Question

我正在通过John Zelle撰写的计算机科学入门书来学习编码。我坚持锻炼5.8。我需要以某种方式修改这个解决方案，其中下一个字符是＆＃34; z＆＃34;是＆＃34; a＆＃34;为了使它循环。任何帮助都会很棒：）

def main():
    print("This program can encode and decode Caesar Ciphers") #e.g. if key value is 2 --> word shifted 2 up e.g. a would be c
    #input from user
    inputText = input("Please enter a string of plaintext:").lower()
    inputValue = eval(input("Please enter the value of the key:"))
    inputEorD = input("Please enter e (to encrypt) or d (to decrypt) ")
    #initate empty list
    codedMessage = ""

    #for character in the string
    if inputEorD == "e":
        for ch in inputText:
            codedMessage += chr(ord(ch) + inputValue) #encode hence plus
    elif inputEorD =="d":
            codedMessage += chr(ord(ch) - inputValue) #decode hence minus
    else:
        print("You did not enter E/D! Try again!!")
    print("The text inputed:", inputText,  ".Is:", inputEorD, ".By the key of",inputValue, ".To make the message", codedMessage)

main()

Answer 1

由于您正在处理.lower() - 大小写字母，因此可以确定其ASCII范围是[97-122]。

制作移动圆形的一个好方法是用[{0}]代表每个字母，由ord(ch) - 97完成，然后添加密钥，然后用26对结果进行模数设置，这样就可以了变为(ord(ch) - 97 + key)%26，然后我们在范围[0-25]中得到一个结果，然后添加97将得到它的ASCII码：

def main():
    print("This program can encode and decode Caesar Ciphers")
    inputText = input("Please enter a string of plaintext:").lower()
    inputValue = int(input("Please enter the value of the key:")) # use int(), don't eval unless you read more about it
    inputEorD = input("Please enter e (to encrypt) or d (to decrypt) ")
    codedMessage = ""

    if inputEorD == "e":
        for ch in inputText:
            codedMessage += chr((ord(ch) - 97 + inputValue)%26 + 97)
    elif inputEorD =="d":
            codedMessage += chr((ord(ch) - 97 - inputValue)%26 + 97)
    else:
        print("You did not enter E/D! Try again!!")
    print("The text inputed:", inputText,  ".Is:", inputEorD, ".By the key of",inputValue, ".To make the message", codedMessage)

main()

编码/解码Caesar密码Python 3 ASCII

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