注意
本文最后更新于 2024-06-30，文中内容可能已过时。
睡醒的时候比赛开始一小时了，队友秒了不少了
个人就做了这些，剩下的部分队友做了，趁大家不熟otp的知识拿了个一血
还得是靠ACM爷带飞
有空可以再补其他题的wp
Misc
签到！(初级)
关注长亭珂兰寺公众号，发送签到即可获取flag。
LitCTF{Welcome_t0_LitCTF2023}
签到成功
What_1s_BASE (初级)
LitCTF{KFC_Cr4zy_Thur3day_V_me_50}
Take me hand (初级)
Tcp 第1流
404notfound (初级)
LitCTF{Its_404_but_1ts_n0t_a_page}
喜欢我的压缩包么 (初级)
压缩包密码114514
这羽毛球怎么只有一半啊（恼 (初级)
直接改高度，然后用画图无视crc校验打开
破损的图片(初级)
png，前八位文件头损坏
89 50 4E 47 0D 0A 1A 0A
修复 打开
（明明.后面还有空格的说)
问卷
NSSCTF{LitCTF_2023?It’s_time_to_g0to_zh1hu!!!}
两仪生四象 (中级)
其实看懂逻辑，就能做出来
就是每个字符 ord转成十位二进制
然后按照reverse_hash的dict，三位二进制一组转成卦
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_hash = {"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"
binary_text = "".join([_hash.get(i) for i in encoded_text])
print("".join([chr(int(binary_text[x:x+10],2)) for x in range(0,len(binary_text),10)]))
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wh1ch_ag4in_pr0duced_the_3ight_Tr1grams
ssvvgg
先把jpg保存出来
就是这段base64
文件尾提示steghide
用stegseek 梭
LitCTF{svg?_base642png!&steghide!}
Easy shark
🐎的key是a
最后一流追踪http
得到个方程组和flag
flag看起来是仿射密码，而且有方程组
不过拿sagemath解了个94出来（？）
用脚本爆破出的flag
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cipher = "gezx{j13p5oznp_1t_z_900y_k3z771h_k001}"
for a in range(100):
for b in range(100):
flag=''
for i in cipher:
if ord(i) >= ord("a") and ord(i) <= ord("z"):
cha=ord(i)-ord('a')
flag+=chr((a*(cha-b)%26)+97)
else:
flag+=i
if 'flag' in flag:
print(flag,(a,b))
flag{w13e5hake_1s_a_900d_t3a771c_t001} (23, 25)
flag{w13e5hake_1s_a_900d_t3a771c_t001} (23, 51)
flag{w13e5hake_1s_a_900d_t3a771c_t001} (23, 77)
flag{w13e5hake_1s_a_900d_t3a771c_t001} (49, 25)
flag{w13e5hake_1s_a_900d_t3a771c_t001} (49, 51)
flag{w13e5hake_1s_a_900d_t3a771c_t001} (49, 77)
flag{w13e5hake_1s_a_900d_t3a771c_t001} (75, 25)
flag{w13e5hake_1s_a_900d_t3a771c_t001} (75, 51)
flag{w13e5hake_1s_a_900d_t3a771c_t001} (75, 77)
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拼起来
NSSCTF{w13e5hake_1s_a_900d_t3a771c_t001_a}
crypto
(校外)md5的破解
一共缺了4位
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import string
import itertools
import hashlib
def md5(str:str):
return hashlib.md5(str.encode()).hexdigest()
for i in itertools.product(string.ascii_lowercase+string.digits,repeat=4):
flag = 'LitCTF{md5can%s%s3de%srypt213thoughcr%ssh}'%i
# print((flag))
if md5(flag) == '496603d6953a15846cd7cc476f146771':
print(flag)
break
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LitCTF{md5can123dexrypt213thoughcrpsh}
The same common divisor (高级)
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from libnum import n2s,invmod,gcd
n1= 9852079772293301283705208653824307027320071498525390578148444258198605733768947108049676831872672654449631852459503049139275329796717506126689710613873813880735666507857022786447784753088176997374711523987152412069255685005264853118880922539048290400078105858759506186417678959028622484823376958194324034590514104266608644398160457382895380141070373685334979803658172378382884352616985632157233900719194944197689860219335238499593658894630966428723660931647038577670614850305719449893199713589368780231046895222526070730152875112477675102652862254926169713030701937231206405968412044029177246460558028793385980934233
