last_line while not __input_last_line: __input_last_line = __input_file.readline().split()[::-1] return __input_last_line.pop() def next_int(): return int(next_token()) def next_float(): return float(next_token()) def next_int_array(n: int) -> List[int]: return [int(next_token()) for _ in range(n)] if IS_DEBUG or ASSERT_IN_PROD: def assert_predicate(p: bool, message: str = ""): if not p: raise AssertionError(message) def assert_not_equal(unexpected, actual): if unexpected == actual: raise AssertionError(f"assert_not_equal: {unexpected} == {actual}") def assert_equal(expected, actual): if expected != actual: raise AssertionError(f"assert_equal: {expected} != {actual}") else: def assert_predicate(p: bool, message: str = ""): pass def assert_not_equal(unexpected, actual): pass def assert_equal(expected, actual): pass if IS_DEBUG: __log_file = open(f"../logs/py_solution_{int(time.time() * 1000)}.log", "w") if LOG_TO_FILE else sys.stdout def log(*args, **kwargs): print(datetime.datetime.now(), "-", *args, **kwargs, flush=True, file=__log_file) @contextmanager def timer(label: str): start_time = time.time() try: yield finally: log(f"Timer[{label}]: {time.time() - start_time:.6f}s") else: def log(*args, **kwargs): pass @contextmanager def timer(label: str): yield if __name__ == "__main__": run()
# from __future__ import annotations import csv import datetime import string import sys import time from collections import defaultdict from contextlib import contextmanager from typing import List, Optional def solve() -> None: s = next_token() t = next_token() ls = len(s) lt = len(t) is_start = [s[i:i + lt] == t for i in range(ls)] d: List[List[Optional[List[int]]]] = [[None for _ in range(j + 1)] for j in range(ls)] for ln in range(1, ls + 1): for j in range(ln - 1, ls): i = j - ln + 1 for k in range(i, j + 1): if k + lt - 1 <= j and is_start[k]: l = (d[k - 1][i] if k - 1 >= i else None) or [0, 1] if l[0] == 0: r = (d[j][k + lt] if j >= k + lt else None) or [0, 1] tt = d[j][i] if tt is None or tt[0] > l[0] + r[0] + 1: {{completion}} elif tt[0] == l[0] + r[0] + 1: tt[1] = tt[1] + r[1] d[j][i] = tt if d[j][i]: d[j][i][1] %= 1000000007 print(*(d[ls - 1][0] or [0, 1])) def global_init() -> None: pass RUN_N_TESTS_IN_PROD = True PRINT_CASE_NUMBER = False ASSERT_IN_PROD = False LOG_TO_FILE = False READ_FROM_CONSOLE_IN_DEBUG = False WRITE_TO_CONSOLE_IN_DEBUG = True TEST_TIMER = False IS_DEBUG = "DEBUG_MODE" in sys.argv __output_file = None __input_file = None __input_last_line = None def run() -> None: global __input_file, __input_last_line, __output_file __output_file = sys.stdout if not IS_DEBUG or WRITE_TO_CONSOLE_IN_DEBUG else open("../output.txt", "w") try: __input_file = sys.stdin if not IS_DEBUG or READ_FROM_CONSOLE_IN_DEBUG else open("../input.txt") try: with timer("total"): global_init() t = next_int() if RUN_N_TESTS_IN_PROD or IS_DEBUG else 1 for i in range(t): if PRINT_CASE_NUMBER: fprint(f"Case #{i + 1}: ") if TEST_TIMER: with timer(f"test #{i + 1}"): solve() else: solve() if IS_DEBUG: __output_file.flush() finally: __input_last_line = None __input_file.close() __input_file = None finally: __output_file.flush() __output_file.close() def fprint(*objects, **kwargs): print(*objects, end="", file=__output_file, **kwargs) def fprintln(*objects, **kwargs): print(*objects, file=__output_file, **kwargs) def next_line() -> str: global __input_last_line __input_last_line = None return __input_file.readline() def next_token() -> str: global __input_last_line while not __input_last_line: __input_last_line = __input_file.readline().split()[::-1] return __input_last_line.pop() def next_int(): return int(next_token()) def next_float(): return float(next_token()) def next_int_array(n: int) -> List[int]: return [int(next_token()) for _ in range(n)] if IS_DEBUG or ASSERT_IN_PROD: def assert_predicate(p: bool, message: str = ""): if not p: raise AssertionError(message) def assert_not_equal(unexpected, actual): if unexpected == actual: raise AssertionError(f"assert_not_equal: {unexpected} == {actual}") def assert_equal(expected, actual): if expected != actual: raise AssertionError(f"assert_equal: {expected} != {actual}") else: def assert_predicate(p: bool, message: str = ""): pass def assert_not_equal(unexpected, actual): pass def assert_equal(expected, actual): pass if IS_DEBUG: __log_file = open(f"../logs/py_solution_{int(time.time() * 1000)}.log", "w") if LOG_TO_FILE else sys.stdout def log(*args, **kwargs): print(datetime.datetime.now(), "-", *args, **kwargs, flush=True, file=__log_file) @contextmanager def timer(label: str): start_time = time.time() try: yield finally: log(f"Timer[{label}]: {time.time() - start_time:.6f}s") else: def log(*args, **kwargs): pass @contextmanager def timer(label: str): yield if __name__ == "__main__": run()
tt = [l[0] + r[0] + 1, r[1]]
[{"input": "8\n\nabababacababa\n\naba\n\nddddddd\n\ndddd\n\nxyzxyz\n\nxyz\n\nabc\n\nabcd\n\nabacaba\n\nabaca\n\nabc\n\ndef\n\naaaaaaaa\n\na\n\naaaaaaaa\n\naa", "output": ["2 2\n1 4\n2 1\n0 1\n1 1\n0 1\n8 1\n3 6"]}]
block_completion_008646
block
python
Complete the code in python to solve this programming problem: Description: Nastya baked $$$m$$$ pancakes and spread them on $$$n$$$ dishes. The dishes are in a row and numbered from left to right. She put $$$a_i$$$ pancakes on the dish with the index $$$i$$$.Seeing the dishes, Vlad decided to bring order to the stacks and move some pancakes. In one move, he can shift one pancake from any dish to the closest one, that is, select the dish $$$i$$$ ($$$a_i &gt; 0$$$) and do one of the following: if $$$i &gt; 1$$$, put the pancake on a dish with the previous index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i - 1} = a_{i - 1} + 1$$$; if $$$i &lt; n$$$, put the pancake on a dish with the following index, after this move $$$a_i = a_i - 1$$$ and $$$a_{i + 1} = a_{i + 1} + 1$$$.Vlad wants to make the array $$$a$$$non-increasing, after moving as few pancakes as possible. Help him find the minimum number of moves needed for this.The array $$$a=[a_1, a_2,\dots,a_n]$$$ is called non-increasing if $$$a_i \ge a_{i+1}$$$ for all $$$i$$$ from $$$1$$$ to $$$n-1$$$. Input Specification: The first line of the input contains two numbers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 250$$$) — the number of dishes and the number of pancakes, respectively. The second line contains $$$n$$$ numbers $$$a_i$$$ ($$$0 \le a_i \le m$$$), the sum of all $$$a_i$$$ is equal to $$$m$$$. Output Specification: Print a single number: the minimum number of moves required to make the array $$$a$$$ non-increasing. Notes: NoteIn the first example, you first need to move the pancake from the dish $$$1$$$, after which $$$a = [4, 4, 2, 3, 3, 3]$$$. After that, you need to move the pancake from the dish $$$2$$$ to the dish $$$3$$$ and the array will become non-growing $$$a = [4, 3, 3, 3, 3, 3]$$$. Code: from functools import cache from math import inf from itertools import accumulate def solve(A, m): n = len(A) A.reverse() @cache def dp(i, val, balance): if abs(balance) > m: # TODO: Your code here if (n - i) * val > m: return inf if i == n: return 0 if balance == 0 else inf curr = A[i] + balance take = abs(curr - val) + dp(i + 1, val, curr - val) skip = dp(i, val + 1, balance) return min(take, skip) return dp(0, 0, 0) n, m = map(int, input().split()) A = list(map(int, input().split())) print(solve(A, m))
from functools import cache from math import inf from itertools import accumulate def solve(A, m): n = len(A) A.reverse() @cache def dp(i, val, balance): if abs(balance) > m: {{completion}} if (n - i) * val > m: return inf if i == n: return 0 if balance == 0 else inf curr = A[i] + balance take = abs(curr - val) + dp(i + 1, val, curr - val) skip = dp(i, val + 1, balance) return min(take, skip) return dp(0, 0, 0) n, m = map(int, input().split()) A = list(map(int, input().split())) print(solve(A, m))
return inf
[{"input": "6 19\n5 3 2 3 3 3", "output": ["2"]}, {"input": "3 6\n3 2 1", "output": ["0"]}, {"input": "3 6\n2 1 3", "output": ["1"]}, {"input": "6 19\n3 4 3 3 5 1", "output": ["3"]}, {"input": "10 1\n0 0 0 0 0 0 0 0 0 1", "output": ["9"]}]
block_completion_003581
block
python
Complete the code in python to solve this programming problem: Description: My orzlers, we can optimize this problem from $$$O(S^3)$$$ to $$$O\left(T^\frac{5}{9}\right)$$$!— Spyofgame, founder of Orzlim religionA long time ago, Spyofgame invented the famous array $$$a$$$ ($$$1$$$-indexed) of length $$$n$$$ that contains information about the world and life. After that, he decided to convert it into the matrix $$$b$$$ ($$$0$$$-indexed) of size $$$(n + 1) \times (n + 1)$$$ which contains information about the world, life and beyond.Spyofgame converted $$$a$$$ into $$$b$$$ with the following rules. $$$b_{i,0} = 0$$$ if $$$0 \leq i \leq n$$$; $$$b_{0,i} = a_{i}$$$ if $$$1 \leq i \leq n$$$; $$$b_{i,j} = b_{i,j-1} \oplus b_{i-1,j}$$$ if $$$1 \leq i, j \leq n$$$. Here $$$\oplus$$$ denotes the bitwise XOR operation.Today, archaeologists have discovered the famous matrix $$$b$$$. However, many elements of the matrix has been lost. They only know the values of $$$b_{i,n}$$$ for $$$1 \leq i \leq n$$$ (note that these are some elements of the last column, not the last row).The archaeologists want to know what a possible array of $$$a$$$ is. Can you help them reconstruct any array that could be $$$a$$$? Input Specification: The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$b_{1,n}, b_{2,n}, \ldots, b_{n,n}$$$ ($$$0 \leq b_{i,n} &lt; 2^{30}$$$). Output Specification: If some array $$$a$$$ is consistent with the information, print a line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$. If there are multiple solutions, output any. If such an array does not exist, output $$$-1$$$ instead. Notes: NoteIf we let $$$a = [1,2,3]$$$, then $$$b$$$ will be: $$$\bf{0}$$$$$$\bf{1}$$$$$$\bf{2}$$$$$$\bf{3}$$$$$$\bf{0}$$$$$$1$$$$$$3$$$$$$0$$$$$$\bf{0}$$$$$$1$$$$$$2$$$$$$2$$$$$$\bf{0}$$$$$$1$$$$$$3$$$$$$1$$$ The values of $$$b_{1,n}, b_{2,n}, \ldots, b_{n,n}$$$ generated are $$$[0,2,1]$$$ which is consistent with what the archaeologists have discovered. Code: a=[*map(int,[*open(0)][1].split())] n=len(a) for k in 0,1: for i in range(19): for j in range(n): l=j^1<<i if k^(l<j)and l<n: # TODO: Your code here print(*reversed(a))
a=[*map(int,[*open(0)][1].split())] n=len(a) for k in 0,1: for i in range(19): for j in range(n): l=j^1<<i if k^(l<j)and l<n: {{completion}} print(*reversed(a))
a[j]^=a[l]
[{"input": "3\n0 2 1", "output": ["1 2 3"]}, {"input": "1\n199633", "output": ["199633"]}, {"input": "10\n346484077 532933626 858787727 369947090 299437981 416813461 865836801 141384800 157794568 691345607", "output": ["725081944 922153789 481174947 427448285 516570428 509717938 855104873 280317429 281091129 1050390365"]}]
block_completion_002117
block
python
Complete the code in python to solve this programming problem: Description: You are given a string $$$s$$$ consisting of $$$n$$$ characters. Each character of $$$s$$$ is either 0 or 1.A substring of $$$s$$$ is a contiguous subsequence of its characters.You have to choose two substrings of $$$s$$$ (possibly intersecting, possibly the same, possibly non-intersecting — just any two substrings). After choosing them, you calculate the value of the chosen pair of substrings as follows: let $$$s_1$$$ be the first substring, $$$s_2$$$ be the second chosen substring, and $$$f(s_i)$$$ be the integer such that $$$s_i$$$ is its binary representation (for example, if $$$s_i$$$ is 11010, $$$f(s_i) = 26$$$); the value is the bitwise OR of $$$f(s_1)$$$ and $$$f(s_2)$$$. Calculate the maximum possible value you can get, and print it in binary representation without leading zeroes. Input Specification: The first line contains one integer $$$n$$$ — the number of characters in $$$s$$$. The second line contains $$$s$$$ itself, consisting of exactly $$$n$$$ characters 0 and/or 1. All non-example tests in this problem are generated randomly: every character of $$$s$$$ is chosen independently of other characters; for each character, the probability of it being 1 is exactly $$$\frac{1}{2}$$$. This problem has exactly $$$40$$$ tests. Tests from $$$1$$$ to $$$3$$$ are the examples; tests from $$$4$$$ to $$$40$$$ are generated randomly. In tests from $$$4$$$ to $$$10$$$, $$$n = 5$$$; in tests from $$$11$$$ to $$$20$$$, $$$n = 1000$$$; in tests from $$$21$$$ to $$$40$$$, $$$n = 10^6$$$. Hacks are forbidden in this problem. Output Specification: Print the maximum possible value you can get in binary representation without leading zeroes. Notes: NoteIn the first example, you can choose the substrings 11010 and 101. $$$f(s_1) = 26$$$, $$$f(s_2) = 5$$$, their bitwise OR is $$$31$$$, and the binary representation of $$$31$$$ is 11111.In the second example, you can choose the substrings 1110010 and 11100. Code: n = int(input()) s=input() b=int(s,2) a=b; mx=a|b for i in range(0,7): a=a>>1 m=a|b if m>mx: # TODO: Your code here st=format(mx ,"b") print(st)
n = int(input()) s=input() b=int(s,2) a=b; mx=a|b for i in range(0,7): a=a>>1 m=a|b if m>mx: {{completion}} st=format(mx ,"b") print(st)
mx=m
[{"input": "5\n11010", "output": ["11111"]}, {"input": "7\n1110010", "output": ["1111110"]}, {"input": "4\n0000", "output": ["0"]}]
block_completion_002156
block
python
Complete the code in python to solve this programming problem: Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock. Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query. Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase. Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$. Code: f=open(0) R=lambda:map(int,next(f).split()) n,q=R();p=[0] for w in sorted(R()): p+=p[-1]+w, for _ in " "*q: # TODO: Your code here
f=open(0) R=lambda:map(int,next(f).split()) n,q=R();p=[0] for w in sorted(R()): p+=p[-1]+w, for _ in " "*q: {{completion}}
x, y=R();print(p[n-x+y]-p[n-x])
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
block_completion_000515
block
python
