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&#

babysign1234567891011121314151617181920212223242526272829303132333435363738394041424344import hashlibimport ecdsafrom Crypto.Util.number import *import randomimport osflag = b"xxx"assert fla

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

picproblem1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556from PIL import Imagefrom Crypto.Util.number import *from numpy import array, zeros, ui

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

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(

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

impossible RSA12345678910111213141516171819from Crypto.Util.number import *from Crypto.PublicKey import RSAe = 65537flag = b'ACTF{...}'while True: p = getPrime(1024) q = inve

n_n1234567891011121314151617181920ucnd fclsmn.rmjy.irdapc jdsncm zpmscjdp, almph_mn_ynizucnd hpfcpm jdsncm uyvzjdsncm zdsl2s = zpmscjdp(1024)x = zpmscjdp(1024)i = s * xp = 0g130u7u3k = zdsl2.jiwpcm(p,

easy_real12345678910111213141516import randomimport hashlibflag = 'xxxxxxxxxxxxxxxxxxxx'key = random.randint(1,10)for i in range(len(flag)): crypto += chr(ord(flag[i])^key)m = crypto的ascii十六