一道水题

下载源文件后点开是一个刮刮乐的图片，没有任何操作空间，查壳后扔进IDApro得到源代码

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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 )
    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(v7, 0, 10);
      memset(v18, 0, 0xFFFFu);
      v18[0] = String[5];
      v18[2] = String[7];
      v18[1] = String[6];
      v4 = (const char *)sub_401000(v18, strlen(v18));
      memset(v18, 0, 0xFFFFu);
      v18[1] = String[3];
      v18[0] = String[2];
      v18[2] = String[4];
      v5 = (const char *)sub_401000(v18, strlen(v18));
      if ( String[0] == v7[0] + 34
        && String[1] == v10
        && 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;
}

注意到sub_4010F0(v7, 0, 10);应该是加密函数，点进去查看：

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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);
    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);
    result = a3;
    ++i;
  }
  return result;
}

发现a2和a3是一段[0,11]区间，可是main函数里v7只有两个下标，于是盲猜v7-v16地址是连在一起的，相当于一个数组。于是main函数的代码逻辑应该是：flag是string，通过

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if ( String[0] == v7[0] + 34
        && String[1] == v10
        && 4 * String[2] - 141 == 3 * v8
        && String[3] / 4 == 2 * (v13 / 9)
        && !strcmp(v4, "ak1w")
        && !strcmp(v5, "V1Ax") )
      {
        MessageBoxA(hDlg, "U g3t 1T!", "@_@", 0);
      }

判断是否正确，于是可以推出

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string[0]=v7[0]+34
string[1]=v10
string[2]=(3*v8+141)/4
string[3]=8*(v13/9)
string[2,3,4],string[5,6,7]经过sub_401000加密后应该是"ak1w","V1Ax"

sub_401000明显是个base64加密，那么只要知道sub_4010F0即可得到flag。

v6 = *(_DWORD *)(4 * i + a1);语句的意思是吧

于是直接开始写exp

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#include<bits/stdc++.h>
#define _DWORD unsigned char
using namespace std;

char v7[15]={90,74,83,69,67,97,78,72,51,110,103};
char flag[20];

int sub(char *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 = i;
   	 	v6 = *(_DWORD *)(i + a1);
    	if ( a2 < result && i < result )
    	{
      	do
      	{
        	if ( v6 > *(_DWORD *)(a1 + result) )
        	{
          	if ( i >= result )
            break;
          	++i;
          	*(_DWORD *)(v5 + a1) = *(_DWORD *)(a1 + result);
          	if ( i >= result )
            break;
          	while ( *(_DWORD *)(a1 + i) <= v6 )
          	{
            	if ( ++i >= result )
             	goto LABEL_13;
          	}
          	if ( i >= result )
            break;
          	v5 = i;
          	*(_DWORD *)(a1 + result) = *(_DWORD *)( i + a1);
        	}
        	--result;
    	}
      	while ( i < result );
	}
	LABEL_13:
    *(_DWORD *)(a1 + result) = v6;
    sub(a1, a2, i - 1);
    result = a3;
    ++i;
  	}
  	return result;
}

int main()
{
	sub(v7,0,10);
	cout<<v7<<endl;
	flag[0]=v7[0]+34;
	flag[1]=v7[4];
	flag[2]=(3*v7[2]+141)/4;
	flag[3]=8*(v7[7]/9);
	cout<<flag<<endl;
}

得到v7字符串：3CEHJNSZagn

得到flag前四位：UJWP

接着对之前的两串字符进行base64转码后分别为jMp和WP1

前后拼接上即可获得flag：UJMP1jMP