Complete the code in python to solve this programming problem: Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs. Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$). Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating. Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$. Code: from bisect import bisect b = [-5001, 1400, 1600, 1900] for i in range(int(input())): # TODO: Your code here
from bisect import bisect b = [-5001, 1400, 1600, 1900] for i in range(int(input())): {{completion}}
print(f'Division {-bisect(b, int(input()))+5}')
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
block_completion_000726
block
python
Complete the code in python to solve this programming problem: Description: We say an infinite sequence $$$a_{0}, a_{1}, a_2, \ldots$$$ is non-increasing if and only if for all $$$i\ge 0$$$, $$$a_i \ge a_{i+1}$$$.There is an infinite right and down grid. The upper-left cell has coordinates $$$(0,0)$$$. Rows are numbered $$$0$$$ to infinity from top to bottom, columns are numbered from $$$0$$$ to infinity from left to right.There is also a non-increasing infinite sequence $$$a_{0}, a_{1}, a_2, \ldots$$$. You are given $$$a_0$$$, $$$a_1$$$, $$$\ldots$$$, $$$a_n$$$; for all $$$i&gt;n$$$, $$$a_i=0$$$. For every pair of $$$x$$$, $$$y$$$, the cell with coordinates $$$(x,y)$$$ (which is located at the intersection of $$$x$$$-th row and $$$y$$$-th column) is white if $$$y&lt;a_x$$$ and black otherwise.Initially there is one doll named Jina on $$$(0,0)$$$. You can do the following operation. Select one doll on $$$(x,y)$$$. Remove it and place a doll on $$$(x,y+1)$$$ and place a doll on $$$(x+1,y)$$$. Note that multiple dolls can be present at a cell at the same time; in one operation, you remove only one. Your goal is to make all white cells contain $$$0$$$ dolls.What's the minimum number of operations needed to achieve the goal? Print the answer modulo $$$10^9+7$$$. Input Specification: The first line of input contains one integer $$$n$$$ ($$$1\le n\le 2\cdot 10^5$$$). The second line of input contains $$$n+1$$$ integers $$$a_0,a_1,\ldots,a_n$$$ ($$$0\le a_i\le 2\cdot 10^5$$$). It is guaranteed that the sequence $$$a$$$ is non-increasing. Output Specification: Print one integer — the answer to the problem, modulo $$$10^9+7$$$. Notes: NoteConsider the first example. In the given grid, cells $$$(0,0),(0,1),(1,0),(1,1)$$$ are white, and all other cells are black. Let us use triples to describe the grid: triple $$$(x,y,z)$$$ means that there are $$$z$$$ dolls placed on cell $$$(x,y)$$$. Initially the state of the grid is $$$(0,0,1)$$$.One of the optimal sequence of operations is as follows: Do the operation with $$$(0,0)$$$. Now the state of the grid is $$$(1,0,1),(0,1,1)$$$. Do the operation with $$$(0,1)$$$. Now the state of the grid is $$$(1,0,1),(1,1,1),(0,2,1)$$$. Do the operation with $$$(1,0)$$$. Now the state of the grid is $$$(1,1,2),(0,2,1),(2,0,1)$$$. Do the operation with $$$(1,1)$$$. Now the state of the grid is $$$(1,1,1),(0,2,1),(2,0,1),(1,2,1),(2,1,1)$$$. Do the operation with $$$(1,1)$$$. Now the state of the grid is $$$(0,2,1),(2,0,1),(1,2,2),(2,1,2)$$$. Now all white cells contain $$$0$$$ dolls, so we have achieved the goal with $$$5$$$ operations. Code: N = 4 * 10**5 + 5 MOD = 10**9 + 7 fact = [1] invf = [1] for i in range(1, N): fact.append(fact[i-1] * i % MOD) invf.append(pow(fact[-1], MOD-2, MOD)) def C(m, n): if # TODO: Your code here: return 0 return fact[m] * invf[n] % MOD * invf[m-n] % MOD n = int(input()) a = list(map(int, input().split())) ans = sum(C(v+i, i+1) for i, v in enumerate(a)) % MOD print(ans)
N = 4 * 10**5 + 5 MOD = 10**9 + 7 fact = [1] invf = [1] for i in range(1, N): fact.append(fact[i-1] * i % MOD) invf.append(pow(fact[-1], MOD-2, MOD)) def C(m, n): if {{completion}}: return 0 return fact[m] * invf[n] % MOD * invf[m-n] % MOD n = int(input()) a = list(map(int, input().split())) ans = sum(C(v+i, i+1) for i, v in enumerate(a)) % MOD print(ans)
n < 0 or m < n
[{"input": "2\n2 2 0", "output": ["5"]}, {"input": "10\n12 11 8 8 6 6 6 5 3 2 1", "output": ["2596"]}]
control_completion_007310
control_fixed
python
Complete the code in python to solve this programming problem: Description: Codeforces separates its users into $$$4$$$ divisions by their rating: For Division 1: $$$1900 \leq \mathrm{rating}$$$ For Division 2: $$$1600 \leq \mathrm{rating} \leq 1899$$$ For Division 3: $$$1400 \leq \mathrm{rating} \leq 1599$$$ For Division 4: $$$\mathrm{rating} \leq 1399$$$ Given a $$$\mathrm{rating}$$$, print in which division the $$$\mathrm{rating}$$$ belongs. Input Specification: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of testcases. The description of each test consists of one line containing one integer $$$\mathrm{rating}$$$ ($$$-5000 \leq \mathrm{rating} \leq 5000$$$). Output Specification: For each test case, output a single line containing the correct division in the format "Division X", where $$$X$$$ is an integer between $$$1$$$ and $$$4$$$ representing the division for the corresponding rating. Notes: NoteFor test cases $$$1-4$$$, the corresponding ratings are $$$-789$$$, $$$1299$$$, $$$1300$$$, $$$1399$$$, so all of them are in division $$$4$$$.For the fifth test case, the corresponding rating is $$$1400$$$, so it is in division $$$3$$$.For the sixth test case, the corresponding rating is $$$1679$$$, so it is in division $$$2$$$.For the seventh test case, the corresponding rating is $$$2300$$$, so it is in division $$$1$$$. Code: x = input() for i in range(int(x)): z = input() if int(z) >= 1900: print('Division 1') elif int(z) >= 1600: print('Division 2') elif int(z) >= 1400: # TODO: Your code here else: print('Division 4')
x = input() for i in range(int(x)): z = input() if int(z) >= 1900: print('Division 1') elif int(z) >= 1600: print('Division 2') elif int(z) >= 1400: {{completion}} else: print('Division 4')
print('Division 3')
[{"input": "7\n-789\n1299\n1300\n1399\n1400\n1679\n2300", "output": ["Division 4\nDivision 4\nDivision 4\nDivision 4\nDivision 3\nDivision 2\nDivision 1"]}]
block_completion_000732
block
python
Complete the code in python to solve this programming problem: Description: You are given a string $$$s$$$. You have to determine whether it is possible to build the string $$$s$$$ out of strings aa, aaa, bb and/or bbb by concatenating them. You can use the strings aa, aaa, bb and/or bbb any number of times and in any order.For example: aaaabbb can be built as aa $$$+$$$ aa $$$+$$$ bbb; bbaaaaabbb can be built as bb $$$+$$$ aaa $$$+$$$ aa $$$+$$$ bbb; aaaaaa can be built as aa $$$+$$$ aa $$$+$$$ aa; abab cannot be built from aa, aaa, bb and/or bbb. Input Specification: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case consists of one line containing the string $$$s$$$ ($$$1 \le |s| \le 50$$$), consisting of characters a and/or b. Output Specification: For each test case, print YES if it is possible to build the string $$$s$$$. Otherwise, print NO. You may print each letter in any case (for example, YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer). Notes: NoteThe first four test cases of the example are described in the statement. Code: t=int(input()) while(t): i=0 s=input() if(len(s)==1): print("NO") t=t-1 continue while(i<len(s)): if(i==0): if(s[0:2]=="ab" or s[0:2]=="ba"): print("NO") t=t-1 break if(i>0 and i<len(s)-1): if(s[i-1:i+2]=="bab" or s[i-1:i+2]=="aba"): print("NO") t=t-1 break if(i==len(s)-1): if# TODO: Your code here: print("NO") t=t-1 break else: print("YES") t=t-1 break i+=1
