IDF实验室-python ByteCode writeup

最新推荐文章于 2025-09-07 01:02:55 发布

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原文链接：http://www.cnblogs.com/zhengjim/p/6105071.html

文章标签：

#python #php

本文介绍了一个Python编写的Crackme挑战，通过反编译获取到加密算法，并使用暴力破解方法找到了正确的Flag输入，最终实现了程序的成功破解。

题目地址：http://ctf.idf.cn/index.php?g=game&m=article&a=index&id=45

下载来发现是crackme.pyc

可以用uncompyle2反编译。也可以直接http://tool.lu/pyc/在这个网站反编译。

得到源代码：

 1 #!/usr/bin/env python
 2 # encoding: utf-8
 3 # 如果觉得不错，可以推荐给你的朋友！http://tool.lu/pyc
 4 
 5 def encrypt(key, seed, string):
 6     rst = []
 7     for v in string:
 8         rst.append((ord(v) + seed ^ ord(key[seed])) % 255)
 9         seed = (seed + 1) % len(key)
10     
11     return rst
12 
13 if __name__ == '__main__':
14     print "Welcome to idf's python crackme"
15     flag = input('Enter the Flag: ')
16     KEY1 = 'Maybe you are good at decryptint Byte Code, have a try!'
17     KEY2 = [
18         124,
19         48,
20         52,
21         59,
22         164,
23         50,
24         37,
25         62,
26         67,
27         52,
28         48,
29         6,
30         1,
31         122,
32         3,
33         22,
34         72,
35         1,
36         1,
37         14,
38         46,
39         27,
40         232]
41     en_out = encrypt(KEY1, 5, flag)
42     if KEY2 == en_out:
43         print 'You Win'
44     else:
45         print 'Try Again !'

程序加密函数：

1 def encrypt(key, seed, string):
2     rst = []
3     for v in string:
4         rst.append((ord(v) + seed ^ ord(key[seed])) % 255)
5         seed = (seed + 1) % len(key)

flag加密后与KEY2比较一样的话输出You Win

本来想逆向，但弄不来，就直接爆破了。

a-z A-Z 0-9 加上符号可以有AscII码遍历，然后编码转换回来，加入数组。

然后加密，与KEY数组的值比较。

代码如下：

#!/usr/bin/env python
# encoding: utf-8

def encrypt(key, seed, string):
    for v in string:
        a = (ord(v) + seed ^ ord(key[seed]) % 255)
        return a

KEY1 = 'Maybe you are good at decryptint Byte Code, have a try!'
KEY2 = [
    124,
    48,
    52,
    59,
    164,
    50,
    37,
    62,
    67,
    52,
    48,
    6,
    1,
    122,
    3,
    22,
    72,
    1,
    1,
    14,
    46,
    27,
    232]
s=[]
seed=5;
key= 'Maybe you are good at decryptint Byte Code, have a try!'
for i in range(33,127):
    j = chr(i)
    s.append(j)
for i in range(23):
    for j in s:
        aa = encrypt(key,seed,j)
        if aa == KEY2[i]:
            print j
    seed = (seed + 1) % len(key)