2021DASCTF八月挑战赛Writeup

MISC

签到

看看公告就有啦(嘿嘿嘿，真-签到题)

flag{welcome_to_dasctf_aug}

寒王’sblog

到给的博客里面看，可以看到写了关于outguess隐写的内容，整个博客就博客头像一个图片，猜测是图片，拿博客内容里面的密码试，不对。仔细观察博客，要找到flag.jpg就行了，因为之前有过搭建博客的经验，就决定到这个博客的搭建的平台gitee去寻找，找到gitee上的仓库

找到最新上传的文件，找到flag.jpg

在装有outguess的lunix系统下使用outguess的命令

outguess -k "hahahahahahaha" -r flag.jpg out.txt

得到flag

flag{50aa7fe02602264e7d8102746416cd74}

stealer

方法一
打开流量包，过滤DNS，发现有很多重复的数据，过滤ip

dns and ip.src==172.27.221.13

将info取出，观察发现是图片的base64编码，将字符串进行编辑方便转码。

字符串的变化如下：
原字符串：
Standard query 0x6a7a A iVBORw0KGgoAAAANSUhEUgAABMoAAAMxCAIAAACVY8g6AAAAAXNSR0IAr-.s4c6QAAAARnQU1BAACxjwv8YQUAAAAJcEhZcwAADsMAAA7DAcdvqGQAAP-.lSURBVHhe7P1HlBzZmd6NUxtqjrShdtzoHB6ttMROswlNp/U9OiGKwd-.f8KbRDtXw3hdQMGWBgjcF1432aDZN04MUySGHMxqvgUajGWg*6fsoffpL-.ctf.com.cn OPT
 
操作：
1、去除多余字符串”Standard query 0x6a7a A”、”ctf.com.cn OPT”、”-.”
2、将“*”替换为“+”
 
转化后字符串：iVBORw0KGgoAAAANSUhEUgAABMoAAAMxCAIAAACVY8g6AAAAAXNSR0IArs4c6QAAAARnQU1BAACxjwv8YQUAAAAJcEhZcwAADsMAAA7DAcdvqGQAAPlSURBVHhe7P1HlBzZmd6NUxtqjrShdtzoHB6ttMROswlNp/U9OiGKwdf8KbRDtXw3hdQMGWBgjcF1432aDZN04MUySGHMxqvgUajGWg+6fsoffpL

然后拿到图片
方法二(别人的做法，没弄懂)
使用tshark命令(linux安装)

tshark -r  dump.pcapng -T fields -e dns.qry.name  -Y "udp and dns.flags.response==0"  > res.txt

from Crypto.Util.number import *
import base64

with open('D:\\桌面\\res.txt','r')as f:
	lines=f.readlines()
tmp=[]
# tmp1=[]
for i,line in enumerate(lines):
	if '-.ctf.com.cn' in line:
		line1=line.strip().replace("-.ctf.com.cn","").replace('-.',"").replace("*","+")
		tmp.append(line1)
		# tmp1.append(len(line1))
b=''.join(tmp)		
# print(b)		
b1=base64.b64decode(b.encode())
open('res.png','wb').write(b1)
# print(tmp1)

拿到图片

flag(有点坑，只要输{}里面的md5就行)（md5里面没有o、l，所以图片里面的就是0和1）

 1d3f729ac02bbc15f00adccd79207ab0

CRYPTO

easymath

题目：

assert(len(open('flag.txt', 'rb').read()) < 50)
assert(str(int.from_bytes(open('flag.txt', 'rb').read(), byteorder='big') << 10000).endswith(
    '1862790884563160582365888530869690397667546628710795031544304378154769559410473276482265448754388655981091313419549689169381115573539422545933044902527020209259938095466283008'))

解题脚本：

import gmpy2
r = 1862790884563160582365888530869690397667546628710795031544304378154769559410473276482265448754388655981091313419549689169381115573539422545933044902527020209259938095466283008
y = r //(2 ** 175)
flag=(y*gmpy2.invert((2 ** 9825),(5 ** 175)))%(5 ** 175)
t=int(flag).to_bytes(50,byteorder='big')
print(t)

