BUUCTF--findKey

题目文件：https://lanzous.com/icwp75g

 

思路

查找字符串，将flag{}处的汇编代码转换为函数（选中区域，按P键）

主要就两个过程

第一个过程，使用0kk`d1a`55k222k2a776jbfgd`06cjjb和SS异或，得到c8837b23ff8aaa8a2dde915473ce0991，md5解密得到String1=123321

接着使用String1与v13同样，异或，得到flag

 

脚本

# -*- coding:utf-8 -*-

a = "0kk`d1a`55k222k2a776jbfgd`06cjjb"
b = "SS"
s = ''

for i in range(len(a)):
    s += chr(ord(a[i]) ^ ord(b[i % 2]))
print (s)

c = [0x57, 0x5E, 0x52, 0x54, 0x49, 0x5F, 0x01, 0x6D, 0x69, 0x46, 0x02, 0x6E, 0x5F, 0x02, 0x6C, 0x57, 0x5B, 0x54, 0x4C,
     0x00, 0x00, 0x00, 0x00, 0x00, 0x53, 0x53, 0x00, 0x00, 0x30, 0x6B, 0x6B, 0x60, 0x64, 0x31, 0x61, 0x60, 0x35, 0x35,
     0x6B, 0x32, 0x32, 0x32, 0x6B, 0x32, 0x61, 0x37, 0x37, 0x36, 0x6A, 0x62]
d = '123321'
flag = ''
for i in range((len(c))):
    flag += chr(c[i]^ord(d[i % len(d)]))
print (flag)

 

get flag！

flag{n0_Zu0_n0_die}