我试图将在暴力破解阶段（3）中的解密限制在ASCII 32和ASCII 126之间。我在前两个阶段都成功了，但是在暴力破解期间实现它有点麻烦，所以我的结果回来准确。所需的输出是：
*** Menu ***
1. Encrypt string
2. Decrypt string
3. Brute force decryption
4. Quit
What would you like to do [1,2,3,4]? 3
Please enter string to decrypt: ykixkz&yw{oxxkr
Offset: 1 = Decrypted string: xjhwjy%xvznwwjq
Offset: 2 = Decrypted string: wigvix$wuymvvip
Offset: 3 = Decrypted string: vhfuhw#vtxluuho
Offset: 4 = Decrypted string: ugetgv"uswkttgn
Offset: 5 = Decrypted string: tfdsfu!trvjssfm
Offset: 6 = Decrypted string: secret squirrel
Offset: 7 = Decrypted string: rdbqds~rpthqqdk
Offset: 8 = Decrypted string: qcapcr}qosgppcj
Offset: 9 = Decrypted string: pb`obq|pnrfoobi
Offset: 10 = Decrypted string: oa_nap{omqennah
它需要产生“秘密松鼠”。
在暴力中，我不知道在哪里实现
for char in stringEncrypt:
x = ord(char)
x = x + offsetValue
while x < 32:
x += 95
while x > 126:
x -= 95
total += chr(x)
所以我还可以实现将ASCII 32解密为ASCII 126的输出。
任何帮助将不胜感激。
我尝试将其放入while循环中，并将其放置在代码中的不同位置。
print("*** Menu ***")
print(" ")
print("1. Encrypt string")
print("2. Decrypt string")
print("3. Brute force decryption")
print("4. Quit")
print(" ")
selection = int(input("What would you like to do [1,2,3,4]? "))
while selection == 1:
stringEncrypt = input("Please enter string to encrypt: ")
offsetValue = int(input("Please enter offset value (1 to 94): "))
total = ""
for char in stringEncrypt:
x = ord(char)
x = x + offsetValue
while x < 32:
x += 95
while x > 126:
x -= 95
total += chr(x)
print(" ")
print("Encrypted string:")
print(total)
print(" ")
print("*** Menu ***")
print(" ")
print("1. Encrypt string")
print("2. Decrypt string")
print("3. Brute force decryption")
print("4. Quit")
print(" ")
selection = int(input("What would you like to do [1,2,3,4]? "))
while selection == 2:
stringDecrypt = input("Please enter string to decrypt: ")
offsetValue = int(input("Please enter offset value (1 to 94): "))
total = ""
for char in stringDecrypt:
x = ord(char)
x = x - offsetValue
while x < 32:
x += 95
while x > 126:
x -= 95
total += chr(x)
print(" ")
print("Decrypted string:")
print(total)
print(" ")
print("*** Menu ***")
print(" ")
print("1. Encrypt string")
print("2. Decrypt string")
print("3. Brute force decryption")
print("4. Quit")
print(" ")
selection = int(input("What would you like to do [1,2,3,4]? "))
while selection == 3:
stringDecrypt = input("Please enter string to decrypt: ")
decryptList = list(stringDecrypt)
offsetValue = 0
decryptIndex = 0
for offsetValue in range(1, 95, 1):
for decryptIndex in range(len(decryptList)):
shifting = (ord(decryptList[decryptIndex]) - ord(" ") - offsetValue) % 95
chrDecrypt = chr(shifting + ord(" "))
decryptList[decryptIndex] = chrDecrypt
decryptIndex += 1
stringDecrypt = ''.join(decryptList)
print("Offset", offsetValue, " = Decrypted string:", stringDecrypt)
print(" ")
print("*** Menu ***")
print(" ")
print("1. Encrypt string")
print("2. Decrypt string")
print("3. Brute force decryption")
print("4. Quit")
print(" ")
selection = int(input("What would you like to do [1,2,3,4]?"))
if selection == 4:
print("Goodbye.")
