简单的re（零基础AI做题）

 签到题一般简单，上来就是IDA（不管了，IDA！！！）

找主函数，这个题类似的做过好几个了

int __fastcall main(int argc, const char **argv, const char **envp)
{
  __int64 v3; // rdx
  __int64 v4; // rcx
  int v6; // [rsp+4h] [rbp-8Ch]
  unsigned int v7; // [rsp+8h] [rbp-88h]
  int v8; // [rsp+Ch] [rbp-84h]
  _DWORD v9[31]; // [rsp+10h] [rbp-80h] BYREF
  _BYTE v10[4]; // [rsp+8Ch] [rbp-4h] BYREF

  v9[0] = 102;
  v9[1] = 109;
  v9[2] = 99;
  v9[3] = 98;
  v9[4] = 127;
  v9[5] = 58;
  v9[6] = 85;
  v9[7] = 106;
  v9[8] = 57;
  v9[9] = 82;
  v9[10] = 122;
  v9[11] = 55;
  v9[12] = 81;
  v9[13] = 19;
  v9[14] = 51;
  v9[15] = 35;
  v9[16] = 67;
  v9[17] = 70;
  v9[18] = 41;
  v9[19] = 61;
  v9[20] = 41;
  v9[21] = 32;
  v9[22] = 127;
  v9[23] = 28;
  v9[24] = 38;
  v9[25] = 77;
  v9[26] = 49;
  v9[27] = 20;
  v9[28] = 80;
  v9[29] = 94;
  v9[30] = -24;
  sub_4007F8(v10, 0LL, 4LL);
  v7 = 0;
  v6 = 0;
  sub_400808(aFlag);
  do
  {
    v8 = sub_400818();
    v6 |= v8 ^ v7 ^ (v7 + (v7 ^ v9[v7]));
    v4 = v7++;
  }
  while ( v8 && v8 != 10 && v8 != -1 );
  if ( v6 )
    sub_400828(aFailed, 0LL, v3, v4);
  else
    sub_400828(aCorrect, 0LL, v3, v4);
  return 0;
}

主函数的代码找到，一看一堆数据，拿AI分析代码逻辑

1. 初始化一个包含 31 个整数的数组v9，这些值是某种加密后的 flag
2. 调用sub_4007F8函数读取用户输入（可能是 4 个字节）
3. 进入一个循环，每次迭代：
   • 调用sub_400818读取一个字符
   • 进行一系列异或和加法运算，更新校验值v6
4. 根据校验值v6判断输入是否正确

依旧异或这一块，那就按照逻辑写脚本呗

v9 = [
    102, 109, 99, 98, 127, 58, 85, 106, 57, 82,
    122, 55, 81, 19, 51, 35, 67, 70, 41, 61,
    41, 32, 127, 28, 38, 77, 49, 20, 80, 94, -24
]

flag = []
for v7 in range(len(v9)):
    # 计算 v8 = v7 ^ (v7 + (v7 ^ v9[v7]))
    x = v7 ^ v9[v7]
    y = v7 + x
    v8 = v7 ^ y
    flag.append(chr(v8 & 0xFF))  # 确保在ASCII范围内

print('Flag:', ''.join(flag))

运行就是：

Flag: flag{A_s1mpLe&E4sy_RE_i5Nt_1t}

overoverover！！！