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：
#!/usr/bin/env python
# encoding: utf-8
# 如果觉得不错，可以推荐给你的朋友！http://tool.lu/pyc def encrypt(key, seed, string):
rst = []
for v in string:
rst.append((ord(v) + seed ^ ord(key[seed])) % 255)
seed = (seed + 1) % len(key) return rst if __name__ == '__main__':
print "Welcome to idf's python crackme"
flag = input('Enter the Flag: ')
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]
en_out = encrypt(KEY1, 5, flag)
if KEY2 == en_out:
print 'You Win'
else:
print 'Try Again !'
程序加密函数：
def encrypt(key, seed, string):
rst = []
for v in string:
rst.append((ord(v) + seed ^ ord(key[seed])) % 255)
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)
要注意的是seed 的改变要在flag与KEY2比较后。