t=int(input()) while(t): i=0 s=input() if(len(s)==1): print("NO") t=t-1 continue while(i<len(s)): if(i==0): if(s[0:2]=="ab" or s[0:2]=="ba"): print("NO") t=t-1 break if(i>0 and i<len(s)-1): if(s[i-1:i+2]=="bab" or s[i-1:i+2]=="aba"): print("NO") t=t-1 break if(i==len(s)-1): if{{completion}}: print("NO") t=t-1 break else: print("YES") t=t-1 break i+=1
(s[i-1:]=="ba" or s[i-1:]=="ab")
[{"input": "8\n\naaaabbb\n\nbbaaaaabbb\n\naaaaaa\n\nabab\n\na\n\nb\n\naaaab\n\nbbaaa", "output": ["YES\nYES\nYES\nNO\nNO\nNO\nNO\nYES"]}]
control_completion_001643
control_fixed
python
Complete the code in python to solve this programming problem: Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest. Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$. Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion. Code: R=lambda:map(int,input().split()) n,m=R() *a,=R() b=[[0],[0]] f=max for x in b: for # TODO: Your code here:x+=x[-1]+f(0,u-v), f=min for _ in[0]*m:s,t=R();l=b[s>t];print(l[t]-l[s])
R=lambda:map(int,input().split()) n,m=R() *a,=R() b=[[0],[0]] f=max for x in b: for {{completion}}:x+=x[-1]+f(0,u-v), f=min for _ in[0]*m:s,t=R();l=b[s>t];print(l[t]-l[s])
u,v in zip([0]+a,a)
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
control_completion_002892
control_fixed
python
Complete the code in python to solve this programming problem: Description: The store sells $$$n$$$ items, the price of the $$$i$$$-th item is $$$p_i$$$. The store's management is going to hold a promotion: if a customer purchases at least $$$x$$$ items, $$$y$$$ cheapest of them are free.The management has not yet decided on the exact values of $$$x$$$ and $$$y$$$. Therefore, they ask you to process $$$q$$$ queries: for the given values of $$$x$$$ and $$$y$$$, determine the maximum total value of items received for free, if a customer makes one purchase.Note that all queries are independent; they don't affect the store's stock. Input Specification: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of items in the store and the number of queries, respectively. The second line contains $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le 10^6$$$), where $$$p_i$$$ — the price of the $$$i$$$-th item. The following $$$q$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ each ($$$1 \le y_i \le x_i \le n$$$) — the values of the parameters $$$x$$$ and $$$y$$$ in the $$$i$$$-th query. Output Specification: For each query, print a single integer — the maximum total value of items received for free for one purchase. Notes: NoteIn the first query, a customer can buy three items worth $$$5, 3, 5$$$, the two cheapest of them are $$$3 + 5 = 8$$$.In the second query, a customer can buy two items worth $$$5$$$ and $$$5$$$, the cheapest of them is $$$5$$$.In the third query, a customer has to buy all the items to receive the three cheapest of them for free; their total price is $$$1 + 2 + 3 = 6$$$. Code: from sys import stdin # t = int(stdin.readline().rstrip()) # while t>0: # t-=1 n,q = map(int,stdin.readline().split()) l = list(map(int,stdin.readline().split())) l.sort() for i in range(1,n): l[i] += l[i-1] # print(l) for i in range(q): x,y = map(int,stdin.readline().split()) actual = n-x+y-1 val = l[actual] if # TODO: Your code here: val -= l[n-x-1] print(val)
from sys import stdin # t = int(stdin.readline().rstrip()) # while t>0: # t-=1 n,q = map(int,stdin.readline().split()) l = list(map(int,stdin.readline().split())) l.sort() for i in range(1,n): l[i] += l[i-1] # print(l) for i in range(q): x,y = map(int,stdin.readline().split()) actual = n-x+y-1 val = l[actual] if {{completion}}: val -= l[n-x-1] print(val)
n-x > 0
[{"input": "5 3\n5 3 1 5 2\n3 2\n1 1\n5 3", "output": ["8\n5\n6"]}]
control_completion_000505
control_fixed
python
Complete the code in python to solve this programming problem: Description: On an $$$8 \times 8$$$ grid, some horizontal rows have been painted red, and some vertical columns have been painted blue, in some order. The stripes are drawn sequentially, one after the other. When the stripe is drawn, it repaints all the cells through which it passes.Determine which color was used last. The red stripe was painted after the blue one, so the answer is R. Input Specification: The first line of the input contains a single integer $$$t$$$ ($$$1 \leq t \leq 4000$$$) — the number of test cases. The description of test cases follows. There is an empty line before each test case. Each test case consists of $$$8$$$ lines, each containing $$$8$$$ characters. Each of these characters is either 'R', 'B', or '.', denoting a red square, a blue square, and an unpainted square, respectively. It is guaranteed that the given field is obtained from a colorless one by drawing horizontal red rows and vertical blue columns. At least one stripe is painted. Output Specification: For each test case, output 'R' if a red stripe was painted last, and 'B' if a blue stripe was painted last (without quotes). Notes: NoteThe first test case is pictured in the statement.In the second test case, the first blue column is painted first, then the first and last red rows, and finally the last blue column. Since a blue stripe is painted last, the answer is B. Code: n = int(input()) for i in range(n) : b = [] j = 0 key = '.' while(j<8) : a = input() if(a != '') : b.append(a) j += 1 for j in range(8) : if(len(set(b[j])) == 1 and b[j][0] == 'R') : # TODO: Your code here if(key!= 'R') : key = 'B' print(key)
n = int(input()) for i in range(n) : b = [] j = 0 key = '.' while(j<8) : a = input() if(a != '') : b.append(a) j += 1 for j in range(8) : if(len(set(b[j])) == 1 and b[j][0] == 'R') : {{completion}} if(key!= 'R') : key = 'B' print(key)
key = 'R' break
[{"input": "4\n\n\n\n\n....B...\n\n....B...\n\n....B...\n\nRRRRRRRR\n\n....B...\n\n....B...\n\n....B...\n\n....B...\n\n\n\n\nRRRRRRRB\n\nB......B\n\nB......B\n\nB......B\n\nB......B\n\nB......B\n\nB......B\n\nRRRRRRRB\n\n\n\n\nRRRRRRBB\n\n.B.B..BB\n\nRRRRRRBB\n\n.B.B..BB\n\n.B.B..BB\n\nRRRRRRBB\n\n.B.B..BB\n\n.B.B..BB\n\n\n\n\n........\n\n........\n\n........\n\nRRRRRRRR\n\n........\n\n........\n\n........\n\n........", "output": ["R\nB\nB\nR"]}]
block_completion_005812
block
python
