NSSCTF-[NCTF 2021]狗狗的秘密

题目链接：NSSCTF

根据题目标签，这道题考了SMC，xtea和base47。

无壳，载入IDA，看main函数可知输入长度是42。然后创造了新线程，进入线程开始地址'StartAddress'。
是一个赋值语句就没别的了，很迷。在进入‘off_405014’看一看发现是个fake_flag。好像这个main函数没有什么太有用的...

那对’off_405014‘交叉引用，进入到下图

主要看if语句，有个反调试和赋值，看一下那个sub_403000函数：

很诡异，看一看sub_403000汇编代码，正好在SMC代码区。所以他应该最后会被解密为一个完成函数代码。回去继续观察

进入这个红框里的函数，是一个xtea加密，并且是对v8操作，而v8和sub_403000有关，所以通过xtea进行解密。所以动态调试，自解密。

因为有反调试，patch掉。

patch这些跳转，使直接执行if的子句。

sub_4011C0函数也有反调试，nop掉

patch反调试完保存一下，在sub_403000处下个断点(smc自解密执行在这个断点之前)，动态调试。

进入sub_403000。把那些数据按热键u未定义，再热键c改为汇编代码。如下但是f5无法识别，可以看到上面有红色警告，堆栈不平衡。在红色提醒那里快捷键alt+p，

修改end address到此sub_403000函数的最后位置就可以f5了，位置如下：

f5查看伪代码

void __cdecl sub_2C3000(const char *flag)
{
  signed int v1; // [esp+0h] [ebp-98h]
  unsigned int v2; // [esp+10h] [ebp-88h]
  signed int v3; // [esp+1Ch] [ebp-7Ch]
  int v4; // [esp+2Ch] [ebp-6Ch]
  int v5; // [esp+2Ch] [ebp-6Ch]
  char v6; // [esp+32h] [ebp-66h]
  signed int Size; // [esp+34h] [ebp-64h]
  unsigned int v8; // [esp+38h] [ebp-60h]
  int k; // [esp+38h] [ebp-60h]
  unsigned __int8 *v10; // [esp+3Ch] [ebp-5Ch]
  int i; // [esp+40h] [ebp-58h]
  signed int j; // [esp+40h] [ebp-58h]
  signed int l; // [esp+40h] [ebp-58h]
  signed int m; // [esp+40h] [ebp-58h]
  signed int n; // [esp+40h] [ebp-58h]
  char v16[62]; // [esp+44h] [ebp-54h]
  int v17; // [esp+82h] [ebp-16h]
  int v18; // [esp+86h] [ebp-12h]
  int v19; // [esp+8Ah] [ebp-Eh]
  int v20; // [esp+8Eh] [ebp-Ah]
  __int16 v21; // [esp+92h] [ebp-6h]
 