我希望输出为ykixkz＆yw {oxxkr：
Offset: 1 = Decrypted string: xjhwjy%xvznwwjq
Offset: 2 = Decrypted string: wigvix$wuymvvip
Offset: 3 = Decrypted string: vhfuhw#vtxluuho
Offset: 4 = Decrypted string: ugetgv"uswkttgn
Offset: 5 = Decrypted string: tfdsfu!trvjssfm
Offset: 6 = Decrypted string: secret squirrel
Offset: 7 = Decrypted string: rdbqds~rpthqqdk
Offset: 8 = Decrypted string: qcapcr}qosgppcj
Offset: 9 = Decrypted string: pb`obq|pnrfoobi
Offset: 10 = Decrypted string: oa_nap{omqennah
但是我得到了：
Offset 1 = Decrypted string: xjhwjy%xvznwwjq
Offset 2 = Decrypted string: vhfuhw#vtxluuho
Offset 3 = Decrypted string: secret squirrel
Offset 4 = Decrypted string: oa_nap{omqennah
Offset 5 = Decrypted string: j\Zi\kvjhl`ii\c
Offset 6 = Decrypted string: dVTcVepdbfZccV]
Offset 7 = Decrypted string: ]OM\O^i][_S\\OV
Offset 8 = Decrypted string: UGETGVaUSWKTTGN
Offset 9 = Decrypted string: L><K>MXLJNBKK>E
Offset 10 = Decrypted string: B42A4CNB@D8AA4;
（最多94个）。
答案 0 :(得分：1)
请注意，您在解密函数（x）上的char解密与蛮力函数（chrDecrypt）之间的区别。稍后，您不确定字符是否正确循环。这是条件所在，基本上确保您在32到128的值上循环。
一种实现方式如下：
shifting = (ord(decryptList[decryptIndex]) - ord(" ") - offsetValue) % 95
chrDecrypt = chr(shifting + ord(" "))
这将是所需字符上的移位模块。
为了处理decryptList数组的覆盖，可以执行以下操作：
...
tempDecrypt = []
for decryptIndex in range(len(decryptList)):
shifting = (ord(decryptList[decryptIndex]) - ord(" ") - offsetValue) % 95
chrDecrypt = chr(shifting + ord(" "))
tempDecrypt.append(chrDecrypt)
decryptIndex += 1
stringDecrypt = ''.join(tempDecrypt)
...
这将解决您在上一个代码中注意到的顺序更改。
答案 1 :(得分：0)
尝试：
cipher_text = input("Enter the Cipher Text = ")
length = len(cipher_text)
plain_text = ""
i = 0
key = 1
while key < 26:
plain_text = ""
while i < length:
if ord(cipher_text[i]) - key < 32:
plain_text += chr(ord(cipher_text[i]) - key + 95)
else:
plain_text += chr(ord(cipher_text[i]) - key)
i += 1
i = 0
print("Decrypting cipher text with key ", key, "is", plain_text)
key += 1
输出：-
Enter the Cipher Text = ykixkz&yw{oxxkr
Decrypting cipher text with key 1 is xjhwjy%xvznwwjq
Decrypting cipher text with key 2 is wigvix$wuymvvip
Decrypting cipher text with key 3 is vhfuhw#vtxluuho
Decrypting cipher text with key 4 is ugetgv"uswkttgn
Decrypting cipher text with key 5 is tfdsfu!trvjssfm
Decrypting cipher text with key 6 is secret squirrel
Decrypting cipher text with key 7 is rdbqds~rpthqqdk
Decrypting cipher text with key 8 is qcapcr}qosgppcj
Decrypting cipher text with key 9 is pb`obq|pnrfoobi
Decrypting cipher text with key 10 is oa_nap{omqennah
Decrypting cipher text with key 11 is n`^m`oznlpdmm`g
Decrypting cipher text with key 12 is m_]l_nymkocll_f
Decrypting cipher text with key 13 is l^\k^mxljnbkk^e
Decrypting cipher text with key 14 is k][j]lwkimajj]d
Decrypting cipher text with key 15 is j\Zi\kvjhl`ii\c
Decrypting cipher text with key 16 is i[Yh[juigk_hh[b
Decrypting cipher text with key 17 is hZXgZithfj^ggZa
Decrypting cipher text with key 18 is gYWfYhsgei]ffY`
Decrypting cipher text with key 19 is fXVeXgrfdh\eeX_
Decrypting cipher text with key 20 is eWUdWfqecg[ddW^
Decrypting cipher text with key 21 is dVTcVepdbfZccV]
Decrypting cipher text with key 22 is cUSbUdocaeYbbU\
Decrypting cipher text with key 23 is bTRaTcnb`dXaaT[
Decrypting cipher text with key 24 is aSQ`Sbma_cW``SZ
Decrypting cipher text with key 25 is `RP_Ral`^bV__RY
PS 。您提供的代码不是凯撒密码，而是修改后的凯撒密码。两者之间的区别在于，Caesar密码使用恒定密钥（密钥= 3），而修改后的Caesar密码可以使用可变密钥（0 <密钥<26）。