【Reverse】每日必逆0x03

BUU-刮开有奖

附件：https://files.buuoj.cn/files/abe6e2152471e1e1cbd9e5c0cae95d29/8f80610b-8701-4c7f-ad60-63861a558a5b.exe

题解

• 查壳
  
  
• 程序分析

INT_PTR __stdcall DialogFunc(HWND hDlg, UINT a2, WPARAM a3, LPARAM a4)
{
  const char *v4; // esi
  const char *v5; // edi
  int v7[2]; // [esp+8h] [ebp-20030h] BYREF
  int v8; // [esp+10h] [ebp-20028h]
  int v9; // [esp+14h] [ebp-20024h]
  int v10; // [esp+18h] [ebp-20020h]
  int v11; // [esp+1Ch] [ebp-2001Ch]
  int v12; // [esp+20h] [ebp-20018h]
  int v13; // [esp+24h] [ebp-20014h]
  int v14; // [esp+28h] [ebp-20010h]
  int v15; // [esp+2Ch] [ebp-2000Ch]
  int v16; // [esp+30h] [ebp-20008h]
  CHAR String[65536]; // [esp+34h] [ebp-20004h] BYREF
  char v18[65536]; // [esp+10034h] [ebp-10004h] BYREF

  if ( a2 == 272 )
    return 1;
  if ( a2 != 273 )                              // a2 = 273
    return 0;
  if ( (_WORD)a3 == 1001 )
  {
    memset(String, 0, 0xFFFFu);
    GetDlgItemTextA(hDlg, 1000, String, 0xFFFF);// ***
    if ( strlen(String) == 8 )
    {
      v7[0] = 90;
      v7[1] = 74;
      v8 = 83;
      v9 = 69;
      v10 = 67;
      v11 = 97;
      v12 = 78;
      v13 = 72;
      v14 = 51;
      v15 = 110;
      v16 = 103;
      sub_4010F0((int)v7, 0, 10);               // sort function- 51,67,69,72,74,78,83,90,97,103,110
      memset(v18, 0, 0xFFFFu);
      v18[0] = String[5];
      v18[2] = String[7];
      v18[1] = String[6];                       // v18 = str[5:6]
      v4 = sub_401000((int)v18, strlen(v18));   // sub_401000() base64加密
      memset(v18, 0, 0xFFFFu);
      v18[1] = String[3];
      v18[0] = String[2];
      v18[2] = String[4];                       // v18 = str[2,4]
      v5 = sub_401000((int)v18, strlen(v18));
      if ( String[0] == v7[0] + 34              // 51+34
        && String[1] == v10                     // 74
        && 4 * String[2] - 141 == 3 * v8
        && String[3] / 4 == 2 * (v13 / 9)
        && !strcmp(v4, "ak1w")
        && !strcmp(
              v5,
              "V1Ax") )
      {
        MessageBoxA(hDlg, "U g3t 1T!", "@_@", 0);
      }
    }
    return 0;
  }
  if ( (_WORD)a3 != 1 && (_WORD)a3 != 2 )
    return 0;
  EndDialog(hDlg, (unsigned __int16)a3);
  return 1;
}

• 关键点

1. sub_4010F0（） sort
2. sub_401000() base64加密算法

sub_4010F0

// a1 = v7;a2 = 0;a3 = 10
int __cdecl sub_4010F0(int a1, int a2, int a3)
{
  int result; // eax
  int i; // esi
  int v5; // ecx
  int v6; // edx

  result = a3;
  for ( i = a2; i <= a3; a2 = i )
  {
    v5 = 4 * i;
    v6 = *(_DWORD *)(4 * i + a1);               // v6 遍历数组中的元素
    if ( a2 < result && i < result )
    {
      do
      {
        if ( v6 > *(_DWORD *)(a1 + 4 * result) )
        {
          if ( i >= result )
            break;
          ++i;
          *(_DWORD *)(v5 + a1) = *(_DWORD *)(a1 + 4 * result);
          if ( i >= result )
            break;
          while ( *(_DWORD *)(a1 + 4 * i) <= v6 )
          {
            if ( ++i >= result )
              goto LABEL_13;
          }
          if ( i >= result )
            break;
          v5 = 4 * i;
          *(_DWORD *)(a1 + 4 * result) = *(_DWORD *)(4 * i + a1);
        }
        --result;
      }
      while ( i < result );
    }
LABEL_13:
    *(_DWORD *)(a1 + 4 * result) = v6;
    sub_4010F0(a1, a2, i - 1);                  // 递归排序
    result = a3;
    ++i;
  }
  return result;
}

sub_401000

// array,length of array
_BYTE *__cdecl sub_401000(int a1, int a2)
{
  int v2; // eax
  int v3; // esi
  size_t v4; // ebx
  _BYTE *v5; // eax
  _BYTE *v6; // edi
  int v7; // eax
  _BYTE *v8; // ebx
  int v9; // edi
  int v10; // edx
  int v11; // edi
  int v12; // eax
  int i; // esi
  _BYTE *result; // eax
  _BYTE *v15; // [esp+Ch] [ebp-10h]
  _BYTE *v16; // [esp+10h] [ebp-Ch]
  int v17; // [esp+14h] [ebp-8h]
  int v18; // [esp+18h] [ebp-4h]

  v2 = a2 / 3;
  v3 = 0;
  if ( a2 % 3 > 0 )
    ++v2;
  v4 = 4 * v2 + 1;
  v5 = malloc(v4);
  v6 = v5;
  v15 = v5;
  if ( !v5 )
    exit(0);
  memset(v5, 0, v4);
  v7 = a2;
  v8 = v6;                                      // v8 = malloc(v4);
  v16 = v6;                                     // v16 = malloc(v4);
  if ( a2 > 0 )
  {
    while ( 1 )
    {
      v9 = 0;
      v10 = 0;
      v18 = 0;
      do
      {
        if ( v3 >= v7 )
          break;
        ++v10;
        v9 = *(unsigned __int8 *)(v3 + a1) | (v9 << 8);
        ++v3;
      }
      while ( v10 < 3 );                        // Recycle for 3 times
      v11 = v9 << (8 * (3 - v10));
      v12 = 0;
      v17 = v3;
      for ( i = 18; i > -6; i -= 6 )
      {
        if ( v10 >= v12 )
        {
          *((_BYTE *)&v18 + v12) = (v11 >> i) & 0x3F;
          v8 = v16;
        }
        else
        {
          *((_BYTE *)&v18 + v12) = 64;
        }
        *v8++ = byte_407830[*((char *)&v18 + v12++)];
        v16 = v8;
      }
      v3 = v17;
      if ( v17 >= a2 )
        break;
      v7 = a2;
    }
    v6 = v15;
  }
  result = v6;
  *v8 = 0;
  return result;
}

• 逆向分析
  
  略，这里就很简单了，主要是能识别出该函数的作用

总结

要提高正向编程能力，熟悉常见加密及代码实现