HexRays CTF Challenge By kao May 22, 2023inReverse Engineering Articles Share More sharing options... Followers1 kao Full Member+ 2.9k Wallis & Futuna PostedMay 22, 2023 Quote Madame de Maintenon (AKA “IDA lady”) is locked in a castle and needs help to escape. Do you think you could...
whereasminelookedlikecrap.ThisisespeciallytruewhenitcomestousingZ3solver.Perhaps,IcouldhavesolvedSsssecretifIspentmoretimegooglingandtryingtofigureoutwhatsecretsharingmethodsexist.Imustpaymoreattentiontothosesmallhintshiddeninthechallengedescriptions.ConclusionRTNCTFwasaverynice,smallcompetitionfeaturingrelativelyeasy...
from Crypto.Util.number import * import crcsolver import zlib from pwn import * sh = remote("IP",端口号) for i in range(100): data = sh.recvuntil(b"Here is my CRC32 value:") c = eval(sh.recvline().decode()) m = crcsolver.solve(b'_'*20, range(8*20), c, zlib.crc32) #...
from z3 import * m = BitVec('m',8) v0,v1,v2,v3,v4,v5,v6,v7,v8,v9= BitVecs('v0 v1 v2 v3 v4 v5 v6 v7 v8 v9', 8) s = Solver()#初始化一个Solver类 s.add(-85 * v8 + 58 * v7 + 97 * v6 + v5 + -45 * v4 + 84 * v3 + 95 * v0 - 20 * v1 + 12 * ...
本文结合许多当下互联网存在的资料整理出了自己对RSA的一份笔记,本版只是初版,对许多东西还有待补充。本文所有的解题脚本都经过本人亲自尝试,环境都是Python3,用到的Python库是pycrypto和gmpy2两个。备注:因平台检测C和4连接词,因此改为C四数论模运算规则:模运算与基
count -= len(tmp)returnbufdefsl(data):returnio.sendline(data)defsn(data):returnio.send(data)definfo(string):returnlog.info(string)defdehex(s):returns.replace(' ','').decode('hex')deflimu8(x):returnc_uint8(x).valuedeflimu16(x):returnc_uint16(x).valuedeflimu32(x):returnc_uint...
Katana - Automatic CTF Challenge Solver in Python3 Resources Readme License Unknown, Unknown licenses found Activity Stars 1.3k stars Watchers 30 watching Forks 183 forks Report repository Releases No releases published Sponsor this project patreon.com/johnhammond010 https://www.paypal....
This repository attempts to offer code and material to automate "running through the check-list" or hitting the "low-hanging fruit" in a Capture the Flag challenge. It is meant to act as a utility to help an individualdo things they mightotherwise forget to do. ...
s = Solver() s.add(rows_c + cols_c + num_c) r = [] if s.check() == sat: m = s.model() r = [[m[X[i][j]].as_long() for j in range(9)] for i in range(9)] print(r) # interactive p = remote('challenge.yuanloo.com', 46427) context.log_level = 'debug' for...
1. challenge time cost: 1 hour 15 min, 9:30-10:45 score: 80PT solver: 49 Welcome 2 N1CTF2023! (^ω^) https://drive.google.com/file/d/1g-c70UHiXSAhTkBKagpPp_sbSnq9ZPDt/view?usp=sharing importhashlibimportecdsafromCrypto.CipherimportAESfromCrypto.Util.PaddingimportpadfromCrypto.Util.nu...