羊城杯-2020-Simple [羊城杯 2020]Simple12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455from Crypto.Util.number import *from Crypto.Cipher import DESimport gmpy2from secre 2022-08-20 Crypto #RSA #Wiener's Attack
2022-巅峰极客-网络安全技能挑战赛 point-power1234567891011121314151617181920212223242526from Crypto.Util.number import *from gmpy2 import *from random import *from secrets import flagassert len(flag)==42p=getPrime(600)a=bytes_to_long( 2022-08-20 Crypto #ECC #RSA-CRT Fault Attack
第六届“强网杯”全国网络安全挑战赛 线上赛 polydiv4位哈希爆破 + 手撸40遍求b(x) (sagemath上下载pwn库报错) 12345678910111213141516171819#4位哈希爆破part = 'dSyotqgg1qNFrKru'hashstr = '3c678965931a61af1a4950bb0da7fc02980bbf4524dbff107a941aabccc8bf8e&# 2022-08-04 Crypto #RSA
DASCTF2022.07赋能赛 babysign1234567891011121314151617181920212223242526272829303132333435363738394041424344import hashlibimport ecdsafrom Crypto.Util.number import *import randomimport osflag = b"xxx"assert fla 2022-07-24 Crypto #ECDSA #NTRU #多项式RSA #LWE
NepCTF 2022 signinp,q极其近似,对n开根号取前后素数求出p,q 123456789101112131415161718192021from Crypto.Util.number import *import gmpy2import sympyn = 19955580242010925349026385826277356862322608500430230515928936214328341334162 2022-07-21 Crypto #Dual_EC
2022 DSCTF picproblem1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556from PIL import Imagefrom Crypto.Util.number import *from numpy import array, zeros, ui 2022-07-19 Crypto #RSA #Pell #LLL #Logistic
2022DASCTF MAY 出题人挑战赛 Yusa的密码学课堂——一见如故12345678910111213141516171819202122232425262728293031323334353637383940414243444546class Myrand(): def __init__(self,seed): self.index = 0 self.isInit = 1 self.MT = [seed] + [0] * 6 2022-07-13 Crypto #Crypto #MT19937
Dest0g3 520迎新赛 很遗憾,由于暑假需要参与数模,暑假每日一练可能很难实现,但仍会陆续复现一些之前的题目。时间跨度上可能会长一些。 babyRSA123456789101112131415from Crypto.Util.number import bytes_to_long, getPrimefrom gmpy2 import next_primep = getPrime(1024)q = next_prime( 2022-07-10 Crypto #RSA #AES #背包密码
SUSCTF - Ez_Pager_Tiper Ez_Pager_Tiper分析magic_box中的代码,测试发现malicious_magic中的now = (-magic & magic)的值取决于magic二进制末尾0的个数,即$now = 2^{二进制末尾0的个数}$,进而每次执行该函数,magic的二进制中最后一个1改为0,故confusion函数中的循环次数为magic二进制中的1的个数。 再来看看confusi 2022-06-28 Crypto #Crypto #流密码
ACTF2022 impossible RSA12345678910111213141516171819from Crypto.Util.number import *from Crypto.PublicKey import RSAe = 65537flag = b'ACTF{...}'while True: p = getPrime(1024) q = inve 2022-06-26 Crypto #Crypto #RSA