推理过程看大佬详细的WP

let’s play with rsa~

题目：

from sympy import isprime,nextprime
from Crypto.Util.number import getPrime as getprime ,long_to_bytes,bytes_to_long,inverse
flag='flag{***************}'
 
def play():
    p=getprime(1024)
    q=getprime(1024)
 
    n=p*q
    e=65537
 
    print "Hello,let's play rsa~\n"
    print 'Now,I make some numbers,wait a second\n'
    n1=getprime(200)
    n2=getprime(200)
    number=n1*n2
    print "Ok,i will send two numbers to you,one of them was encoded.\n"
    print "Encode n1:%d,\n"%(pow(n1,e,n))
    print "And n2:%d.\n"%n2
 
    print "Information that can now be made public:the public key (n,e):(%d,%d)\n"%(n,e)
    while True:
        try:
            c=int(raw_input("ok,now,tell me the value of the number (encode it for safe):"))
        except:
            print "Sorry,the input is illeagal, and the integer is accept~"
        else:
            break
    d=inverse(e,(p-1)*(q-1))
    m=pow(c,d,n)
    if m==number:
        print "It's easy and interesting,didn't it?\n"
        print "This is the gift for you :"+flag
    else:
        print "Emmmmm,there is something wrong, bye~\n"
 
if __name__ == '__main__':
    play()

题目给出n、e、pow(n1,e,n)、n2，求c
$$c=numbe{r}^{e}modn=(n1\ast n2{)}^{e}modn=((n{1}^{e}modn)\ast (n{2}^{e}modn))modn$$
解题脚本：
(nc连上后会给出n、e、pow(n1,e,n)、n2）(这边以我nc得到的为例解题)

a = 12961409879680531683887750211274207457513312084569696075205671038512045983337591135848947097185699604453906986636309421174482607699320759296109315315716510712493223473758909665964696575970829701893860144925111613206750829054110563245759651636500800410800611799341091945238575572204744180794490600917575137957033704047292694479180193784052662788188456092611151703198998693787287001909323881137356776611226888433795185583110831437797736663143463934550216639249647171621868213923613785097805958460050341741859668930711065658867650077543948915545547210086665007538174470035036590197762464220663633877458187978934543610137
n2 = 975655394741265791809418303766264964352618962129532647048963
n=17984939304898650803158254840695698991976109768815074321859078838424106854968841553367705813676890566467526226777393337854287622085939699972674300671017434385268427211627149875201145197397614084981293689781191198014025392826496825995772507143839566355038037041002531894155725475951426118850290776380334951977592054772067742756171095427402698085736400771503751576111160577981596580675760373492947131376839865309631659266553436954094643785398989702090587019292927772255810184788705872695851653803767757272932373185881635106333541090761339331432007675440553128966093172385018789133381793617410838502288664808530049842759
e = 65537
c = (a * pow(n2, e, n)) % n
print(c)

提交c返回得到flag

ezRSA

题目：

from secret import flag
from Crypto.Util.number import *
from random import getrandbits
from hashlib import sha256
 
 
class EzRsa:
    def __init__(self):
        self.E = 0x10001
        self.P = getPrime(1024)
        self.Q = getPrime(1024)
        while GCD((self.P-1)*(self.Q-1), self.E) != 1:
            self.Q = getPrime(1024)
        self.N = self.P*self.Q
 
    def encrypt(self):
        f = getrandbits(32)
        c = pow(f, self.E, self.N)
        return (f, c)
 
    def encrypt_flag(self, flag):
        f = bytes_to_long(flag)
        c = pow(f, self.E, self.N)
        return c
 
 
def proof():
    seed = getrandbits(32)
    print(seed)
    sha = sha256(str(seed).encode()).hexdigest()
    print(f"sha256({seed>>18}...).hexdigest() = {sha}")
    sha_i = input("plz enter seed: ")
    if sha256(sha_i.encode()).hexdigest() != sha:
        exit(0)
 
 
if __name__ == "__main__":
    proof()
    print("welcome to EzRsa")
    print("""
    1. Get flag
    2. Encrypt
    3. Insert
    4. Exit
    """)
    A = EzRsa()
    coin = 5
    while coin > 0:
        choose = input("> ")
        if choose == "1":
            print(
                f"pow(flag,e,n) = {A.encrypt_flag(flag)}\ne = 0x10001")
            exit(0)
        elif choose == "2":
            f, c = A.encrypt()
            print(f"plain = {f}\ncipher = {c}")
            coin -= 1
        elif choose == "3":
            q = getrandbits(1024)
            n = A.P*q
            f = getrandbits(32)
            c = pow(f, 0x10001, n)
            print(f"plain = {f}\ncipher = {c}")
            coin -= 1
        elif choose == "4":
            print("bye~")
        else:
            print("wrong input")
    print("Now you get the flag right?")

