Question

我有一个代码而我无法将其设为交互式。这是问题所在：

“”“编写一个名为rot13的函数，它使用Caesar密码来加密消息。凯撒密码的工作方式类似于替换密码，但每个字符在字母表中由字符13个字符替换为”其右边“。所以对于例如，字母“a”变为字母“n”。如果字母超过字母表的中间，则计数再次包围字母“a”，因此“n”变为“a”，“o”变为“ b“依此类推。提示：每当你谈到围绕它的好主意思考模运算时（使用余数运算符）。”“

以下是此问题的代码：

def rot13(mess, char):
    alphabet = 'abcdefghijklmnopqrstuvwxyz'
    encrypted = ''
    for char in mess:
        if char == ' ':
            encrypted = encrypted + ' '
        else:
            rotated_index = alphabet.index(char) + mess
            if rotated_index < 26:
                encrypted = encrypted + alphabet[rotated_index]
            else:
                encrypted = encrypted + alphabet[rotated_index % 26]
    return encrypted

def main():
    messy_shit = input("Rotate by: ")
    the_message = input("Type a message")
    print(rot13(the_message, messy_shit))


if __name__ == "__main__":
    main()

我希望将其设置为交互式，您可以输入任何消息，并且可以根据需要多次旋转字母。这是我的尝试。我知道你需要传递两个参数，但我对如何替换一些项目毫无头绪。

这是我的尝试：

while()

我不知道我的输入应该在函数中发生什么。我有一种感觉可以加密吗？

Answer 1

这可能就是你要找的东西。它通过messy_shit输入旋转消息。

def rot13(mess, rotate_by):
        alphabet = 'abcdefghijklmnopqrstuvwxyz'
        encrypted = ''
        for char in mess:
            if char == ' ':
                encrypted = encrypted + ' '
            else:
                rotated_index = alphabet.index(char) + int(rotate_by)
                if rotated_index < 26:
                    encrypted = encrypted + alphabet[rotated_index]
                else:
                    encrypted = encrypted + alphabet[rotated_index % 26]
        return encrypted

    def main():
        messy_shit = input("Rotate by: ")
        the_message = input("Type a message")
        print(rot13(the_message, messy_shit))

Answer 2

def rot(message, rotate_by):
    '''
    Creates a string as the result of rotating the given string 
    by the given number of characters to rotate by

    Args:
        message: a string to be rotated by rotate_by characters
        rotate_by: a non-negative integer that represents the number
            of characters to rotate message by

    Returns:
        A string representing the given string (message) rotated by
        (rotate_by) characters. For example:

        rot('hello', 13) returns 'uryyb'
    '''
    assert isinstance(rotate_by, int) == True
    assert (rotate_by >= 0) == True

    alphabet = 'abcdefghijklmnopqrstuvwxyz'
    rotated_message = []
    for char in message:
        if char == ' ':
            rotated_message.append(char)
        else:
            rotated_index = alphabet.index(char) + rotate_by
            if rotated_index < 26:
                rotated_message.append(alphabet[rotated_index])
            else:
                rotated_message.append(alphabet[rotated_index % 26])

    return ''.join(rotated_message)

if __name__ == '__main__':
    while True:
        # Get user input necessary for executing the rot function
        message = input("Enter message here: ")
        rotate_by = input("Enter your rotation number: ")

        # Ensure the user's input is valid
        if not rotate_by.isdigit():
            print("Invalid! Expected a non-negative integer to rotate by")
            continue

        rotated_message = rot(message, int(rotate_by))
        print("rot-ified message:", rotated_message)

        # Allow the user to decide if they want to continue
        run_again = input("Continue? [y/n]: ").lower()
        while run_again != 'y' and run_again != 'n':
            print("Invalid! Expected 'y' or 'n'")
            run_again = input("Continue? [y/n]: ")

        if run_again == 'n':
            break

注意：创建一个字符列表然后加入它们以生成字符串而不是使用string = string + char会更有效。请参阅方法6 here和Python文档here。另外，请注意我们的rot函数仅适用于字母表的小写字母。如果您尝试使用大写字符或alphabet中不包含的任何字符来破坏邮件，则会中断。

交互式rot13密码

2 个答案: