Question

我正试图在家里复制那些警戒密码当这个运行时，结果关键字比它应该移动少一个地方？任何帮助??

虽然是真的： Alphabet =＆＃39; ABCDEFGHIJKLMNOPQRSTUVWXYZ＆＃39; #these是可以输入的字母

def main():
    Mode = input("Press 1 for encryption or 2 for decyrption:")#this will allow the user to select whether to encode or decode


    if Mode == '1':#this will allow user to encrypt when 1 is pressed
        myMessage = input ("Enter a message to encrypt:")#this allows the user to enter the message they would like to be encrypted
        myKey = input("Enter a keyword to use:")#this allows the user to enter what keyword they would like to encypt by
        x = encryptM(myKey, myMessage)
    elif Mode == '2':#this will allow user to decrypt when 2 is pressed
        myMessage = input ("Enter a message to decrypt:")#this allows the user to enter the message they would like to be encrypted
        myKey = input("Enter a keyword to use:")#this allows the user to enter what keyword they would like to encypt by
        x = decryptM(myKey, myMessage)



    print("Your encrypted message is: " +x ) #this will print the encrypted or decrypted message to the user 

def encryptM(key, message):
    return translateM(key, message, 'encrypt')

def decryptM(key, message):
    return translateM(key, message, 'decrypt')

def translateM(key, message, mode):
    x = []

    keyIndex = 0
    key = key.upper()

    for symbol in message:
        num = Alphabet.find(symbol.upper())
        if num != -1:
            if mode == 'encrypt':
                num += Alphabet.find(key[keyIndex])#this will add the number from the keyword with the encryption number
            elif mode == 'decrypt':
                num -= Alphabet.find(key[keyIndex])#this will take away the number from the keyword from the decryption number

            num %= len(Alphabet)#this converts the alphabet string into numbers
            #this function will add the alphabet string to the end of the key so they can be added
            if symbol.isupper():
                x.append(Alphabet[num])
            elif symbol.islower():
                x.append(Alphabet[num].lower())

            keyIndex += 1
            if keyIndex == len(key):
                keyIndex = 0
    return "".join(x)

if __name__ == '__main__':
    main( )

print (' ')#this adds a line between the loops

Answer 1

问题是Alphabet.find返回数组中的索引（即你的Alphabet字符串），并且Python中字符的索引是从零开始的。

繁殖：

symbol = 'a'
Alphabet = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'     
num = Alphabet.find(symbol.upper())     
print num

http://ideone.com/cUD1uz

Answer 2

我建议您使用字典、If 循环和列表来包含信息。我会告诉你如何制作这个代码。

首先，我们必须创建一个移位字典函数，稍后你会发现原因：

def shift_dict(Caesar, Shift):
  dic_len = len(Caesar)
  Shift = Shift % dic_len
  list_dic = [(k,v) for k, v in iter(Caesar.items())]
  Shifted = {
    list_dic[x][0]: list_dic[(x - Shift) % dic_len][1]
    for x in range(dic_len)
  }
  return Shifted

好的，现在我们将创建字典，“Shift-ee's”，这些列表将有助于代码左右：将不断变化的转变：

Viginere = {
  "A":0,
  "B":1,
  "C":2,
  "D":3,
  "E":4,
  "F":5,
  "G":6,
  "H":7,
  "I":8,
  "J":9,
  "K":10,
  "L":11,
  "M":12,
  "N":13,
  "O":14,
  "P":15,
  "Q":16,
  "R":17,
  "S":18,
  "T":19,
  "U":20,
  "V":21,
  "W":22,
  "X":23,
  "Y":24,
  "Z":25
}

这是班次，班次的代码。这是 VFU，即 Viginere - For - Upper，它将返回大写的大写加密文本：

VFU = {
  0:"A",
  1:"B",
  2:"C",
  3:"D",
  4:"E",
  5:"F",
  6:"G",
  7:"H",
  8:"I",
  9:"J",
  10:"K",
  11:"L",
  12:"M",
  13:"N",
  14:"O",
  15:"P",
  16:"Q",
  17:"R",
  18:"S",
  19:"T",
  20:"U",
  21:"V",
  22:"W",
  23:"X",
  24:"Y",
  25:"Z"
}

接下来，VFL，Viginere - For - Lower：

VFL = {
  0:"a",
  1:"b",
  2:"c",
  3:"d",
  4:"e",
  5:"f",
  6:"g",
  7:"h",
  8:"i",
  9:"j",
  10:"k",
  11:"l",
  12:"m",
  13:"n",
  14:"o",
  15:"p",
  16:"q",
  17:"r",
  18:"s",
  19:"t",
  20:"u",
  21:"v",
  22:"w",
  23:"x",
  24:"y",
  25:"z"
}

现在，我们将创建主体：变量：

Asker = int(input("Do you want to... 1. Encode, or 2. Decode? "))
X = 0
Lister = []
Text = list(str(input("")))
Key = list(str(input("")).upper().replace(" ", "")
Z = 0