给你5个coin，1次选择选项1，剩下四次选择选项2、3，当然是平均分配啦。得到四组f、c，两组同q解n，两组不同q解p

2选项
k1n=m21^e-c21
k2n=m22^e-c22
如果gcd(k1,k2)!=1:
那么gcd(k1n,k2n)=k5n

3选项
k3p*q=m31^e-c31
k4p*q=m32^e-c32
如果gcd(k3,k4)!=1:
那么gcd(k3p*q,k4p*q)=k6p

注意！！ 要考虑不互素的情况
解题脚本：

import gmpy2
import libnum

m21 = 432540985
c21 = 7166105329146126000346846799091650898966999155559872071334165373085824639643357340387736867673948949437295645705541881374251463627954867215153754248479607455314501625059368006554456541594829525164594977575051605373580652734948771955132736124144931483976254038698037906320691536102595142817254635128316548640526742258396037736624974201452775566498749686521125689117077825614676728737889758051136033004868110506865885780994503922363995103685835021430586212695942148526088098869737679103317241698921778369133664568044970132074560250492688801256203642144910896398777358726349100245228945110134579964713795561704675625836

m22 = 832433238
c22 = 7068072774444874580843417684352862167509575799447044790925098273022805484339636628726051282308361580345803607470642836938247714174670185418844448110396620415640990336788728326764477161069207367942449417003955828316228120602384920985161057720856713726319102264142124823852191253035614776899528170161081527818980079567071942336953307505627425190268269896232927663142207335803294770668174075896648763132547865986794466490963711592538821527241565132434315309203043850884330880471384804577858687121008366255392040964137615612622718249221451373050550035988762430971866909043093506143514830259205676124914411640454949769514

m31 = 803092604
c31 = 9858061312093654496768968596097501224678614520625887945733170012236020650884606662567525945230720496908398041384708999618217120424456835837520254472794298782801785322426682540037835690902209518506529355172493693012753783133447141964001801908108246725869203422002782991435705743052764243289474944794105647964810575201516329006631623431476942605223169157252417903703894306636035587745490699125350078705206470856005724867576219806201586124846417599897712607698931227973794077078199208851048569386781383768930174636782623527857434298933738057735468619183445620008223850132708634891907499854374991187011529197042230646959

m32 = 407827977
c32 = 9863511828345122655318417789758326362972155249380521046094363106330914667719541369853161354832638229300504131579973818373987565381237053514335756148865347534946328612468008346912307466621874348634289579602255609545387098973079063492763912905225505720948250925345458151757932596246692755570051379007452621160936151870226301905454456709756228915895779952880287559823982391786349058583700592960020679432834444698462657381016169068254820248864870315358313005721370177632775087124268837165293488573472585505563114178522926629394131843183761480379187251213034958089397342815187352111820752209064084639988490442466085392363

c = 7381125776751959446383348016580510257704798119845665945201944033052555935656713698467247959233790127327214355429648255631628985300458608740078555958883489917944617439105827815224630836947733765684747756636698321010977136609738673905945624585893659056271171662069018034510708480585607431573246586211145591759329437967929837683542135674353508920055645757133194288396748112704213294483816147806206043194476131284869098739640171544518373288980710485452989822448431998651248525720045903096843189714992580406453445524466346573496924330678123163145497572043186951795043533663444724588959680284679456217293084234000618759022
e = 0x10001

# 3选项，求p
def p_def(p):
    k = 1
    while 1:
        if gmpy2.is_prime(p):
            return p
        elif p % (k) == 0 and k != 1:
            p = p // k
        else:
            k += 1

# 2选项求q
def q_def(n, p):
    k = 1
    q = n // p
    while 1:
        if gmpy2.is_prime(q):
            return q
        elif p % (k) == 0 and k != 1:
            q = q // k
        else:
            k += 1

kp = gmpy2.gcd(m31 ** e - c31, m32 ** e - c32)
p=p_def(kp)
kn = gmpy2.gcd(m21 ** e - c21, m22 ** e - c22)
#print(kn)
q=q_def(kn,p)
n=p*q

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