Complete the code in python to solve this programming problem: Description: A row of $$$n$$$ cells is given, all initially white. Using a stamp, you can stamp any two neighboring cells such that one becomes red and the other becomes blue. A stamp can be rotated, i.e. it can be used in both ways: as $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}$$$ and as $$$\color{red}{\texttt{R}}\color{blue}{\texttt{B}}$$$.During use, the stamp must completely fit on the given $$$n$$$ cells (it cannot be partially outside the cells). The stamp can be applied multiple times to the same cell. Each usage of the stamp recolors both cells that are under the stamp.For example, one possible sequence of stamps to make the picture $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\color{blue}{\texttt{B}}\texttt{W}$$$ could be $$$\texttt{WWWWW} \to \texttt{WW}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\texttt{W} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\texttt{W} \to \color{blue}{\texttt{B}}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}\texttt{W}$$$. Here $$$\texttt{W}$$$, $$$\color{red}{\texttt{R}}$$$, and $$$\color{blue}{\texttt{B}}$$$ represent a white, red, or blue cell, respectively, and the cells that the stamp is used on are marked with an underline.Given a final picture, is it possible to make it using the stamp zero or more times? Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the length of the picture. The second line of each test case contains a string $$$s$$$ — the picture you need to make. It is guaranteed that the length of $$$s$$$ is $$$n$$$ and that $$$s$$$ only consists of the characters $$$\texttt{W}$$$, $$$\texttt{R}$$$, and $$$\texttt{B}$$$, representing a white, red, or blue cell, respectively. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$. Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. As an answer, output "YES" if it possible to make the picture using the stamp zero or more times, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer). Notes: NoteThe first test case is explained in the statement.For the second, third, and fourth test cases, it is not possible to stamp a single cell, so the answer is "NO".For the fifth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{blue}{\texttt{B}}$$$.For the sixth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}$$$.For the seventh test case, you don't need to use the stamp at all. Code: for # TODO: Your code here: l = int(input()) print("NO" if any (len(set(x)) == 1 for x in input().split('W') ) else "YES")
for {{completion}}: l = int(input()) print("NO" if any (len(set(x)) == 1 for x in input().split('W') ) else "YES")
_ in range(int(input()))
[{"input": "12\n5\nBRBBW\n1\nB\n2\nWB\n2\nRW\n3\nBRB\n3\nRBB\n7\nWWWWWWW\n9\nRBWBWRRBW\n10\nBRBRBRBRRB\n12\nBBBRWWRRRWBR\n10\nBRBRBRBRBW\n5\nRBWBW", "output": ["YES\nNO\nNO\nNO\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO"]}]
control_completion_000906
control_fixed
python
Complete the code in python to solve this programming problem: Description: While searching for the pizza, baby Hosssam came across two permutations $$$a$$$ and $$$b$$$ of length $$$n$$$.Recall that a permutation is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array) and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).Baby Hosssam forgot about the pizza and started playing around with the two permutations. While he was playing with them, some elements of the first permutation got mixed up with some elements of the second permutation, and to his surprise those elements also formed a permutation of size $$$n$$$.Specifically, he mixed up the permutations to form a new array $$$c$$$ in the following way. For each $$$i$$$ ($$$1\le i\le n$$$), he either made $$$c_i=a_i$$$ or $$$c_i=b_i$$$. The array $$$c$$$ is a permutation. You know permutations $$$a$$$, $$$b$$$, and values at some positions in $$$c$$$. Please count the number different permutations $$$c$$$ that are consistent with the described process and the given values. Since the answer can be large, print it modulo $$$10^9+7$$$.It is guaranteed that there exists at least one permutation $$$c$$$ that satisfies all the requirements. Input Specification: The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1\le n\le 10^5$$$) — the length of the permutations. The next line contains $$$n$$$ distinct integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1\le a_i\le n$$$) — the first permutation. The next line contains $$$n$$$ distinct integers $$$b_1,b_2,\ldots,b_n$$$ ($$$1\le b_i\le n$$$) — the second permutation. The next line contains $$$n$$$ distinct integers $$$d_1,d_2,\ldots,d_n$$$ ($$$d_i$$$ is either $$$0$$$, $$$a_i$$$, or $$$b_i$$$) — the description of the known values of $$$c$$$. If $$$d_i=0$$$, then there are no requirements on the value of $$$c_i$$$. Otherwise, it is required that $$$c_i=d_i$$$. It is guaranteed that there exists at least one permutation $$$c$$$ that satisfies all the requirements. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$5 \cdot 10^5$$$. Output Specification: For each test case, print the number of possible permutations $$$c$$$, modulo $$$10^9+7$$$. Notes: NoteIn the first test case, there are $$$4$$$ distinct permutation that can be made using the process: $$$[2,3,1,4,5,6,7]$$$, $$$[2,3,1,7,6,5,4]$$$, $$$[2,3,1,4,6,5,7]$$$, $$$[2,3,1,7,5,6,4]$$$.In the second test case, there is only one distinct permutation that can be made using the process: $$$[1]$$$.In the third test case, there are $$$2$$$ distinct permutation that can be made using the process: $$$[6,5,2,1,4,3]$$$, $$$[6,5,3,1,4,2]$$$.In the fourth test case, there are $$$2$$$ distinct permutation that can be made using the process: $$$[1,2,8,7,4,3,6,5]$$$, $$$[1,6,4,7,2,3,8,5]$$$.In the fifth test case, there is only one distinct permutation that can be made using the process: $$$[1,9,2,3,4,10,8,6,7,5]$$$. Code: import sys input = sys.stdin.readline class Solver1670C: def __init__(self): self.n = int(input()) self.a = list(map(int, input().split(' '))) self.b = list(map(int, input().split(' '))) self.d = list(map(int, input().split(' '))) self.pos_a = [0 for _ in range(self.n+1)] self.pos_b = [0 for _ in range(self.n+1)] self.been = [0 for _ in range(self.n+1)] self.mod = (10**9)+7 def solve(self): for i in range(0, self.n): self.pos_a[self.a[i]] = i self.pos_b[self.b[i]] = i for i in range(0, self.n): if self.d[i] and self.been[i] == 0: if self.d[i] == self.a[i]: j = i while # TODO: Your code here: self.been[j] = 1 j = self.pos_a[self.b[j]] else: j = i while self.been[j] == 0: self.been[j] = 1 j = self.pos_b[self.a[j]] outp = 1 for i in range(0, self.n): if self.been[i] == 0: cnt = 0 j = i while self.been[j] == 0: self.been[j] = 1 j = self.pos_a[self.b[j]] cnt += 1 if cnt >= 2: outp = outp * 2 % self.mod return outp t = int(input()) while t: t -= 1 cur = Solver1670C() print(cur.solve())
