文章分类 - CyberSpace Security / CTF / Crypto
摘要：chall.py from AITMCLAB.libnum import s2n, invmod cof_t = [[353, -1162, 32767], [206, -8021, 42110], [262, -7088, 31882], [388, -6394, 21225], [295, -9
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posted @ 2023-10-22 12:29
N0zoM1z0
摘要：chall.py from Crypto.Util.number import bytes_to_long from FLAG import flag p1 = 401327687854144602104262478345650053155149834850813791388612732559616
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posted @ 2023-10-18 12:38
N0zoM1z0
摘要：Final Consensus meet in middle chall.py from Crypto.Cipher import AES import random from Crypto.Util.Padding import pad a = b"" b = b"" FLAG = b"TCP1P
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posted @ 2023-10-17 08:30
N0zoM1z0
摘要：简单的有限域上的多项式求根 题目代码 from Crypto.Util.number import * flag = b'NSSCTF{******}' m = bytes_to_long(flag) a = getPrime(512) b = getPrime(512) c = getPrime(
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posted @ 2023-10-15 17:40
N0zoM1z0
摘要：题目给的Java代码: public class Main { int kmatrix[][]; int tmatrix[]; int rmatrix[]; public void div(String temp, int size) { while (temp.length() > size) {
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posted @ 2023-10-15 15:15
N0zoM1z0
摘要：考察了威尔逊定理 题目代码: import sympy import random def myGetPrime(): A= getPrime(513) print(A) B=A-random.randint(1e3,1e5) print(B) return sympy.nextPrime((B!)
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posted @ 2023-10-15 14:30
N0zoM1z0
摘要：题目给的是标准的RSA加密格式 #!/usr/bin/env python3 import gmpy2 from Crypto.Util.number import getPrime from Crypto.PublicKey import RSA from Crypto.Cipher import
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摘要：题目代码: from gmpy2 import lcm , powmod , invert , gcd , mpz from Crypto.Util.number import getPrime from sympy import nextprime from random import randi
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posted @ 2023-10-15 11:10
N0zoM1z0
摘要：题目代码: import sympy import random from gmpy2 import gcd, invert from Crypto.Util.number import getPrime, isPrime, getRandomNBitInteger, bytes_to_long,
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posted @ 2023-10-15 10:26
N0zoM1z0
摘要：题目代码 #!/usr/bin/env python # -*- coding: utf-8 -*- from Crypto.Util.number import * import random n = 2 ** 512 m = random.randint(2, n-1) | 1 c = pow(
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posted @ 2023-10-15 10:12
N0zoM1z0
摘要：题目代码 import subprocess p = subprocess.check_output('openssl prime -generate -bits 2048 -hex') q = subprocess.check_output('openssl prime -generate -bi
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posted @ 2023-10-14 16:17
N0zoM1z0
摘要：题目给了很大很大一个n 而e只有65537 这种情况直接对c开e次方即可 这里学习一下提取数据的简便方法 exp: import gmpy2 from Crypto.Util.number import * data = open(r'.\rsa_16m').read().split('\n') m
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posted @ 2023-10-13 11:36
N0zoM1z0
摘要：这题我求出flag后代回去反复验证没有任何问题然而为什么这个flag long_to_bytes乱码啊！！！ 先记录一下解题过程 题目代码: from Crypto.Util.number import * with open("key.txt", "r") as fs: key = int(fs.
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摘要：from Crypto.Util.number import * p = getPrime(512) q = getPrime(512) e = 65537 print(p) # 685349523826215539197501105792931452370615902047808406102012
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posted @ 2023-10-11 00:14
N0zoM1z0
摘要：低加密指数广播攻击 这里更新一下以前的CRT脚本(这题好像要用十几组才行) 首先 手动nc 获取20组n,c CRT求出m^e mod(N1N2N3...)后开e次方即可 exp: from functools import * def CRT(mi, ai): M = reduce(lambda
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posted @ 2023-10-10 11:06
N0zoM1z0
摘要：很巧妙的一道题 题目源码： import random from secret import flag ror = lambda x, l, b: (x >> l) | ((x & ((1<<l)-1)) << (b-l)) N = 1 for base in [2, 3, 7]: N *= pow
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posted @ 2023-10-08 22:39
N0zoM1z0
摘要：from Crypto.Util.number import * from gmpy2 import * from secret import flag p = getPrime(25) e = '# Hidden' q = getPrime(25) n = p * q m = bytes_to_l
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摘要：lfsr题目 flag = "flag{xxxxxxxxxxxxxxxx}" assert flag.startswith("flag{") assert flag.endswith("}") assert len(flag)==14 def lfsr(R,mask): output = (R <<
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摘要：给出了 RSA的标准public key生成 直接用标准库 from Crypto.PublicKey.RSA import * with open(r'D:\浏览器下载\attachment\safety_in_numbers\pubkey.pem','r+') as f: key = f.rea
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posted @ 2023-10-06 12:23
N0zoM1z0
摘要：本来以为可以好好练一练流密码的逆向分析 结果练习到的反而是python3对hex byte 类型的处理转换 题目源码 import sys from binascii import unhexlify if(len(sys.argv)<4): print("Usage: python Encrypt
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