运行
1.查壳
32位exe文件,没有壳
2.用32位IDA打开
找到main函数,F5查看伪代码
int __cdecl __noreturn main(int argc, const char **argv, const char **envp)
{
_BYTE v3[29]; // [esp+17h] [ebp-35h] BYREF
int v4; // [esp+34h] [ebp-18h]
int v5; // [esp+38h] [ebp-14h] BYREF
int i; // [esp+3Ch] [ebp-10h]
_BYTE v7[12]; // [esp+40h] [ebp-Ch] BYREF
__main();
v3[26] = 0;
*(_WORD *)&v3[27] = 0;
v4 = 0;
strcpy(v3, "*11110100001010000101111#");
while ( 1 )
{
puts("you can choose one action to execute");
puts("1 up");
puts("2 down");
puts("3 left");
printf("4 right\n:");
scanf("%d", &v5);
if ( v5 == 2 )
{
++*(_DWORD *)&v3[25];
}
else if ( v5 > 2 )
{
if ( v5 == 3 )
{
--v4;
}
else
{
if ( v5 != 4 )
LABEL_13:
exit(1);
++v4;
}
}
else
{
if ( v5 != 1 )
goto LABEL_13;
--*(_DWORD *)&v3[25];
}
for ( i = 0; i <= 1; ++i )
{
if ( *(_DWORD *)&v3[4 * i + 25] >= 5u )
exit(1);
}
if ( v7[5 * *(_DWORD *)&v3[25] - 41 + v4] == 49 )
exit(1);
if ( v7[5 * *(_DWORD *)&v3[25] - 41 + v4] == 35 )
{
puts("\nok, the order you enter is the flag!");
exit(0);
}
}
}
分析:
49对应的ASCII是1
,35对应的ascii是#
遇到1
时程序退出,遇到#
时,才是flag
这是一个迷宫题,可以把v3分成五行五列
*1111
01000
01010
00010
1111#
flag{222441144222}