import sys input = sys.stdin.readline class Solver1670C: def __init__(self): self.n = int(input()) self.a = list(map(int, input().split(' '))) self.b = list(map(int, input().split(' '))) self.d = list(map(int, input().split(' '))) self.pos_a = [0 for _ in range(self.n+1)] self.pos_b = [0 for _ in range(self.n+1)] self.been = [0 for _ in range(self.n+1)] self.mod = (10**9)+7 def solve(self): for i in range(0, self.n): self.pos_a[self.a[i]] = i self.pos_b[self.b[i]] = i for i in range(0, self.n): if self.d[i] and self.been[i] == 0: if self.d[i] == self.a[i]: j = i while {{completion}}: self.been[j] = 1 j = self.pos_a[self.b[j]] else: j = i while self.been[j] == 0: self.been[j] = 1 j = self.pos_b[self.a[j]] outp = 1 for i in range(0, self.n): if self.been[i] == 0: cnt = 0 j = i while self.been[j] == 0: self.been[j] = 1 j = self.pos_a[self.b[j]] cnt += 1 if cnt >= 2: outp = outp * 2 % self.mod return outp t = int(input()) while t: t -= 1 cur = Solver1670C() print(cur.solve())
self.been[j] == 0
[{"input": "9\n7\n1 2 3 4 5 6 7\n2 3 1 7 6 5 4\n2 0 1 0 0 0 0\n1\n1\n1\n0\n6\n1 5 2 4 6 3\n6 5 3 1 4 2\n6 0 0 0 0 0\n8\n1 6 4 7 2 3 8 5\n3 2 8 1 4 5 6 7\n1 0 0 7 0 3 0 5\n10\n1 8 6 2 4 7 9 3 10 5\n1 9 2 3 4 10 8 6 7 5\n1 9 2 3 4 10 8 6 7 5\n7\n1 2 3 4 5 6 7\n2 3 1 7 6 5 4\n0 0 0 0 0 0 0\n5\n1 2 3 4 5\n1 2 3 4 5\n0 0 0 0 0\n5\n1 2 3 4 5\n1 2 3 5 4\n0 0 0 0 0\n3\n1 2 3\n3 1 2\n0 0 0", "output": ["4\n1\n2\n2\n1\n8\n1\n2\n2"]}]
control_completion_005923
control_fixed
python
Complete the code in python to solve this programming problem: Description: A row of $$$n$$$ cells is given, all initially white. Using a stamp, you can stamp any two neighboring cells such that one becomes red and the other becomes blue. A stamp can be rotated, i.e. it can be used in both ways: as $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}$$$ and as $$$\color{red}{\texttt{R}}\color{blue}{\texttt{B}}$$$.During use, the stamp must completely fit on the given $$$n$$$ cells (it cannot be partially outside the cells). The stamp can be applied multiple times to the same cell. Each usage of the stamp recolors both cells that are under the stamp.For example, one possible sequence of stamps to make the picture $$$\color{blue}{\texttt{B}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\color{blue}{\texttt{B}}\texttt{W}$$$ could be $$$\texttt{WWWWW} \to \texttt{WW}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\texttt{W} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{red}{\texttt{R}}\color{blue}{\texttt{B}}\texttt{W} \to \color{blue}{\texttt{B}}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}\texttt{W}$$$. Here $$$\texttt{W}$$$, $$$\color{red}{\texttt{R}}$$$, and $$$\color{blue}{\texttt{B}}$$$ represent a white, red, or blue cell, respectively, and the cells that the stamp is used on are marked with an underline.Given a final picture, is it possible to make it using the stamp zero or more times? Input Specification: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the length of the picture. The second line of each test case contains a string $$$s$$$ — the picture you need to make. It is guaranteed that the length of $$$s$$$ is $$$n$$$ and that $$$s$$$ only consists of the characters $$$\texttt{W}$$$, $$$\texttt{R}$$$, and $$$\texttt{B}$$$, representing a white, red, or blue cell, respectively. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$. Output Specification: Output $$$t$$$ lines, each of which contains the answer to the corresponding test case. As an answer, output "YES" if it possible to make the picture using the stamp zero or more times, and "NO" otherwise. You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer). Notes: NoteThe first test case is explained in the statement.For the second, third, and fourth test cases, it is not possible to stamp a single cell, so the answer is "NO".For the fifth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{blue}{\texttt{B}}\color{red}{\texttt{R}}}}\color{blue}{\texttt{B}}$$$.For the sixth test case, you can use the stamp as follows: $$$\texttt{WWW} \to \texttt{W}\color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}} \to \color{brown}{\underline{\color{red}{\texttt{R}}\color{blue}{\texttt{B}}}}\color{blue}{\texttt{B}}$$$.For the seventh test case, you don't need to use the stamp at all. Code: t = int(input()) Ans = [-1]*t for z in range(t): n = int(input()) l = input().split('W') bad = False for s in l: b1 = 'R' in s b2 = 'B' in s if (b1 ^ b2): # TODO: Your code here print("NO" if bad else "YES")
t = int(input()) Ans = [-1]*t for z in range(t): n = int(input()) l = input().split('W') bad = False for s in l: b1 = 'R' in s b2 = 'B' in s if (b1 ^ b2): {{completion}} print("NO" if bad else "YES")
bad = True
[{"input": "12\n5\nBRBBW\n1\nB\n2\nWB\n2\nRW\n3\nBRB\n3\nRBB\n7\nWWWWWWW\n9\nRBWBWRRBW\n10\nBRBRBRBRRB\n12\nBBBRWWRRRWBR\n10\nBRBRBRBRBW\n5\nRBWBW", "output": ["YES\nNO\nNO\nNO\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO"]}]
block_completion_000931
block
python
Complete the code in python to solve this programming problem: Description: You are given a rooted tree of $$$2^n - 1$$$ vertices. Every vertex of this tree has either $$$0$$$ children, or $$$2$$$ children. All leaves of this tree have the same distance from the root, and for every non-leaf vertex, one of its children is the left one, and the other child is the right one. Formally, you are given a perfect binary tree.The vertices of the tree are numbered in the following order: the root has index $$$1$$$; if a vertex has index $$$x$$$, then its left child has index $$$2x$$$, and its right child has index $$$2x+1$$$. Every vertex of the tree has a letter written on it, either A or B. Let's define the character on the vertex $$$x$$$ as $$$s_x$$$.Let the preorder string of some vertex $$$x$$$ be defined in the following way: if the vertex $$$x$$$ is a leaf, then the preorder string of $$$x$$$ be consisting of only one character $$$s_x$$$; otherwise, the preorder string of $$$x$$$ is $$$s_x + f(l_x) + f(r_x)$$$, where $$$+$$$ operator defines concatenation of strings, $$$f(l_x)$$$ is the preorder string of the left child of $$$x$$$, and $$$f(r_x)$$$ is the preorder string of the right child of $$$x$$$. The preorder string of the tree is the preorder string of its root.Now, for the problem itself...You have to calculate the number of different strings that can be obtained as the preorder string of the given tree, if you are allowed to perform the following operation any number of times before constructing the preorder string of the tree: choose any non-leaf vertex $$$x$$$, and swap its children (so, the left child becomes the right one, and vice versa). Input Specification: The first line contains one integer $$$n$$$ ($$$2 \le n \le 18$$$). The second line contains a sequence of $$$2^n-1$$$ characters $$$s_1, s_2, \dots, s_{2^n-1}$$$. Each character is either A or B. The characters are not separated by spaces or anything else. Output Specification: Print one integer — the number of different strings that can be obtained as the preorder string of the given tree, if you can apply any number of operations described in the statement. Since it can be very large, print it modulo $$$998244353$$$. Code: def dfs(tree,i,h): if i>=2**(h-1)-1: return [tree[i],1] ls,li=dfs(tree,i*2+1,h) rs,ri=dfs(tree,i*2+2,h) res=li*ri if # TODO: Your code here: res*=2 if ls>rs: return [tree[i]+rs+ls,res] else: return [tree[i]+ls+rs,res] h=int(input()) tree=input() print(dfs(tree,0,h)[1]%998244353 )
