buu-re-11

查壳&脱壳

Exeinfope 一下
1

IDA64 静态分析

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int __cdecl __noreturn main(int argc, const char **argv, const char **envp)
{
int v3; // eax
char v4; // [rsp+Fh] [rbp-1h]

while ( 1 )
{
while ( 1 )
{
printf("Welcome to CTF game!\nPlease input d/D to start or input q/Q to quit this program: ", argv, envp);
v4 = getchar();
if ( v4 != 100 && v4 != 68 )
break;
Decry();
}
if ( v4 == 113 || v4 == 81 )
Exit();
puts("Input fault format!");
v3 = getchar();
putchar(v3);
}
}

进入关键函数Decry()

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unsigned __int64 Decry()
{
char v1; // [rsp+Fh] [rbp-51h]
int v2; // [rsp+10h] [rbp-50h]
int v3; // [rsp+14h] [rbp-4Ch]
int i; // [rsp+18h] [rbp-48h]
int v5; // [rsp+1Ch] [rbp-44h]
char src[8]; // [rsp+20h] [rbp-40h]
__int64 v7; // [rsp+28h] [rbp-38h]
int v8; // [rsp+30h] [rbp-30h]
__int64 v9; // [rsp+40h] [rbp-20h]
__int64 v10; // [rsp+48h] [rbp-18h]
int v11; // [rsp+50h] [rbp-10h]
unsigned __int64 v12; // [rsp+58h] [rbp-8h]

v12 = __readfsqword(0x28u);
*(_QWORD *)src = 357761762382LL;
v7 = 0LL;
v8 = 0;
v9 = 512969957736LL;
v10 = 0LL;
v11 = 0;
text = join(key3, (const char *)&v9);
strcpy(key, key1);
strcat(key, src);
v2 = 0;
v3 = 0;
getchar();
v5 = strlen(key);
for ( i = 0; i < v5; ++i )
{
if ( key[v3 % v5] > 64 && key[v3 % v5] <= 90 )
key[i] = key[v3 % v5] + 32; // key 转 小写
++v3;
}
printf("Please input your flag:", src);
while ( 1 )
{
v1 = getchar();
if ( v1 == 10 )
break;
if ( v1 == 32 )
{
++v2;
}
else
{
if ( v1 <= 96 || v1 > 122 )
{
if ( v1 > 64 && v1 <= 90 )
str2[v2] = (v1 - 39 - key[v3++ % v5] + 97) % 26 + 97;
}
else
{
str2[v2] = (v1 - 39 - key[v3++ % v5] + 97) % 26 + 97;
}
if ( !(v3 % v5) )
putchar(32);
++v2;
}
}
if ( !strcmp(text, str2) )
puts("Congratulation!\n");
else
puts("Try again!\n");
return __readfsqword(0x28u) ^ v12;
}

逆向

偷个懒,配置一下环境动态分析,直接得到textkey
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text=’killshadow’

key=’adsfkndcls’

分析后,按理说逆向结果应该是不唯一的,因为大小写的处理是一样的,但是根据提交结果和网上wp发现是全大写(所以我觉得这题出得不太好),那脚本就简单了:

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#include<iostream>
using namespace std;

int main()
{
int i;
char key[] = "adsfkndcls";
char flag[] = "";
char text[] = "killshadow";
for (i = 0; i < 10; i++)
{
flag[i] = text[i] - 97 - 6 + key[i] % 26;
while (flag[i] < 65) flag[i] = flag[i] + 26;
}
for (i = 0; i < 10; i++)
cout << flag[i];
return 0;
}

flag{KLDQCUDFZO}

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