交互式 rot13 密码

Question

我有一个代码，但我在使其具有交互性时遇到问题。 问题是这样的：

""编写一个名为 rot13 的函数，该函数使用 Caesar 密码来加密消息。凯撒密码的工作方式类似于替换密码，但每个字符都被字母表中"其右侧"的 13 个字符替换。例如，字母"a"变成了字母"n"。如果一个字母超过字母表的中间，那么计数将再次绕到字母"a"，所以"n"变成"a"，"o"变成"b"，依此类推。提示：每当你谈论围绕的事情时，最好考虑模算术(使用余数运算符(。

这是此问题的代码：

def rot13(mess):
alphabet = 'abcdefghijklmnopqrstuvwxyz'
encrypted = ''
for char in mess:
if char == ' ':
encrypted = encrypted + ' '
else:
rotated_index = alphabet.index(char) + 13
if rotated_index < 26:
encrypted = encrypted + alphabet[rotated_index]
else:
encrypted = encrypted + alphabet[rotated_index % 26]
return encrypted
def main():
print(rot13('abcde'))
print(rot13('nopqr'))
print(rot13(rot13('since rot thirteen is symmetric you should see this message')))
if __name__ == "__main__":
main()

我想让它具有交互性，您可以在其中输入任何消息，并且可以根据需要多次旋转字母。这是我的尝试。我知道您需要两个参数才能传递，但我对如何替换一些项目一无所知。

这是我的尝试：

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()

我不知道我的输入应该在函数中的哪个位置进行。我有一种感觉它可以加密？

Answer 1

这可能是你要找的。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：创建字符列表然后连接它们以生成字符串比使用string = string + char更有效。请参阅此处的方法 6 和此处的 Python 文档。另外，请注意，我们的 rot 函数仅适用于字母表中的小写字母。如果您尝试腐烂带有大写字符或任何不在alphabet中的字符的消息，它会中断。