n3= 4940268030889181135441311597961813780480775970170156650560367030148383674257975796516865571557828263935532335958510269356443566533284856608454193676600884849913964971291145182724888816164723930966472329604608512023988191536173112847915884014445539739070437180314205284883149421228744714989392788108329929896637182055266508625177260492776962915873036873839946591259443753924970795669864031580632650140641456386202636466624658715315856453572441182758855085077441336516178544978457053552156714181607801760605521338788424464551796638531143900048375037218585999440622490119344971822707261432953755569507740550277088437182
c1= 7066425618980522033304943700150361912772559890076173881522840300333719222157667104461410726444725540513601550570478331917063911791020088865705346188662290524599499769112250751103647749860198318955619903728724860941709527724500004142950768744200491448875522031555564384426372047270359602780292587644737898593450148108629904854675417943165292922990980758572264063039172969633878015560735737699147707712154627358077477591293746136250207139049702201052305840453700782016480965369600667516646007546442708862429431724013679189842300429421340122052682391471347471758814138218632022564279296594279507382548264409296929401260
c2= 854668035897095127498890630660344701894030345838998465420605524714323454298819946231147930930739944351187708040037822108105697983018529921300277486094149269105712677374751164879455815185393395371001495146490416978221501351569800028842842393448555836910486037183218754013655794027528039329299851644787006463456162952383099752894635657833907958930587328480492546831654755627949756658554724024525108575961076341962292900510328611128404001877137799465932130220386963518903892403159969133882215092783063943679288192557384595152566356483424061922742307738886179947575613661171671781544283180451958232826666741028590085269
e = 65537
n2 = n3^n1
q = gcd(n1, n2)
p = n1 // q
r = n2 // q
print(n2s(pow(c1, int(invmod(e, (p - 1) * (q - 1))), n1)).decode())
# LitCTF{TH3_Tw0_nUmb3rs_H@v3_The_sAme_D1v1s0r!!}
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easy_math (中级)
p**3 >> q**5
所以hint近似于，p的三次根方
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import libnum
n = 2230791374046346835775433548641067593691369485828070649075162141394476183565187654365131822111419512477883295758461313983481545182887415447403634720326639070667688614534290859200753589300443797
c = 2168563038335029902089976057856861885635845445863841607485310134441400500612435296818745930370268060353437465666224400129105788787423156958336380480503762222278722770240792709450637433509537280
hint = 392490868359411675557103683163021977774935163924606169241731307258226973701652855448542714274348304997416149742779376023311152228735117186027560227613656229190807480010615064372521942836446425717660375242197759811804760170129768647414717571386950790115746414735411766002368288743086845078803312201707960465419405926186622999423245762570917629351110970429987377475979058821154568001902541710817731089463915930932142007312230897818177067675996751110894377356758932
e = 65537
p = libnum.nroot(hint,3) + 1
q = n//p
print(libnum.n2s(pow(c, int(libnum.invmod(e, (p - 1) * (q - 1))), n)).decode())
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(校外)babyLCG
lcg板子题，seed恢复
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# MMI = lambda A, n,s=1,t=0,N=0: (n < 2 and t%N or MMI(n, A%n, t, s-A//n*t, N or n),-1)[n<1] #逆元计算
from functools import reduce
from gmpy2 import invert,gcd
from Crypto.Util.number import long_to_bytes
results = [699175025435513913222265085178805479192132631113784770123757454808149151697608216361550466652878, 193316257467202036043918706856603526262215679149886976392930192639917920593706895122296071643390, 1624937780477561769577140419364339298985292198464188802403816662221142156714021229977403603922943, 659236391930254891621938248429619132720452597526316230221895367798170380093631947248925278766506, 111407194162820942281872438978366964960570302720229611594374532025973998885554449685055172110829, 1415787594624585063605356859393351333923892058922987749824214311091742328340293435914830175796909, 655057648553921580727111809001898496375489870757705297406250204329094679858718932270475755075698, 1683427135823894785654993254138434580152093609545092045940376086714124324274044014654085676620851, 492953986125248558013838257810313149490245209968714980288031443714890115686764222999717055064509, 70048773361068060773257074705619791938224397526269544533030294499007242937089146507674570192265]
def crack_unknown_increment(states, modulus, multiplier):
increment = (states[1] - states[0]*multiplier) % modulus
seed = (invert(multiplier,modulus)*(states[0]-increment))%modulus
return crack_unknown_seed(modulus, multiplier, increment, states[0])
def crack_unknown_multiplier(states, modulus):
multiplier = (states[2] - states[1]) * invert(states[1] - states[0], modulus) % modulus
return crack_unknown_increment(states, modulus, multiplier)
def crack_unknown_modulus(states):
diffs = [s1 - s0 for s0, s1 in zip(states, states[1:])]
zeroes = [t2*t0 - t1*t1 for t0, t1, t2 in zip(diffs, diffs[1:], diffs[2:])]
modulus = abs(reduce(gcd, zeroes))
return crack_unknown_multiplier(states, modulus)
def crack_unknown_seed(modulus, multiplier, increment, states0):
return (invert(multiplier,modulus)*(states0-increment))%modulus
seed = crack_unknown_modulus(results)
print(seed)
print(long_to_bytes(seed))
# 637558173724466419277009432838902680292766683894338509150988265759228898373642016332098881873533
# b'LitCTF{31fcd7832029a87f6c9f760fcf297b2f}'
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(校外)我测你vva
flag.txt就是flag
HZCTF{Java666}
NSSCTF{Java666}
(校外)Virginia
维吉尼亚，分成两段
数组前7个，每个和LitCTF{相减，得到10 11 12 13 14 15 16
也就是按位+10再加index
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flag=[86, 116, 128, 80, 98, 85, 139, 122, 134, 114, 125, 136, 117, 123, 129, 127, 128, 128, 142, 130, 140, 147, 127, 132, 131, 136, 151, 134, 152, 164]
# print(len(flag))
print("".join(chr(v-10-i) for i,v in enumerate(flag)))
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LitCTF{it_is_different_caesar}
(校外)隐晦的聊天记录（一血）
otp被称为完美加密，就是在key的长度大于等于明文长度的前提下
每次一密，不复用
算法就是xor，返璞归真
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from Crypto.Util.strxor import strxor
import binascii
key = strxor((b'attack at dawn') ,binascii.unhexlify(b'6c73d5240a948c86981bc294814d'))
flag = strxor(key,b'Monday or Thur')
print(binascii.hexlify(flag).decode())
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NSSCTF{4068cf2108868c889e1bf29d8351}
(校外)Where is P?
p**3是2052bit 比n 2048bit只差一点
先爆破，得到P
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import libnum
n= 24479907029118467064460793139240403258697681144532146836881997837526487637306591893357774423547391867013441147680031968367449693796015901951120514250935018725570026327610524687128709707340727799633444550317834481416507364804274266363478822257132586592232042108076935945436358397787891169163821061005102693505011197453089873909085170776511350713452580692963748763166981047023704528272230392479728897831538235554137129584665886878574314566549330671483636900134584707867654841021494106881794644469229030140144595938886437242375435914268001721437309283611088568191856208951867342004280893021653793820874747638264412653721
c= 6566517934961780069851397787369134601399136324586682773286046135297104713708615112015588908759927424841719937322574766875308296258325687730658550956691921018605724308665345526807393669538103819281108643141723589363068859617542807984954436567078438099854340705208503317269397632214274507740533638883597409138972287275965697689862321166613821995226000320597560745749780942467497435742492468670016480112957715214640939272457886646483560443432985954141177463448896521810457886108311082101521263110578485768091003174683555938678346359150123350656418123918738868598042533211541966786594006129134087145798672161268647536724
a= 22184346235325197613876257964606959796734210361241668065837491428527234174610482874427139453643569493268653377061231169173874401139203757698022691973395609028489121048788465356158531144787135876251872262389742175830840373281181905217510352227396545981674450409488394636498629147806808635157820030290630290808150235068140864601098322473572121965126109735529553247807211711005936042322910065304489093415276688746634951081501428768318098925390576594162098506572668709475140964400043947851427774550253257759990959997691631511262768785787474750441024242552456956598974533625095249106992723798354594261566983135394923063605
for i in range(100):
P3 = n*i + a
P = libnum.nroot(P3,3)
if pow(P,3,n) == a:
print('P =',P)
break
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P知道860bit，高位泄露恢复p
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from sage.all import *
import gmpy2
p4 = 66302204855869216148926460265779698576660998574555407124043768605865908069722142097621926304390549253688814246272903647124801382742681337653915017783954290069842646020090511605930590064443141710086879668946
n = 24479907029118467064460793139240403258697681144532146836881997837526487637306591893357774423547391867013441147680031968367449693796015901951120514250935018725570026327610524687128709707340727799633444550317834481416507364804274266363478822257132586592232042108076935945436358397787891169163821061005102693505011197453089873909085170776511350713452580692963748763166981047023704528272230392479728897831538235554137129584665886878574314566549330671483636900134584707867654841021494106881794644469229030140144595938886437242375435914268001721437309283611088568191856208951867342004280893021653793820874747638264412653721
e = 0x10001
c = 6566517934961780069851397787369134601399136324586682773286046135297104713708615112015588908759927424841719937322574766875308296258325687730658550956691921018605724308665345526807393669538103819281108643141723589363068859617542807984954436567078438099854340705208503317269397632214274507740533638883597409138972287275965697689862321166613821995226000320597560745749780942467497435742492468670016480112957715214640939272457886646483560443432985954141177463448896521810457886108311082101521263110578485768091003174683555938678346359150123350656418123918738868598042533211541966786594006129134087145798672161268647536724
pbits = 1024
kbits = pbits - p4.nbits()
print(p4.nbits())
p4 = p4 << kbits
PR.<x> = PolynomialRing(Zmod(n))
f = x + p4
roots = f.small_roots(X=2^kbits, beta=0.4)
if roots:
p = p4+int(roots[0])
print ("p: ", p)
d = gmpy2.invert(e, (p - 1) * ((n//p) - 1))
print(long_to_bytes(int(pow(c, int(d),n))))
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b’LitCTF{Y0U_hAV3_g0T_Th3_r1ghT_AnsW3r}
(校外)baby_xor(赛后复现)
和ctfshow愚人杯的的xor差不多
不过泄露了只有一半bit
可以通过LitCTF{头补56位
然后coppersmith
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from Crypto.Util.number import *
n = 139167681803392690594490403105432649693546256181767408269202101512534988406137879788255103631885736461742577594980136624933914700779445704490217419248411578290305101891222576080645870988658334799437317221565839991979543660824098367011942169305111105129234902517835649895908656770416774539906212596072334423407
c1 = 11201139662236758800406931253538295757259990870588609533820056210585752522925690049252488581929717556881067021381940083808024384402885422258545946243513996
c2 = 112016152270171196606652761990170033221036025260883289104273504703557624964071464062375228351458191745141525003775876044271210498526920529385038130932141551598616579917681815276713386113932345056134302042399379895915706991873687943357627747262597883603999621939794450743982662393955266685255577026078256473601
t = bytes_to_long(b'LitCTF{')
c0 = c1^^(t<<(8*25))
P.<x> = PolynomialRing(Zmod(n))
f = c0+x
a = f.monic().small_roots(X=2^(256-(7*8)), beta=0.4)
p = f(a)
m = int(p)^^c1
print(long_to_bytes(m))
# b'LitCTF{oh!!!!coppersmith_is_fun}'
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web
我Flag呢？
Ctrl+u看源码
flag在最下面
导弹迷踪
看源码找到game.js
Follow me and hack me
GET和POST传参
PHP是世界上最好的语言！！
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<?php system('cat /*');
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Vimyyds
扫得index.php.swp
Vim -r恢复
可知传password=R2l2ZV9NZV9Zb3VyX0ZsYWc可以命令执行
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password=R2l2ZV9NZV9Zb3VyX0ZsYWc=&cmd=cat /f*
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作业管理系统
admin admin登录
然后传个马
这是什么？SQL ！注一下 ！
Sqlmap 梭
Http pro max plus
分别传
然后得到
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借一步说话--->> /wtfwtfwtfwtf.php
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得到flag
Ping
;截断
1zjs
眼都找瞎了
就当无事发生
https://github.com/ProbiusOfficial/ProbiusOfficial.github.io/commit/main
直接看commit
pwn
只需要nc一下~
感觉，是题目环境问题
反正flag在环境变量里，没写入文件
输入env看到flag
口算题卡
交互题
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from pwn import *
context.log_level = 'debug'
p = remote('node5.anna.nssctf.cn','28533')
r = lambda : p.recv()
rx = lambda x: p.recv(x)
ru = lambda x: p.recvuntil(x)
rud = lambda x: p.recvuntil(x, drop=True)
s = lambda x: p.send(x)
sl = lambda x: p.sendline(x)
sa = lambda x, y: p.sendafter(x, y)
sla = lambda x, y: p.sendlineafter(x, y)
close = lambda : p.close()
debug = lambda : gdb.attach(p)
shell = lambda : p.interactive()
rud(b'Good luck & Have fun!\n')
# print(p.recvline())
while True:
calc = p.recvline().decode()[8:-2]
sl(str(eval(calc)))
p.recvline()
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Re
世界上最棒的程序员
看字符串
LitCTF{I_am_the_best_programmer_ever}
ez_XOR
异或
LitCTF{XOR_1s_3asy_to_OR}
enbase64
base64变表
basecheck下个断点，动调
把source提出来
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basetable = [ 0x67, 0x4A, 0x31, 0x42, 0x52, 0x6A, 0x51, 0x69, 0x65, 0x2F,
0x46, 0x49, 0x57, 0x68, 0x45, 0x73, 0x6C, 0x71, 0x37, 0x47,
0x78, 0x62, 0x6E, 0x4C, 0x32, 0x36, 0x4D, 0x34, 0x2B, 0x48,
0x58, 0x55, 0x74, 0x63, 0x70, 0x6D, 0x56, 0x54, 0x4B, 0x61,
0x79, 0x64, 0x4F, 0x50, 0x33, 0x38, 0x6F, 0x66, 0x35, 0x76,
0x39, 0x30, 0x5A, 0x53, 0x77, 0x72, 0x6B, 0x59, 0x7A, 0x43,
0x41, 0x75, 0x4E, 0x44]
basetable = "".join([chr(x) for x in basetable])
print(basetable)
# gJ1BRjQie/FIWhEslq7GxbnL26M4+HXUtcpmVTKaydOP38of5v90ZSwrkYzCAuND
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然后变表
LitCTF{B@5E64_l5_tooo0_E3sy!!!!!}
snake
pyc缺magic文件头
文件名是37，找个3.7的pyc头补上去
uncompyle6
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flag = [
30, 196, 52, 252, 49, 220, 7, 243, 3, 241, 24, 224, 40, 230, 25, 251, 28, 233, 40, 237, 4, 225, 4, 215, 40, 231, 22, 237, 14, 251, 10, 169]
for i in range(0, len(flag), 2):
flag[i], flag[i + 1] = flag[(i + 1)] ^ 136, flag[i] ^ 119
print(''.join([chr(i) for i in flag]))
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LitCTF{python_snake_is_so_easy!}
shenghuo2