buu-re-13

刮开有奖

这题还挺有意思，考察的点不少，学到的技巧也不少

本题做法

Exeinfope检查，无壳，32位

拖入ida查看，进入关键函数

BOOL __stdcall DialogFunc(HWND hDlg, UINT a2, WPARAM a3, LPARAM a4)
{
  const char *v4; // esi
  const char *v5; // edi
  int v7; // [esp+8h] [ebp-20030h]
  int v8; // [esp+Ch] [ebp-2002Ch]
  int v9; // [esp+10h] [ebp-20028h]
  int v10; // [esp+14h] [ebp-20024h]
  int v11; // [esp+18h] [ebp-20020h]
  int v12; // [esp+1Ch] [ebp-2001Ch]
  int v13; // [esp+20h] [ebp-20018h]
  int v14; // [esp+24h] [ebp-20014h]
  int v15; // [esp+28h] [ebp-20010h]
  int v16; // [esp+2Ch] [ebp-2000Ch]
  int v17; // [esp+30h] [ebp-20008h]
  CHAR String; // [esp+34h] [ebp-20004h]
  char v19; // [esp+35h] [ebp-20003h]
  char v20; // [esp+36h] [ebp-20002h]
  char v21; // [esp+37h] [ebp-20001h]
  char v22; // [esp+38h] [ebp-20000h]
  char v23; // [esp+39h] [ebp-1FFFFh]
  char v24; // [esp+3Ah] [ebp-1FFFEh]
  char v25; // [esp+3Bh] [ebp-1FFFDh]
  char v26; // [esp+10034h] [ebp-10004h]
  char v27; // [esp+10035h] [ebp-10003h]
  char v28; // [esp+10036h] [ebp-10002h]

  if ( a2 == 272 )
    return 1;
  if ( a2 != 273 )
    return 0;
  if ( (_WORD)a3 == 1001 )
  {
    memset(&String, 0, 0xFFFFu);
    GetDlgItemTextA(hDlg, 1000, &String, 0xFFFF);
    if ( strlen(&String) == 8 )
    {
      v7 = 90;
      v8 = 74;
      v9 = 83;
      v10 = 69;
      v11 = 67;
      v12 = 97;
      v13 = 78;
      v14 = 72;
      v15 = 51;
      v16 = 110;
      v17 = 103;
      sub_4010F0(&v7, 0, 10);   //对v7所组成的数组升序排序
      memset(&v26, 0, 0xFFFFu);
      v26 = v23;
      v28 = v25;
      v27 = v24;
      v4 = (const char *)sub_401000(&v26, strlen(&v26));  //对v23~v25进行base64编码
      memset(&v26, 0, 0xFFFFu);
      v27 = v21;
      v26 = v20;
      v28 = v22;
      v5 = (const char *)sub_401000(&v26, strlen(&v26));  //对v20~v22进行base64编码

      //分析知 string（v18）~v25 就是flag
      if ( String == v7 + 34           //U
        && v19 == v11                   //J    
        && 4 * v20 - 141 == 3 * v9      //W
        && v21 / 4 == 2 * (v14 / 9)     //P
        && !strcmp(v4, "ak1w")          //jMp
        && !strcmp(v5, "V1Ax") )        //WP1
                                        //v5修改与前面结果有重叠
      {
        MessageBoxA(hDlg, "U g3t 1T!", "@_@", 0);
      }
    }
    return 0;
  }
  if ( (_WORD)a3 != 1 && (_WORD)a3 != 2 )
    return 0;
  EndDialog(hDlg, (unsigned __int16)a3);
  return 1;
}

关于升序排序函数sub_4010f0

这里的是关于这个排序函数的伪代码分析，但是我没具体去分析算法意思

因为没耐心。。不过对伪代码分析还是有帮助的

int __cdecl sub_4410F0(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=a1[i]
    if ( a2 < result && i < result )
    {
      do
      {
        if ( v6 > *(_DWORD *)(a1 + 4 * result) )  //v6>a1[result]
        {
          if ( i >= result )
            break;
          ++i;
          *(_DWORD *)(v5 + a1) = *(_DWORD *)(a1 + 4 * result);  //a1[v5]=a1[result]
          if ( i >= result )
            break;
          while ( *(_DWORD *)(a1 + 4 * i) <= v6 )   //a1[i]<=v6
          {
            if ( ++i >= result )
              goto LABEL_13;
          }
          if ( i >= result )
            break;
          v5 = 4 * i;
          *(_DWORD *)(a1 + 4 * result) = *(_DWORD *)(4 * i + a1); 
                        //a1[result]=a1[i]
        }
        --result;
      }
      while ( i < result );
    }
LABEL_13:
    *(_DWORD *)(a1 + 4 * result) = v6;  //a1[result]=v6
    sub_4410F0(a1, a2, i - 1);
    result = a3;
    ++i;
  }
  return result;
}