[BUUOJ]刮开有奖reverse

刮开有奖

这是一个赌博程序，快去赚钱吧！！！！！！！！！！！！！！！！！！！！！！！！！！！(在编辑框中的输入值，即为flag，提交即可) 注意：得到的 flag 请包上 flag{} 提交

1.查壳

2.分析（IDA32）

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; // 90,74,83,69,67,97,78,72,51,110,103 v17 = 103;
sub_4010F0((int)&v7, 0, 10);
memset(&v26, 0, 0xFFFFu);
v26 = v23;
v28 = v25;
v27 = v24;
v4 = sub_401000((int)&v26, strlen(&v26));
memset(&v26, 0, 0xFFFFu);
v27 = v21;
v26 = v20;
v28 = v22;
v5 = sub_401000((int)&v26, strlen(&v26));
if ( String == v7 + 34
&& v19 == v11
&& 4 * v20 - 141 == 3 * v9
&& v21 / 4 == 2 * (v14 / 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;
}

可以知道flag长度为8

sub_4010F0对v7字符串进行操作

// 90,74,83,69,67,97,78,72,51,110,103//0//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);
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;
}

研究不太明白，先编译一下

#include <stdio.h>
#include <stdlib.h>

int sub_4010F0(int *a1, int a2, int a3);
int main()
{
int i=0;
int a1[20]={90,74,83,69,67,97,78,72,51,110,103};
sub_4010F0(a1,0,10);
for(i=0;i<11;i++)
printf("%c ",a1[i]);
}
int 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 = i;
v6 = a1[i];
if ( a2 < result && i < result )
{
do
{
if ( v6 > a1[result] )
{
if ( i >= result )
break;
++i;
a1[v5] = a1[result];
if ( i >= result )
break;
while ( a1[i] <= v6 )
{
if ( ++i >= result )
goto LABEL_13;
}
if ( i >= result )
break;
v5 =i;
a1[result]= a1[i];
}
--result;
}
while ( i < result );
}
LABEL_13:
a1[result] = v6;
sub_4010F0(a1, a2, i - 1);
result = a3;
++i;
}
return result;
}

得到3 C E H J N S Z a g n

返回主函数通过观察变量v7到v25定义时的地址（或者双击变量名），我们知道它们在地址中是相邻的，实际上如下图：

memset(&v26, 0, 65535u);
v26 = v23;
v28 = v25;
v27 = v24;
v4 = sub_401000((int)&v26, strlen(&v26));
memset(&v26, 0, 0177777u);
v27 = v21;
v26 = v20;
v28 = v22;
v5 = sub_401000((int)&v26, strlen(&v26));

我们可以知道，v4使用sub_4010F0函数后的字符串的6,7,8位，调用sub_401000函数，v5使用sub_4010F0函数后的字符串的3,4,5位，调用sub_401000函数。

进入sub_401000

_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
signed int v10; // edx
int v11; // edi
signed int v12; // eax
signed int v13; // 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;
v16 = v6;
if ( a2 > 0 )
{
while ( 1 )
{
v9 = 0;
v10 = 0;
v18 = 0;
do
{
if ( v3 >= v7 )
break;
++v10;
v9 = *(unsigned __int8 *)(v3++ + a1) | (v9 << 8);
}
while ( v10 < 3 );
v11 = v9 << 8 * (3 - v10);
v12 = 0;
v17 = v3;
v13 = 18;
do
{
if ( v10 >= v12 )
{
*((_BYTE *)&v18 + v12) = (v11 >> v13) & 0x3F;
v8 = v16;
}
else
{
*((_BYTE *)&v18 + v12) = 64;
}
*v8++ = byte_407830[*((char *)&v18 + v12)];
v13 -= 6;
++v12;
v16 = v8;
}
while ( v13 > -6 );
v3 = v17;
if ( v17 >= a2 )
break;
v7 = a2;
}
v6 = v15;
}
result = v6;
*v8 = 0;
return result;
}

查看byte_407830

猜想可能是base64加密

3.拼接flag

if ( String == v7 + 34 //String=='3'+34='U'即flag第一位为U
&& v19 == v11 //v19==v11=='J'第二位为J
&& 4 * v20 - 141 == 3 * v9
&& v21 / 4 == 2 * (v14 / 9)
&& !strcmp(v4, "ak1w")//将ak1w base64解密后得到三位flag
&& !strcmp(v5, "V1Ax") )//将V1Ax base64解密后得到三位flag，这在v4前
{
MessageBoxA(hDlg, "U g3t 1T!", "@_@", 0);
}
//组合UJWP1jMp

参考博客：

https://www.cnblogs.com/dyhaohaoxuexi/p/11421263.html

https://www.cnblogs.com/hardcoreYutian/p/12602027.html

https://blog.csdn.net/weixin_30607659/article/details/101533297