def dfs(tree,i,h): if i>=2**(h-1)-1: return [tree[i],1] ls,li=dfs(tree,i*2+1,h) rs,ri=dfs(tree,i*2+2,h) res=li*ri if {{completion}}: res*=2 if ls>rs: return [tree[i]+rs+ls,res] else: return [tree[i]+ls+rs,res] h=int(input()) tree=input() print(dfs(tree,0,h)[1]%998244353 )
ls!=rs
[{"input": "4\nBAAAAAAAABBABAB", "output": ["16"]}, {"input": "2\nBAA", "output": ["1"]}, {"input": "2\nABA", "output": ["2"]}, {"input": "2\nAAB", "output": ["2"]}, {"input": "2\nAAA", "output": ["1"]}]
control_completion_001667
control_fixed
python
Complete the code in python to solve this programming problem: Description: Given the string $$$s$$$ of decimal digits (0-9) of length $$$n$$$.A substring is a sequence of consecutive characters of a string. The substring of this string is defined by a pair of indexes — with its left and right ends. So, each pair of indexes ($$$l, r$$$), where $$$1 \le l \le r \le n$$$, corresponds to a substring of the string $$$s$$$. We will define as $$$v(l,r)$$$ the numeric value of the corresponding substring (leading zeros are allowed in it).For example, if $$$n=7$$$, $$$s=$$$"1003004", then $$$v(1,3)=100$$$, $$$v(2,3)=0$$$ and $$$v(2,7)=3004$$$.You are given $$$n$$$, $$$s$$$ and an integer $$$w$$$ ($$$1 \le w &lt; n$$$).You need to process $$$m$$$ queries, each of which is characterized by $$$3$$$ numbers $$$l_i, r_i, k_i$$$ ($$$1 \le l_i \le r_i \le n; 0 \le k_i \le 8$$$).The answer to the $$$i$$$th query is such a pair of substrings of length $$$w$$$ that if we denote them as $$$(L_1, L_1+w-1)$$$ and $$$(L_2, L_2+w-1)$$$, then: $$$L_1 \ne L_2$$$, that is, the substrings are different; the remainder of dividing a number $$$v(L_1, L_1+w-1) \cdot v(l_i, r_i) + v(L_2, L_2 + w - 1)$$$ by $$$9$$$ is equal to $$$k_i$$$. If there are many matching substring pairs, then find a pair where $$$L_1$$$ is as small as possible. If there are many matching pairs in this case, then minimize $$$L_2$$$.Note that the answer may not exist. Input Specification: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — number of input test cases. The first line of each test case contains a string $$$s$$$, which contains only the characters 0-9 and has a length $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$). The second line contains two integers $$$w, m$$$ ($$$1 \le w &lt; n, 1 \le m \le 2 \cdot 10^5$$$), where $$$n$$$ — is the length of the given string $$$s$$$. The number $$$w$$$ denotes the lengths of the substrings being searched for, and $$$m$$$ is the number of queries to be processed. The following $$$m$$$ lines contain integers $$$l_i, r_i, k_i$$$ ($$$1 \le l_i \le r_i \le n$$$, $$$0 \le k_i \le 8$$$) — $$$i$$$th query parameters. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. It is also guaranteed that the sum of $$$m$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. Output Specification: For each request, print in a separate line: left borders of the required substrings: $$$L_1$$$ and $$$L_2$$$; -1 -1 otherwise, if there is no solution. If there are several solutions, minimize $$$L_1$$$ first, and minimize $$$L_2$$$ second. Notes: NoteConsider the first test case of example inputs. In this test case $$$n=7$$$, $$$s=$$$"1003004", $$$w=4$$$ and one query $$$l_1=1$$$, $$$r_1=2$$$, $$$k_1=1$$$. Note that $$$v(1,2)=10$$$. We need to find a pair of substrings of length $$$4$$$ such that $$$v(L_1, L_1+3)\cdot10+v(L_2,L_2+3)$$$ has a remainder of $$$k_1=1$$$ when divided by $$$9$$$. The values $$$L_1=2, L_2=4$$$ actually satisfy all the requirements: $$$v(L_1, L_1+w-1)=v(2,5)=30$$$, $$$v(L_2, L_2+w-1)=v(4,7)=3004$$$. Indeed, $$$30\cdot10+3004=3304$$$, which has a remainder of $$$1$$$ when divided by $$$9$$$. It can be shown that $$$L_1=2$$$ is the minimum possible value, and $$$L_2=4$$$ is the minimum possible with $$$L_1=2$$$. Code: from __future__ import annotations import csv import datetime import string import sys import time from collections import defaultdict from contextlib import contextmanager from typing import List def solve() -> None: a = [int(c) % 9 for c in next_token()] n = len(a) sa = list(a) for i in range(1, n): sa[i] += sa[i - 1] w = next_int() indices = defaultdict(list) for i in range(w, n + 1): vlr = ((sa[i - 1] - (sa[i - w - 1] if i - w > 0 else 0)) % 9 + 9) % 9 indices[vlr].append(i - w + 1) cache = dict() INF = (n + 1, n) for _ in range(next_int()): l = next_int() - 1 r = next_int() - 1 k = next_int() vlr = ((sa[r] - (sa[l - 1] if l > 0 else 0)) % 9 + 9) % 9 if (vlr, k) not in cache: res = INF for v1 in range(9): for v2 in range(9): if ((v1 * vlr + v2) % 9 + 9) % 9 == k: if v1 == v2: if len(indices[v1]) > 1: res = min(res, tuple(indices[v1][:2])) else: if len(indices[v1]) > 0 and len(indices[v2]) > 0: # TODO: Your code here cache[(vlr, k)] = res if res != INF else (-1, -1) print(*cache[(vlr, k)]) def global_init() -> None: pass RUN_N_TESTS_IN_PROD = True PRINT_CASE_NUMBER = False ASSERT_IN_PROD = False LOG_TO_FILE = False READ_FROM_CONSOLE_IN_DEBUG = False WRITE_TO_CONSOLE_IN_DEBUG = True TEST_TIMER = False IS_DEBUG = "DEBUG_MODE" in sys.argv __output_file = None __input_file = None __input_last_line = None def run() -> None: global __input_file, __input_last_line, __output_file __output_file = sys.stdout if not IS_DEBUG or WRITE_TO_CONSOLE_IN_DEBUG else open("../output.txt", "w") try: __input_file = sys.stdin if not IS_DEBUG or READ_FROM_CONSOLE_IN_DEBUG else open("../input.txt") try: with timer("total"): global_init() t = next_int() if RUN_N_TESTS_IN_PROD or IS_DEBUG else 1 for i in range(t): if PRINT_CASE_NUMBER: fprint(f"Case #{i + 1}: ") if TEST_TIMER: with timer(f"test #{i + 1}"): solve() else: solve() if IS_DEBUG: __output_file.flush() finally: __input_last_line = None __input_file.close() __input_file = None finally: __output_file.flush() __output_file.close() def fprint(*objects, **kwargs): print(*objects, end="", file=__output_file, **kwargs) def fprintln(*objects, **kwargs): print(*objects, file=__output_file, **kwargs) def next_line() -> str: global __input_last_line __input_last_line = None return __input_file.readline() def next_token() -> str: global __input_last_line while not __input_last_line: __input_last_line = __input_file.readline().split()[::-1] return __input_last_line.pop() def next_int(): return int(next_token()) def next_float(): return float(next_token()) def next_int_array(n: int) -> List[int]: return [int(next_token()) for _ in range(n)] if IS_DEBUG or ASSERT_IN_PROD: def assert_predicate(p: bool, message: str = ""): if not p: raise AssertionError(message) def assert_not_equal(unexpected, actual): if unexpected == actual: raise AssertionError(f"assert_not_equal: {unexpected} == {actual}") def assert_equal(expected, actual): if expected != actual: raise AssertionError(f"assert_equal: {expected} != {actual}") else: def assert_predicate(p: bool, message: str = ""): pass def assert_not_equal(unexpected, actual): pass def assert_equal(expected, actual): pass if IS_DEBUG: __log_file = open(f"../logs/py_solution_{int(time.time() * 1000)}.log", "w") if LOG_TO_FILE else sys.stdout def log(*args, **kwargs): print(datetime.datetime.now(), "-", *args, **kwargs, flush=True, file=__log_file) @contextmanager def timer(label: str): start_time = time.time() try: yield finally: log(f"Timer[{label}]: {time.time() - start_time:.6f}s") else: def log(*args, **kwargs): pass @contextmanager def timer(label: str): yield if __name__ == "__main__": run()
