


from statistics import mean

import numpy as np







def        SCREAMING_SNAKE_CASE_						(			__A :			int  ,    __A :			Dict  ,    __A :			Tuple  ,    __A :			Any				)	->		Any:

       """simple docstring"""

       a_					:		Any							=  0
       # Number of processes finished
       a_					:		Any							=  0
       # Displays the finished process.
       # If it is 0, the performance is completed if it is 1, before the performance.
       a_					:		List[str]							=  [0] * no_of_process
       # List to include calculation results
       a_					:		List[Any]							=  [0] * no_of_process

       # Sort by arrival time.
       a_					:		Optional[int]							=  [burst_time[i] for i in np.argsort(lowerCamelCase__				)]
       a_					:		Tuple							=  [process_name[i] for i in np.argsort(lowerCamelCase__				)]
       arrival_time.sort()

       while no_of_process > finished_process_count:
              a_					:		Dict							=  0
              while finished_process[i] == 1:
                     i += 1
              if current_time < arrival_time[i]:
                     a_					:		Optional[Any]							=  arrival_time[i]

              a_					:		Dict							=  0
              # Index showing the location of the process being performed
              a_					:		Any							=  0
              # Saves the current response ratio.
              a_					:		int							=  0
              for i in range(0  ,    lowerCamelCase__				):
                     if finished_process[i] == 0 and arrival_time[i] <= current_time:
                            a_					:		List[Any]							=  (burst_time[i] + (current_time - arrival_time[i])) / burst_time[
                                i
                            ]
                     if response_ratio < temp:
                            a_					:		Tuple							=  temp
                            a_					:		Tuple							=  i

        # Calculate the turn around time
              a_					:		int							=  current_time + burst_time[loc] - arrival_time[loc]
              current_time += burst_time[loc]
              # Indicates that the process has been performed.
              a_					:		Dict							=  1
              # Increase finished_process_count by 1
              finished_process_count += 1

       return turn_around_time







def        SCREAMING_SNAKE_CASE_						(			__A :			List[str]  ,    __A :			Tuple  ,    __A :			Any  ,    __A :			List[Any]				)	->		Tuple:

       """simple docstring"""

       a_					:		List[str]							=  [0] * no_of_process
       for i in range(0  ,    lowerCamelCase__				):
              a_					:		int							=  turn_around_time[i] - burst_time[i]
       return waiting_time


if __name__ == "__main__":
   UpperCAmelCase_      : List[Any]  					=       5
   UpperCAmelCase_      : List[Any]  					=       ['''A''', '''B''', '''C''', '''D''', '''E''']
   UpperCAmelCase_      : List[str]  					=       [1, 2, 3, 4, 5]
   UpperCAmelCase_      : List[str]  					=       [1, 2, 3, 4, 5]

   UpperCAmelCase_      : Dict  					=       calculate_turn_around_time(
       process_name, arrival_time, burst_time, no_of_process
   )
   UpperCAmelCase_      : Dict  					=       calculate_waiting_time(
       process_name, turn_around_time, burst_time, no_of_process
   )

   print('Process name \tArrival time \tBurst time \tTurn around time \tWaiting time')
   for i in range(0, no_of_process):
      print(
          F'{process_name[i]}\t\t{arrival_time[i]}\t\t{burst_time[i]}\t\t'
          F'{turn_around_time[i]}\t\t\t{waiting_time[i]}'
      )

   print(F'average waiting time : {mean(waiting_time):.5f}')
   print(F'average turn around time : {mean(turn_around_time):.5f}')







'''simple docstring'''




from math import pow



def 	a			( lowerCamelCase__						,      lowerCamelCase__						,      lowerCamelCase__						,      lowerCamelCase__						,      lowerCamelCase__						,      ):
			'''simple docstring'''





			if current_sum == needed_sum:
						# If the sum of the powers is equal to needed_sum, then we have a solution.
						solutions_count += 1
						return current_sum, solutions_count

			A_	:						Optional[int]	=     int(pow(lowerCamelCase__						,      lowerCamelCase__    )    )
			if current_sum + i_to_n <= needed_sum:
						# If the sum of the powers is less than needed_sum, then continue adding powers.
						current_sum += i_to_n
						A_,    A_	:						int	=     backtrack(
						    lowerCamelCase__						,      lowerCamelCase__						,      current_number + 1						,      lowerCamelCase__						,      lowerCamelCase__    )
						current_sum -= i_to_n
			if i_to_n < needed_sum:
						# If the power of i is less than needed_sum, then try with the next power.
						A_,    A_	:						int	=     backtrack(
						    lowerCamelCase__						,      lowerCamelCase__						,      current_number + 1						,      lowerCamelCase__						,      lowerCamelCase__    )
			return current_sum, solutions_count



def 	a			( lowerCamelCase__						,      lowerCamelCase__    ):
			'''simple docstring'''





			if not (1 <= needed_sum <= 10_00 and 2 <= power <= 10):
						raise ValueError(
						    """Invalid input\n"""
						    """needed_sum must be between 1 and 1000, power between 2 and 10."""    )

			return backtrack(lowerCamelCase__						,      lowerCamelCase__						,      1						,      0						,      0    )[1]  # Return the solutions_count


if __name__ == "__main__":
						import doctest

						doctest.testmod()