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  • 攻防世界 reverse 进阶 12 ReverseMe-120

    程序流程很清晰

     1 int __cdecl main(int argc, const char **argv, const char **envp)
     2 {
     3   unsigned int v3; // edx
     4   unsigned int i; // ecx
     5   __m128i v5; // xmm1
     6   unsigned int v6; // esi
     7   const __m128i *v7; // eax
     8   __m128i v8; // xmm0
     9   int v9; // eax
    10   char sc; // [esp+0h] [ebp-CCh]
    11   char str; // [esp+1h] [ebp-CBh]
    12   char s_; // [esp+64h] [ebp-68h]
    13   char v14; // [esp+65h] [ebp-67h]
    14   unsigned int de_s_len; // [esp+C8h] [ebp-4h]
    15 
    16   printf("please input your flah:");
    17   sc = 0;
    18   memset(&str, 0, 0x63u);
    19   scanf("%s", &sc);
    20   s_ = 0;
    21   memset(&v14, 0, 0x63u);
    22   sub_401000(&de_s_len, &s_, (unsigned __int8 *)&sc, strlen(&sc));// base64解码
    23   v3 = de_s_len;                                // 解码后长度
    24   i = 0;
    25   if ( de_s_len )
    26   {
    27     if ( de_s_len >= 0x10 )
    28     {
    29       v5 = _mm_load_si128((const __m128i *)&xmmword_414F20);
    30       v6 = de_s_len - (de_s_len & 0xF);
    31       v7 = (const __m128i *)&s_;
    32       do
    33       {
    34         v8 = _mm_loadu_si128(v7);
    35         i += 16;
    36         ++v7;
    37         _mm_storeu_si128((__m128i *)&v7[-1], _mm_xor_si128(v8, v5));
    38       }
    39       while ( i < v6 );
    40     }
    41     for ( ; i < v3; ++i )
    42       *(&s_ + i) ^= 0x25u;                      // 异或
    43   }
    44   v9 = strcmp(&s_, "you_know_how_to_remove_junk_code");
    45   if ( v9 )
    46     v9 = -(v9 < 0) | 1;
    47   if ( v9 )
    48     printf("wrong
    ");
    49   else
    50     printf("correct
    ");
    51   system("pause");
    52   return 0;
    53 }

    关键比较

    strcmp(&s_, "you_know_how_to_remove_junk_code")向上跟踪,发现sub_401000(&de_s_len, &s_, (unsigned __int8 *)&sc, strlen(&sc));

    进入函数分析可以发现是base64解码

      1 signed int __usercall sub_401000@<eax>(unsigned int *a1@<edx>, _BYTE *a2@<ecx>, unsigned __int8 *sc, unsigned int size)
      2 {
      3   int j; // ebx
      4   unsigned int k; // eax
      5   int v6; // ecx
      6   unsigned __int8 *v7; // edi
      7   int v8; // edx
      8   bool v9; // zf
      9   unsigned __int8 v10; // cl
     10   char v11; // cl
     11   _BYTE *v12; // esi
     12   unsigned int v13; // ecx
     13   int v14; // ebx
     14   unsigned __int8 v15; // cl
     15   char v16; // dl
     16   _BYTE *v18; // [esp+Ch] [ebp-Ch]
     17   unsigned int *v19; // [esp+10h] [ebp-8h]
     18   int v20; // [esp+14h] [ebp-4h]
     19   unsigned int v21; // [esp+14h] [ebp-4h]
     20   int sizea; // [esp+24h] [ebp+Ch]
     21 
     22   j = 0;
     23   v18 = a2;
     24   k = 0;
     25   v6 = 0;
     26   v19 = a1;
     27   v20 = 0;
     28   if ( !size )
     29     return 0;
     30   v7 = sc;
     31   do
     32   {
     33     v8 = 0;
     34     v9 = k == size;
     35     if ( k < size )
     36     {
     37       do
     38       {
     39         if ( sc[k] != ' ' )
     40           break;
     41         ++k;                                    // 不含空格
     42         ++v8;
     43       }
     44       while ( k < size );
     45       v9 = k == size;
     46     }
     47     if ( v9 )
     48       break;
     49     if ( size - k >= 2 && sc[k] == '
    ' && sc[k + 1] == '
    ' || (v10 = sc[k], v10 == '
    ') )
     50     {
     51       v6 = v20;
     52     }
     53     else
     54     {
     55       if ( v8 )
     56         return 0xFFFFFFD4;
     57       if ( v10 == '=' && (unsigned int)++j > 2 )
     58         return 0xFFFFFFD4;
     59       if ( v10 > 0x7Fu )
     60         return 0xFFFFFFD4;
     61       v11 = byte_414E40[v10];
     62       if ( v11 == 0x7F || (unsigned __int8)v11 < '@' && j )
     63         return 0xFFFFFFD4;
     64       v6 = v20++ + 1;
     65     }
     66     ++k;
     67   }
     68   while ( k < size );
     69   if ( !v6 )
     70     return 0;
     71   v12 = v18;
     72   v13 = ((unsigned int)(6 * v6 + 7) >> 3) - j;
     73   if ( v18 && *v19 >= v13 )
     74   {
     75     v21 = 3;
     76     v14 = 0;
     77     for ( sizea = 0; k; --k )
     78     {
     79       v15 = *v7;
     80       if ( *v7 != '
    ' && v15 != '
    ' && v15 != ' ' )
     81       {
     82         v16 = byte_414E40[v15];                 // 关键处理
     83         v21 -= v16 == '@';
     84         v14 = v16 & 0x3F | (v14 << 6);
     85         if ( ++sizea == 4 )
     86         {
     87           sizea = 0;
     88           if ( v21 )
     89             *v12++ = BYTE2(v14);
     90           if ( v21 > 1 )
     91             *v12++ = BYTE1(v14);
     92           if ( v21 > 2 )
     93             *v12++ = v14;
     94         }
     95       }
     96       ++v7;
     97     }
     98     *v19 = v12 - v18;
     99     return 0;
    100   }
    101   *v19 = v13;
    102   return -42;
    103 }
    View Code

    识别base64解码函数是这题主要的考点,之后的操作就很简单

    流程:

    base64解码-->异或-->strcmp(&s_, "you_know_how_to_remove_junk_code")

    1 import base64
    2 
    3 s = 'you_know_how_to_remove_junk_code'
    4 tmp = ''
    5 for i in range(len(s)):
    6     tmp += chr(ord(s[i]) ^ 0x25)
    7 print(base64.b64encode(tmp.encode('utf-8')))
    XEpQek5LSlJ6TUpSelFKeldASEpTQHpPUEtOekZKQUA=
    
    
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  • 原文地址:https://www.cnblogs.com/DirWang/p/11447515.html
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