from __future__ import annotations import csv import datetime import string import sys import time from collections import defaultdict from contextlib import contextmanager from typing import List def solve() -> None: a = [int(c) % 9 for c in next_token()] n = len(a) sa = list(a) for i in range(1, n): sa[i] += sa[i - 1] w = next_int() indices = defaultdict(list) for i in range(w, n + 1): vlr = ((sa[i - 1] - (sa[i - w - 1] if i - w > 0 else 0)) % 9 + 9) % 9 indices[vlr].append(i - w + 1) cache = dict() INF = (n + 1, n) for _ in range(next_int()): l = next_int() - 1 r = next_int() - 1 k = next_int() vlr = ((sa[r] - (sa[l - 1] if l > 0 else 0)) % 9 + 9) % 9 if (vlr, k) not in cache: res = INF for v1 in range(9): for v2 in range(9): if ((v1 * vlr + v2) % 9 + 9) % 9 == k: if v1 == v2: if len(indices[v1]) > 1: res = min(res, tuple(indices[v1][:2])) else: if len(indices[v1]) > 0 and len(indices[v2]) > 0: {{completion}} cache[(vlr, k)] = res if res != INF else (-1, -1) print(*cache[(vlr, k)]) def global_init() -> None: pass RUN_N_TESTS_IN_PROD = True PRINT_CASE_NUMBER = False ASSERT_IN_PROD = False LOG_TO_FILE = False READ_FROM_CONSOLE_IN_DEBUG = False WRITE_TO_CONSOLE_IN_DEBUG = True TEST_TIMER = False IS_DEBUG = "DEBUG_MODE" in sys.argv __output_file = None __input_file = None __input_last_line = None def run() -> None: global __input_file, __input_last_line, __output_file __output_file = sys.stdout if not IS_DEBUG or WRITE_TO_CONSOLE_IN_DEBUG else open("../output.txt", "w") try: __input_file = sys.stdin if not IS_DEBUG or READ_FROM_CONSOLE_IN_DEBUG else open("../input.txt") try: with timer("total"): global_init() t = next_int() if RUN_N_TESTS_IN_PROD or IS_DEBUG else 1 for i in range(t): if PRINT_CASE_NUMBER: fprint(f"Case #{i + 1}: ") if TEST_TIMER: with timer(f"test #{i + 1}"): solve() else: solve() if IS_DEBUG: __output_file.flush() finally: __input_last_line = None __input_file.close() __input_file = None finally: __output_file.flush() __output_file.close() def fprint(*objects, **kwargs): print(*objects, end="", file=__output_file, **kwargs) def fprintln(*objects, **kwargs): print(*objects, file=__output_file, **kwargs) def next_line() -> str: global __input_last_line __input_last_line = None return __input_file.readline() def next_token() -> str: global __input_last_line while not __input_last_line: __input_last_line = __input_file.readline().split()[::-1] return __input_last_line.pop() def next_int(): return int(next_token()) def next_float(): return float(next_token()) def next_int_array(n: int) -> List[int]: return [int(next_token()) for _ in range(n)] if IS_DEBUG or ASSERT_IN_PROD: def assert_predicate(p: bool, message: str = ""): if not p: raise AssertionError(message) def assert_not_equal(unexpected, actual): if unexpected == actual: raise AssertionError(f"assert_not_equal: {unexpected} == {actual}") def assert_equal(expected, actual): if expected != actual: raise AssertionError(f"assert_equal: {expected} != {actual}") else: def assert_predicate(p: bool, message: str = ""): pass def assert_not_equal(unexpected, actual): pass def assert_equal(expected, actual): pass if IS_DEBUG: __log_file = open(f"../logs/py_solution_{int(time.time() * 1000)}.log", "w") if LOG_TO_FILE else sys.stdout def log(*args, **kwargs): print(datetime.datetime.now(), "-", *args, **kwargs, flush=True, file=__log_file) @contextmanager def timer(label: str): start_time = time.time() try: yield finally: log(f"Timer[{label}]: {time.time() - start_time:.6f}s") else: def log(*args, **kwargs): pass @contextmanager def timer(label: str): yield if __name__ == "__main__": run()
res = min(res, (indices[v1][0], indices[v2][0]))
[{"input": "5\n\n1003004\n\n4 1\n\n1 2 1\n\n179572007\n\n4 2\n\n2 7 3\n\n2 7 4\n\n111\n\n2 1\n\n2 2 6\n\n0000\n\n1 2\n\n1 4 0\n\n1 4 1\n\n484\n\n1 5\n\n2 2 0\n\n2 3 7\n\n1 2 5\n\n3 3 8\n\n2 2 6", "output": ["2 4\n1 5\n1 2\n-1 -1\n1 2\n-1 -1\n1 3\n1 3\n-1 -1\n-1 -1\n-1 -1"]}]
block_completion_008664
block
python
Complete the code in python to solve this programming problem: Description: You are beta testing the new secret Terraria update. This update will add quests to the game!Simply, the world map can be represented as an array of length $$$n$$$, where the $$$i$$$-th column of the world has height $$$a_i$$$.There are $$$m$$$ quests you have to test. The $$$j$$$-th of them is represented by two integers $$$s_j$$$ and $$$t_j$$$. In this quest, you have to go from the column $$$s_j$$$ to the column $$$t_j$$$. At the start of the quest, you are appearing at the column $$$s_j$$$.In one move, you can go from the column $$$x$$$ to the column $$$x-1$$$ or to the column $$$x+1$$$. In this version, you have Spectre Boots, which allow you to fly. Since it is a beta version, they are bugged, so they only allow you to fly when you are going up and have infinite fly duration. When you are moving from the column with the height $$$p$$$ to the column with the height $$$q$$$, then you get some amount of fall damage. If the height $$$p$$$ is greater than the height $$$q$$$, you get $$$p - q$$$ fall damage, otherwise you fly up and get $$$0$$$ damage.For each of the given quests, determine the minimum amount of fall damage you can get during this quest. Input Specification: The first line of the input contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 10^5; 1 \le m \le 10^5$$$) — the number of columns in the world and the number of quests you have to test, respectively. The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), where $$$a_i$$$ is the height of the $$$i$$$-th column of the world. The next $$$m$$$ lines describe quests. The $$$j$$$-th of them contains two integers $$$s_j$$$ and $$$t_j$$$ ($$$1 \le s_j, t_j \le n; s_j \ne t_j$$$), which means you have to move from the column $$$s_j$$$ to the column $$$t_j$$$ during the $$$j$$$-th quest. Note that $$$s_j$$$ can be greater than $$$t_j$$$. Output Specification: Print $$$m$$$ integers. The $$$j$$$-th of them should be the minimum amount of fall damage you can get during the $$$j$$$-th quest completion. Code: (n,m),(*a,),*r=(map(int,s.split())for s in open(0)) b=[[0],[0]] for x in b: for u,v in zip([0]+a,a):# TODO: Your code here max=min for s,t in r:l=b[s>t];print(l[t]-l[s])
(n,m),(*a,),*r=(map(int,s.split())for s in open(0)) b=[[0],[0]] for x in b: for u,v in zip([0]+a,a):{{completion}} max=min for s,t in r:l=b[s>t];print(l[t]-l[s])
x+=x[-1]+max(0,u-v),
[{"input": "7 6\n10 8 9 6 8 12 7\n1 2\n1 7\n4 6\n7 1\n3 5\n4 2", "output": ["2\n10\n0\n7\n3\n1"]}]
block_completion_002946
block