  v2 = strlen(flag);
  Size = 0x92 * v2 / 0x64 + 1;
  v3 = 0;
  v10 = (unsigned __int8 *)malloc(Size);
  v16[0] = 82;
  v16[1] = -61;
  v16[2] = 26;
  v16[3] = -32;
  v16[4] = 22;
  v16[5] = 93;
  v16[6] = 94;
  v16[7] = -30;
  v16[8] = 103;
  v16[9] = 31;
  v16[10] = 31;
  v16[11] = 6;
  v16[12] = 6;
  v16[13] = 31;
  v16[14] = 23;
  v16[15] = 6;
  v16[16] = 15;
  v16[17] = -7;
  v16[18] = 6;
  v16[19] = 103;
  v16[20] = 88;
  v16[21] = -78;
  v16[22] = -30;
  v16[23] = -116;
  v16[24] = 15;
  v16[25] = 42;
  v16[26] = 6;
  v16[27] = -119;
  v16[28] = -49;
  v16[29] = 42;
  v16[30] = 6;
  v16[31] = 31;
  v16[32] = -104;
  v16[33] = 26;
  v16[34] = 62;
  v16[35] = 23;
  v16[36] = 103;
  v16[37] = 31;
  v16[38] = -9;
  v16[39] = 58;
  v16[40] = 68;
  v16[41] = -61;
  v16[42] = 22;
  v16[43] = 51;
  v16[44] = 105;
  v16[45] = 26;
  v16[46] = 117;
  v16[47] = 22;
  v16[48] = 62;
  v16[49] = 23;
  v16[50] = -43;
  v16[51] = 105;
  v16[52] = 122;
  v16[53] = 27;
  v16[54] = 68;
  v16[55] = 68;
  v16[56] = 62;
  v16[57] = 103;
  v16[58] = -9;
  v16[59] = -119;
  v16[60] = 103;
  v16[61] = -61;
  v17 = 0;
  v18 = 0;
  v19 = 0;
  v20 = 0;
  v21 = 0;
  memset(v10, 0, Size);
  v8 = 0;
  for ( i = 0; i < 256; ++i )
  {
    v6 = byte_2C5018[i];
    byte_2C5018[i] = byte_2C5018[(i + *((unsigned __int8 *)&dword_2C5168 + i % 4)) % 256];
    byte_2C5018[(i + *((unsigned __int8 *)&dword_2C5168 + i % 4)) % 256] = v6;
  }
  while ( v8 < strlen(flag) )
  {
    v4 = flag[v8];
    for ( j = 0x92 * v2 / 0x64; ; --j )
    {
      v5 = v4 + (v10[j] << 8);
      v10[j] = v5 % 47;                //47进制
      v4 = v5 / 47;
      if ( j < v3 )
        v3 = j;
      if ( !v4 && j <= v3 )
        break;
    }
    ++v8;
  }
  for ( k = 0; !v10[k]; ++k )
    ;
  for ( l = 0; l < Size; ++l )
    v10[l] = byte_2C5118[v10[k++]];    #关键
  while ( l < Size )
    v10[l++] = 0;
  v1 = strlen((const char *)v10);
  for ( m = 0; m < v1; ++m )
    v10[m] ^= byte_2C5018[v10[m]];     #关键
  for ( n = 0; n < v1; ++n )
  {
    if ( v10[n] != (unsigned __int8)v16[n] )  //比较
    {
      sub_2C1510("Wrong!\n", v1);
      exit(0);
    }
  }
  sub_2C1510("Right!\n", v1);
}

