BUU[ACTF新生赛2020]easyre

查壳后发现有个upx壳，那我们先脱壳去

之后他还会提示你用32位的ida打开，打开之后在main主函数里面找到了关键函数

 1  __main();
 2   qmemcpy(v4, "*F'\"N,\"(I?+@", sizeof(v4));
 3   printf("Please input:");
 4   scanf("%s", v6);
 5   if ( v6[0] != 65 || v6[1] != 67 || v6[2] != 84 || v6[3] != 70 || v6[4] != 123 || v10 != 125 )
 6     return 0;
 7   v5[0] = v7;
 8   v5[1] = v8;
 9   v5[2] = v9;
10   for ( i = 0; i <= 11; ++i )
11   {
12     if ( v4[i] != _data_start__[*((char *)v5 + i) - 1] )
13       return 0;
14   }
15   printf("You are correct!");
16   return 0;
17 }

意思判断v4v5v6....v15这么多的东西是否都等于上面这串东西相对应减1的数。

如果正确的的话关键语句就可以理解成这种  v4[i]!=byte_402000[flag[i]-1]

 1 byte_402000 = '~}|{zyxwvutsrqponmlkjihgfedcba`_^]\[ZYXWVUTSRQPONMLKJIHGFEDCBA@?>=<;:9876543210/.-,+*)(\'&%$# !"'
 2 
 3 v4 = [42,70,39,34,78,44,34,40,73,63,43,64]
 4 
 5 flag = ''
 6 
 7 for i in v4:
 8     flag += chr(byte_402000.find(chr(i)) + 1)
 9 
10 print(flag)

代码就单的就这样就行，

套个flag就是答案

标签：402000,easyre,v4,v5,v6,2020,byte,main,ACTF
来源： https://www.cnblogs.com/lixin666/p/16103137.html