2020年9月22号-----27号做题集

[GXYCTF2019]luck_guy

这道题是一个64位elf文件，分析起来不是很难，只是我没学switch函数，导致磨蹭了些时间
放入64位ida分析伪代码简单明了，输入幸运数字储存进v4执行patch_me函数

如果输入的幸运数字为偶数，则进入get_flag函数

伪代码如下：

unsigned __int64 get_flag()
{
    
    
  unsigned int v0; // eax
  char v1; // al
  signed int i; // [rsp+4h] [rbp-3Ch]
  signed int j; // [rsp+8h] [rbp-38h]
  __int64 s; // [rsp+10h] [rbp-30h]
  char v6; // [rsp+18h] [rbp-28h]
  unsigned __int64 v7; // [rsp+38h] [rbp-8h]

  v7 = __readfsqword(0x28u);
  v0 = time(0LL);
  srand(v0);
  for ( i = 0; i <= 4; ++i )
  {
    
    
    switch ( rand() % 200 )
    {
    
    
      case 1:
        puts("OK, it's flag:");
        memset(&s, 0, 0x28uLL);
        strcat((char *)&s, f1);
        strcat((char *)&s, &f2);
        printf("%s", &s);
        break;
      case 2:
        printf("Solar not like you");
        break;
      case 3:
        printf("Solar want a girlfriend");
        break;
      case 4:
        v6 = 0;
        s = 9180147350284624745LL;             //转换成16进制为0x69,0x63,0x75,0x67,0x60,0x6f,0x66,0x7f
        strcat(&f2, (const char *)&s);
        break;
      case 5:
        for ( j = 0; j <= 7; ++j )
        {
    
    
          if ( j % 2 == 1 )
            v1 = *(&f2 + j) - 2;
          else
            v1 = *(&f2 + j) - 1;
          *(&f2 + j) = v1;
        }
        break;
      default:
        puts("emmm,you can't find flag 23333");
        break;
    }
  }
  return __readfsqword(0x28u) ^ v7;
}

经过查阅知道switch函数使用需要排序，大概分析后顺序为case4，case5，case1
f1中的值为：GXY{do_not_

写个python3脚本跑出flag：GXY{do_not_hate_me}

flag='GXY{do_not_'
f2=[0x69,0x63,0x75,0x67,0x60,0x6f,0x66,0x7f,0]
v1=[]
for i in range(8):
    if i%2==1:
        v1.append(f2[i]+i-2)
    else:
        v1.append(f2[i]+i-1)
    f2[i]=v1[i]-i
    flag+=chr(f2[i])
print(flag)