Jarvis easycrackme

记一次糟心的经历,

printf((unsigned __int64)"Input your password:");  scanf((unsigned __int64)&unk_4A572B);  if ( strlen((const char *)&v11) == 26 ) //flag长度是26  {    v3 = 0LL;    if ( (v11 ^ 0xAB) == list1 )    {      while ( (v12[v3] ^ (unsigned __int8)*(&v5 + ((signed int)v3 + 1) % 6)) == byte_6B41D1[v3] )      {        if ( ++v3 == 25 )        {          printf((unsigned __int64)"Congratulations!");          return 0;        }      }    }  }  printf((unsigned __int64)"Password Wrong!! Please try again.");

简单理一下思路:

和以往的思路几乎一样,就是通过比较的字符串推flag, 该程序先比较第一个字符是不是’P’, 是继续比较, 不是就退出. 继续, 输入的字符串从第二位开始与一串固定6位长度的字符串进行异或运算.然后将异或的结果与另一固定字符串比较, 相等就输出Congratulations!, 否则输出Password Wrong!! Please try again.

解密脚本:

import binasciil1=(0x9e,0x67,0x12,0x4e,0x9d,0x98,0xab,0x00,0x06,0x46,0x8a,0xf4,0xb4,0x06,0x0b,0x43,0xdc,0xd9,0xa4,0x6c,0x31,0x74,0x9c,0xd2,0xa0)l2=(0xab,0xdd,0x33,0x54,0x35,0xef)j=0s=str()for char in l1:    s = s + chr(char^l2[(j+1)%6])    j = j + 1print s

别忘了加上第一个字符p
flag:PCTF{r3v3Rse_i5_v3ry_eAsy}
糟心点:忘记了异或的固定字符串是六位,结果出现不可见字符,必定出错了.这是在分析程序时犯的错误, 长记性了.