Python 必须输入字母A和B的代码，然后将其全部设为A'；通过翻动它们来改变它不'；我不适合所有的选择

我编写了一个代码，从用户那里输入一行a或B，他们还输入翻页器可以翻动的一行中的多少个煎饼，并输出使a或B的一行全部为a所需的翻动次数

如果字母行不能均匀翻转，代码应输出“这无法完成”

示例：如果用户输入的BBBB的翻页大小为2，则程序将输出翻页2次

翻页1来自字母1-2，它是AABB，翻页2来自字母3-4，所以现在需要两个翻页的地方是AAAA

我已经为此编写了代码，在大多数情况下它都能工作，但有一个问题是，当我输入BBAABB时，它说它不能用4号翻页纸来完成，而实际上可以用字母1-4来完成第一个翻页，所以现在是aabbb，第二个翻页是用字母3-6来完成的，所以现在是aaaa，它成功了并进行了2次翻转，我不确定如何用我的代码修复此问题

while True:
    pancakes = input('Enter the row of the pancakes (A/B): ') 
    flipper = int(input('How many pancakes can be flipped at one time? ')) 

    flips, possible = 0, True 
    for row in pancakes.split('A'):
        count, remainder = divmod(len(row), flipper) 
        if remainder != 0:
            possible = False 
            break 
        flips += count 

    if possible: 
      if flips == 1:
        print('It took 1 flip.') #print out how many flips it took
        play = input('Would you like to run this program again? ') 
      else:
        print('It took', flips, 'flips.')
        play = input('Would you like to run this program again? ') 
    else: #if possible is false
      print("IMPOSSIBLE.") #printing that the burgers can't be flipped with the flipper the size inputted in 'flipper'
      play = input("Would you like to run this program again? ") 
    if play not in ['Yes', 'yes', 'Y', 'y']: 
      exit()

非常感谢您的帮助，因为我对编程相当陌生

---

James

你的代码的一个问题是，你必须得到翻转逻辑来将A翻转到B，就像BBAABB一样，在BBAABB中可以用2来完成，因为你只需要将B转换为A，所以它将是AAAAA，但是当你输入4时，你必须将A翻转到B，而B目前没有包含在你的程序中。

你的代码有一个问题代码是，你必须获得翻转逻辑，也可以像BBAABB一样将A翻转到B，在BBAABB中可以使用2，因为你只需要将B转换为A，所以它将是AAAAA，但当你输入4时，你必须将A翻转到B，而B目前不包括在你的程序中。

关键是要寻找一个必须翻转的“B”，然后看看在第一个“B”之后是否有足够多的煎饼可以按照翻页大小翻动。如果可以，可以进行翻转并增加翻转次数。你只需不断重复这一点，直到你不能从a“B”开始做一个完整的翻转，在这一点上你不能得到所有的a。但如果只要你能找到一个“B”，你就可以翻转“flipper”煎饼，你最终就能得到所有的“a”：

def flip(str, pos, count):
    r = str[:pos]
    for i in range(pos, pos + count):
        r += 'A' if str[i] == 'B' else 'B'
    r += str[pos + count:]
    return r

while True:
    pancakes = input('Enter the row of the pancakes (A/B): ')
    flipper = int(input('How many pancakes can be flipped at one time? '))

    flips, possible = 0, True
    while True:
        try:
            i = pancakes.index('B')
        except ValueError:
            # no 'B's, so we did it!
            break
        if (i > len(pancakes) - flipper):
            # not enough pancakes left to flip starting with a 'B', so we can't do it.
            possible = False
            break
        else:
            # Can do a flip, starting with a 'B', so do it
            pancakes = flip(pancakes, i, flipper)
            flips += 1

    if possible:
        if flips == 1:
            print('It took 1 flip.')  # print out how many flips it took
        else:
            print('It took', flips, 'flips.')
    else:  # if possible is false
        print("IMPOSSIBLE.")  # printing that the burgers can't be flipped with the flipper the size inputted in 'flipper'
    play = input("Would you like to run this program again? ")
    if play not in ['Yes', 'yes', 'Y', 'y']:
        exit()

结果:

Enter the row of the pancakes (A/B): BBAABB
How many pancakes can be flipped at one time? 4
It took 2 flips.
Would you like to run this program again? y
Enter the row of the pancakes (A/B): ABBAABBAAA
How many pancakes can be flipped at one time? 4
It took 2 flips.
Would you like to run this program again? y
Enter the row of the pancakes (A/B): AAAA
How many pancakes can be flipped at one time? 2
It took 0 flips.
Would you like to run this program again? y
Enter the row of the pancakes (A/B): BBBB
How many pancakes can be flipped at one time? 2
It took 2 flips.
Would you like to run this program again? y
Enter the row of the pancakes (A/B): BBBB
How many pancakes can be flipped at one time? 4
It took 1 flip.
Would you like to run this program again? y
Enter the row of the pancakes (A/B): ABBB
How many pancakes can be flipped at one time? 2
IMPOSSIBLE.
Would you like to run this program again? y
Enter the row of the pancakes (A/B): ABBB
How many pancakes can be flipped at one time? 4
IMPOSSIBLE.
Would you like to run this program again? n

Process finished with exit code 0

---

关键是要寻找一个必须翻转的“B”，然后看看在第一个“B”之后是否有足够的煎饼按照翻转器的大小翻转。如果可以，可以进行翻转并增加翻转次数。你只需不断重复这一点，直到你不能从a“B”开始做一个完整的翻转，在这一点上你不能得到所有的a。但如果只要你能找到一个“B”，你就可以翻转“flipper”煎饼，你最终就能得到所有的“a”：

def flip(str, pos, count):
    r = str[:pos]
    for i in range(pos, pos + count):
        r += 'A' if str[i] == 'B' else 'B'
    r += str[pos + count:]
    return r

while True:
    pancakes = input('Enter the row of the pancakes (A/B): ')
    flipper = int(input('How many pancakes can be flipped at one time? '))

    flips, possible = 0, True
    while True:
        try:
            i = pancakes.index('B')
        except ValueError:
            # no 'B's, so we did it!
            break
        if (i > len(pancakes) - flipper):
            # not enough pancakes left to flip starting with a 'B', so we can't do it.
            possible = False
            break
        else:
            # Can do a flip, starting with a 'B', so do it
            pancakes = flip(pancakes, i, flipper)
            flips += 1

    if possible:
        if flips == 1:
            print('It took 1 flip.')  # print out how many flips it took
        else:
            print('It took', flips, 'flips.')
    else:  # if possible is false
        print("IMPOSSIBLE.")  # printing that the burgers can't be flipped with the flipper the size inputted in 'flipper'
    play = input("Would you like to run this program again? ")
    if play not in ['Yes', 'yes', 'Y', 'y']:
        exit()

结果:

Enter the row of the pancakes (A/B): BBAABB
How many pancakes can be flipped at one time? 4
It took 2 flips.
Would you like to run this program again? y
Enter the row of the pancakes (A/B): ABBAABBAAA
How many pancakes can be flipped at one time? 4
It took 2 flips.
Would you like to run this program again? y
Enter the row of the pancakes (A/B): AAAA
How many pancakes can be flipped at one time? 2
It took 0 flips.
Would you like to run this program again? y
Enter the row of the pancakes (A/B): BBBB
How many pancakes can be flipped at one time? 2
It took 2 flips.
Would you like to run this program again? y
Enter the row of the pancakes (A/B): BBBB
How many pancakes can be flipped at one time? 4
It took 1 flip.
Would you like to run this program again? y
Enter the row of the pancakes (A/B): ABBB
How many pancakes can be flipped at one time? 2
IMPOSSIBLE.
Would you like to run this program again? y
Enter the row of the pancakes (A/B): ABBB
How many pancakes can be flipped at one time? 4
IMPOSSIBLE.
Would you like to run this program again? n

Process finished with exit code 0

---

如果余数！=0:看起来不正确。抱歉，本来不应该在那里的！这是一个错误。如果它是4，那么JamesIt就不起作用了，因为它实际上是带BBAABB的can，第一个翻转是AAABB，第二个翻转是AAAAA。我已经检查过了，它可以使用2而不是4。谢谢。我的错，锯2而不是4，因为你的不可能条件太基本，所以它不起作用，它只考虑把B变成A，谢谢你的建议。我是编程新手，我真的不知道如何用我的代码将A翻转到B，如果我能做到这一点，我认为这将解决问题。詹姆士如果余数！=0:看起来不正确。抱歉，本来不应该在那里的！这是一个错误。如果它是4，那么JamesIt就不起作用了，因为它实际上是带BBAABB的can，第一个翻转是AAABB，第二个翻转是AAAAA。我已经检查过了，它可以使用2而不是4。谢谢。我的错，锯2而不是4，因为你的不可能条件太基本，所以它不起作用，它只考虑把B变成A，谢谢你的建议。我是编程新手，我真的不知道如何用我的代码将A翻转到B，如果我能做到这一点，我认为这将解决问题。詹姆斯