Question

我最近在与Project Euler Problem 59一起玩之后开始使用加密技术，并且我制作了一个基本的XOR加密系统。在这种情况下，您的输出和随机密钥的连接将被创建并保存到文本文件中（仅作为测试，我会在我修复此错误后使其更好），但我已经注意到了某些邮件不会正确加密或解密。我已将其缩小到长度为＆gt;的消息。 54.我的代码如下：

#encrypt.py

import random

msg = raw_input("Enter your message: ")
out = ""
key = ""

for i in xrange(len(msg)):
    key += chr(random.randint(0,255))

k = 0

for char in msg:
    out += chr(ord(msg[k]) ^ ord(key[k]))
    k += 1

print "\nYour output is:", out
print "Your key is:", key

raw_input()

with open("output.txt","r+") as f:
    f.truncate()
    f.write(out+key)

解密：

#decrypt.py

import sys

if not len(sys.argv) > 1: exit()

with open(sys.argv[1],"r+") as f:
    content = f.read()

msg, key = content[:len(content)/2], content[len(content)/2:]

out = ""

k = 0

for char in msg:
    out += chr(ord(char) ^ ord(key[k]))
    k += 1

print out
raw_input()

对于较长的消息（超过54个字符），当它们被解密时，它们将给出一串随机字符。有没有人知道为什么会这样？

Answer 1

由于使用随机位制作字符可能会在屏幕上产生看起来很难看的字符，因此您可以在打印前尝试对输出进行十六进制编码。

#encrypt.py

import random

msg = raw_input("Enter your message: ")
out = ""
key = ""

for i in xrange(len(msg)):
    key += chr(random.randint(0,255))

k = 0

for char in msg:
    out += chr(ord(msg[k]) ^ ord(key[k]))
    k += 1
out = ":".join("{:02x}".format(ord(c)) for c in out)
key = ":".join("{:02x}".format(ord(c)) for c in key)
print "\nYour output is:", out
print "Your key is:", key

raw_input()

with open("output.txt","w+") as f:
    f.truncate()
    f.write(out+key)

然后，对于解密，您必须先删除冒号和十六进制解码才能继续进行常规解密。

#decrypt.py

import sys

if not len(sys.argv) > 1: exit()

with open(sys.argv[1],"r+") as f:
    content = f.read()

content = content.replace(':', '')
msg, key = content[:len(content)/2], content[len(content)/2:]
msg = msg.decode('hex')
key = key.decode('hex')
out = ""

k = 0

for char in msg:
    out += chr(ord(char) ^ ord(key[k]))
    k += 1

print out
raw_input()

在控制台上，结果如下所示：

$ python encrypt.py           
Enter your message: onetwothreefourfivesixseveneightnineteneleventwelvethirteen
Your output is: 7f:09:d0:8c:2e:1a:e0:a0:84:e7:bf:3b:e8:cc:f3:4e:35:e4:4f:20:c8:00:72:9d:f0:bc:de:54:88:30:a6:3d:93:3f:c1:6d:46:4a:68:ca:96:2e:16:50:43:0f:30:fa:27:f3:f2:8d:35:4b:6a:11:cc:02:04
Your key is: 10:67:b5:f8:59:75:94:c8:f6:82:da:5d:87:b9:81:28:5c:92:2a:53:a1:78:01:f8:86:d9:b0:31:e1:57:ce:49:fd:56:af:08:32:2f:06:af:fa:4b:60:35:2d:7b:47:9f:4b:85:97:f9:5d:22:18:65:a9:67:6a
$ cat output.txt 
7f:09:d0:8c:2e:1a:e0:a0:84:e7:bf:3b:e8:cc:f3:4e:35:e4:4f:20:c8:00:72:9d:f0:bc:de:54:88:30:a6:3d:93:3f:c1:6d:46:4a:68:ca:96:2e:16:50:43:0f:30:fa:27:f3:f2:8d:35:4b:6a:11:cc:02:0410:67:b5:f8:59:75:94:c8:f6:82:da:5d:87:b9:81:28:5c:92:2a:53:a1:78:01:f8:86:d9:b0:31:e1:57:ce:49:fd:56:af:08:32:2f:06:af:fa:4b:60:35:2d:7b:47:9f:4b:85:97:f9:5d:22:18:65:a9:67:6a
$ python decrypt.py output.txt
onetwothreefourfivesixseveneightnineteneleventwelvethirteen

XOR加密算法不适用于＆gt; 54个字符

1 个答案: