这道题是无壳32位的程序,利用ida打开,找到main函数。
int __stdcall WinMain(HINSTANCE hInstance, HINSTANCE hPrevInstance, LPSTR lpCmdLine, int nShowCmd)
{
DialogBoxParamA(hInstance, (LPCSTR)0x67, 0, (DLGPROC)DialogFunc, 0);
return 0;
}
之后进入DialogFunc中查看。
BOOL __userpurge DialogFunc@<eax>(int a1@<edi>, int a2@<esi>, HWND hDlg, UINT a4, WPARAM a5, LPARAM a6)
{
const char *v6; // esi
const char *v7; // edi
int v9; // [esp+4h] [ebp-20030h]
int v10; // [esp+8h] [ebp-2002Ch]
int v11; // [esp+Ch] [ebp-20028h]
int v12; // [esp+10h] [ebp-20024h]
int v13; // [esp+14h] [ebp-20020h]
int v14; // [esp+18h] [ebp-2001Ch]
int v15; // [esp+1Ch] [ebp-20018h]
int v16; // [esp+20h] [ebp-20014h]
int v17; // [esp+24h] [ebp-20010h]
int v18; // [esp+28h] [ebp-2000Ch]
int v19; // [esp+2Ch] [ebp-20008h]
CHAR String; // [esp+30h] [ebp-20004h]
char v21; // [esp+31h] [ebp-20003h]
char v22; // [esp+32h] [ebp-20002h]
char v23; // [esp+33h] [ebp-20001h]
char v24; // [esp+34h] [ebp-20000h]
char v25; // [esp+10030h] [ebp-10004h]
char v26; // [esp+10031h] [ebp-10003h]
char v27; // [esp+10032h] [ebp-10002h]
int v28; // [esp+20028h] [ebp-Ch]
int v29; // [esp+2002Ch] [ebp-8h]
__alloca_probe();
if ( a4 == 272 )
return 1;
v29 = a2;
v28 = a1;
if ( a4 != 273 )
return 0;
if ( (_WORD)a5 == 1001 )
{
memset(&String, 0, 0xFFFFu);
GetDlgItemTextA(hDlg, 1000, &String, 0xFFFF);
if ( strlen(&String) == 8 )
{
v9 = 90;
v10 = 74;
v11 = 83;
v12 = 69;
v13 = 67;
v14 = 97;
v15 = 78;
v16 = 72;
v17 = 51;
v18 = 110;
v19 = 103;
sub_4010F0(&v9, 0, 10);
memset(&v25, 0, 0xFFFFu);
v6 = (const char *)sub_401000(&v25, strlen(&v25));
memset(&v25, 0, 0xFFFFu);
v26 = v23;
v25 = v22;
v27 = v24;
v7 = (const char *)sub_401000(&v25, strlen(&v25));
if ( String == v9 + 34
&& v21 == v13
&& 4 * v22 - 141 == 3 * v11
&& v23 / 4 == 2 * (v16 / 9)
&& !strcmp(v6, "ak1w")
&& !strcmp(v7, "V1Ax") )
{
MessageBoxA(hDlg, "U g3t 1T!", "@_@", 0);
}
}
return 0;
}
if ( (_WORD)a5 != 1 && (_WORD)a5 != 2 )
return 0;
EndDialog(hDlg, (unsigned __int16)a5);
return 1;
}
之后开始分析逻辑,看图。
之后往下跟进sub_4010F0函数。
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;
}
利用c语言,进行改写,之后直接运行。
#include
#include
int __cdecl sub_4010F0(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 = 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;
}
int main() {
char str[] = "ZJSECaNH3ng";
sub_4010F0(str, 0, 10);
printf("%s", str);
return 0;
}
可以获得结果是
3CEHJNSZagn
之后继续往下看sub_401000函数。
_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;
}
之后知道
.rdata:00407830 byte_407830 db 41h ; DATA XREF: sub_401000+C0↑r
.rdata:00407831 aBcdefghijklmno db 'BCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=',0
所以这说明,这是一个base64
回到主函数,看这里。
可以算出来,第一位是U,第二位是J,第三位W,第四位P,之后base64出来,ak1w是jMp,V1Ax是WP1,之后就知道谁前谁后了,之后答案就出来了是UJWP1jMp
flag{UJWP1jMp}