这里倒着观察，已知v10数组已知，并且是int8类型，所以从0-256爆破，然后在看他在byte_2C5118中的位置就可以求出47进制形式。

#byte_F5018
key1=[0x21, 0x43, 0x65, 0x87, 0x09, 0x21, 0x43, 0x65, 0xA2, 0x9B,
  0xF4, 0xDF, 0xAC, 0x7C, 0xA1, 0xC6, 0x16, 0xD0, 0x0F, 0xDD,
  0xDC, 0x73, 0xC5, 0x6B, 0xD1, 0x96, 0x47, 0xC2, 0x26, 0x67,
  0x4E, 0x41, 0x82, 0x20, 0x56, 0x9A, 0x6E, 0x33, 0x92, 0x88,
  0x29, 0xB5, 0xB4, 0x71, 0xA9, 0xCE, 0xC3, 0x34, 0x50, 0x59,
  0xBF, 0x2D, 0x57, 0x22, 0xA6, 0x30, 0x04, 0xB2, 0xCD, 0x36,
  0xD5, 0x68, 0x4D, 0x5B, 0x45, 0x9E, 0x85, 0xCF, 0x9D, 0xCC,
  0x61, 0x78, 0x32, 0x76, 0x31, 0xE3, 0x80, 0xAD, 0x39, 0x4F,
  0xFA, 0x72, 0x83, 0x4C, 0x86, 0x60, 0xB7, 0xD7, 0x63, 0x0C,
  0x44, 0x35, 0xB3, 0x7B, 0x19, 0xD4, 0x69, 0x08, 0x0B, 0x1F,
  0x3D, 0x11, 0x79, 0xD3, 0xEE, 0x93, 0x42, 0xDE, 0x23, 0x3B,
  0x5D, 0x8D, 0xA5, 0x77, 0x5F, 0x58, 0xDB, 0x97, 0xF6, 0x7A,
  0x18, 0x52, 0x15, 0x74, 0x25, 0x62, 0x2C, 0x05, 0xE8, 0x0D,
  0x98, 0x2A, 0x43, 0xE2, 0xEF, 0x48, 0x87, 0x49, 0x1C, 0xCA,
  0x2B, 0xA7, 0x8A, 0x09, 0x81, 0xE7, 0x53, 0xAA, 0xFF, 0x6F,
  0x8E, 0x91, 0xF1, 0xF0, 0xA4, 0x46, 0x3A, 0x7D, 0x54, 0xEB,
  0x2F, 0xC1, 0xC0, 0x0E, 0xBD, 0xE1, 0x6C, 0x64, 0xBE, 0xE4,
  0x02, 0x3C, 0x5A, 0xA8, 0x9F, 0x37, 0xAF, 0xA0, 0x13, 0xED,
  0x1B, 0xEC, 0x8B, 0x3E, 0x7E, 0x27, 0x99, 0x75, 0xAB, 0xFE,
  0xD9, 0x3F, 0xF3, 0xEA, 0x70, 0xF7, 0x95, 0xBA, 0x1D, 0x40,
  0xB0, 0xF9, 0xE5, 0xF8, 0x06, 0xBC, 0xB6, 0x03, 0xC9, 0x10,
  0x9C, 0x2E, 0x89, 0x5C, 0x7F, 0xB1, 0x1A, 0xD6, 0x90, 0xAE,
  0xDA, 0xE6, 0x5E, 0xB9, 0x84, 0xE9, 0x55, 0xBB, 0xC7, 0x0A,
  0xE0, 0x66, 0xF2, 0xD8, 0xCB, 0x00, 0x12, 0xB8, 0x17, 0x94,
  0x6A, 0x4A, 0x01, 0x24, 0x14, 0x51, 0x07, 0x65, 0x21, 0xC8,
  0x38, 0xFD, 0x8F, 0xC4, 0xF5, 0xFC]
#byte_F5118
key2=[ 0xA7, 0x1C, 0x7E, 0xAF, 0xD9, 0xC2, 0xC0, 0xBE, 0x1F, 0x45,
  0x9A, 0x85, 0x26, 0xE3, 0x87, 0xC3, 0x21, 0xE0, 0x95, 0x10,
  0x71, 0x70, 0x02, 0x75, 0x35, 0xA5, 0x1D, 0x0D, 0x2F, 0xEE,
  0x25, 0x7B, 0xB5, 0x82, 0x66, 0x8D, 0xDB, 0x53, 0x3A, 0x29,
  0xD4, 0x43, 0x99, 0x97, 0x9D, 0xE8, 0x49, 0x00]

v10=[82, 195, 26, 224, 22, 93, 94, 226, 103, 31, 31, 6, 6, 31, 23, 6, 15, 249, 6, 103, 88, 178, 226, 140, 15, 42, 6, 137, 207, 42, 6, 31, 152, 26, 62, 23, 103, 31, 247, 58, 68, 195, 22, 51, 105, 26, 117, 22, 62, 23, 213, 105, 122, 27, 68, 68, 62, 103, 247, 137, 103, 195]

for c in v10:
  for i in range(256):
    if c==key1[i]^i:
      #print(i)
      if i in key2:
        print(key2.index(i),end=' ')
  print(' ')

这里会求到不止一个解，观看别的wp发现就是一个一个试的，根据v10这些余数可以求处flag，最终代码如下