编码：

if Asker == 1:
  for i in range(len(Text)):
    Shift = Viginere[Key[(Z % len(Key))]]
    if Text[X].isalpha():
      LetterNum = Viginere[Text[X].upper()]
      Helper = (LetterNum + Shift) % 26
      if Text[X].isupper():
        Lister.append(VFU[Helper])
      else:
        Lister.append(VFL[Helper])
    else:
      Lister.append(Text[X])
      Z -= 1
    X += 1
    Z += 1
  print(*Lister, sep = "")

现在让我们分解这段代码，看看它是如何工作的： for 循环运行整个文本长度：

Shift = Viginere[Key[(Z % len(Key))]]

这提供了转变，您可以轻松地分解它。现在我们对字典进行移位，然后将其恢复为正常，如下代码所示：

    if Text[X].isalpha():
      LetterNum = Viginere[Text[X].upper()]
      Helper = (LetterNum + Shift) % 26
      if Text[X].isupper():
        Lister.append(VFU[Helper])
      else:
        Lister.append(VFL[Helper])
    else:
      Lister.append(Text[X])
      Z -= 1
    X += 1
    Z += 1
  print(*Lister, sep = "")

这是解码的代码：

elif Asker == 2:
  for i in range(len(Text)):
    Shift = Viginere[Key[(Z % len(Key))]]
    if Text[X].isalpha():
      LetterNum = Viginere[Text[X].upper()]
      Helper = (LetterNum - Shift) % 26
      if Text[X].isupper():
        Lister.append(VFU[Helper])
      else:
        Lister.append(VFL[Helper])
    else:
      Lister.append(Text[X])
      Z -= 1
    X += 1
    Z += 1
  print(*Lister, sep = "")

现在，当我们运行组合代码时：

def shift_dict(Caesar, Shift):
  dic_len = len(Caesar)
  Shift = Shift % dic_len
  list_dic = [(k,v) for k, v in iter(Caesar.items())]
  Shifted = {
    list_dic[x][0]: list_dic[(x - Shift) % dic_len][1]
    for x in range(dic_len)
  }
  return Shifted

Viginere = {
  "A":0,
  "B":1,
  "C":2,
  "D":3,
  "E":4,
  "F":5,
  "G":6,
  "H":7,
  "I":8,
  "J":9,
  "K":10,
  "L":11,
  "M":12,
  "N":13,
  "O":14,
  "P":15,
  "Q":16,
  "R":17,
  "S":18,
  "T":19,
  "U":20,
  "V":21,
  "W":22,
  "X":23,
  "Y":24,
  "Z":25
}
VFU = {
  0:"A",
  1:"B",
  2:"C",
  3:"D",
  4:"E",
  5:"F",
  6:"G",
  7:"H",
  8:"I",
  9:"J",
  10:"K",
  11:"L",
  12:"M",
  13:"N",
  14:"O",
  15:"P",
  16:"Q",
  17:"R",
  18:"S",
  19:"T",
  20:"U",
  21:"V",
  22:"W",
  23:"X",
  24:"Y",
  25:"Z"
}
VFL = {
  0:"a",
  1:"b",
  2:"c",
  3:"d",
  4:"e",
  5:"f",
  6:"g",
  7:"h",
  8:"i",
  9:"j",
  10:"k",
  11:"l",
  12:"m",
  13:"n",
  14:"o",
  15:"p",
  16:"q",
  17:"r",
  18:"s",
  19:"t",
  20:"u",
  21:"v",
  22:"w",
  23:"x",
  24:"y",
  25:"z"
}

Asker = int(input("Do you want to... 1. Encode, or 2. Decode? "))
X = 0
Lister = []
Text = list(str(input("")))
Key = list(str(input("")).upper().replace(" ", "")
Z = 0

if Asker == 1:
  for i in range(len(Text)):
    Shift = Viginere[Key[(Z % len(Key))]]
    if Text[X].isalpha():
      LetterNum = Viginere[Text[X].upper()]
      Helper = (LetterNum + Shift) % 26
      if Text[X].isupper():
        Lister.append(VFU[Helper])
      else:
        Lister.append(VFL[Helper])
    else:
      Lister.append(Text[X])
      Z -= 1
    X += 1
    Z += 1
  print(*Lister, sep = "")
elif Asker == 2:
  for i in range(len(Text)):
    Shift = Viginere[Key[(Z % len(Key))]]
    if Text[X].isalpha():
      LetterNum = Viginere[Text[X].upper()]
      Helper = (LetterNum - Shift) % 26
      if Text[X].isupper():
        Lister.append(VFU[Helper])
      else:
        Lister.append(VFL[Helper])
    else:
      Lister.append(Text[X])
      Z -= 1
    X += 1
    Z += 1
  print(*Lister, sep = "")

这变成了：

Do you want to... 1. Encode, or 2. Decode? 1
Please encode, as you always have worked!
Python is Awesome
Ejxhgr mfckhw, oe cds tskngk hwzw kavzcw!

还有：

Do you want to... 1. Encode, or 2. Decode? 2
Ejxhgr mfckhw, oe cds tskngk hwzw kavzcw!
Python is awesome
Please encode, as you always have worked!

感谢这个非常好的问题。我意识到这是在创建它的 5 年后，但我刚刚加入 StackOverflow，我在中学，我刚刚找到并解决了这个问题。

请告诉这是否有帮助，感谢您提出这个惊人的问题！

需要Viginere ciphere帮助

2 个答案: