Crypto - 随笔分类(第6页) - sevensnight

Schemidt-Samoa密码例题2

摘要：题目: from Crypto.Util.number import * from secret import flag def genKey(nbits): p = getPrime(nbits) q = getPrime(nbits) N = p*p*q d = inverse(N, (p-1) 阅读全文

posted @ 2025-03-11 21:52 sevensnight 阅读(56) 评论(0) 推荐(0)

Schemidt-Samoa密码例题

摘要：题目: from Crypto.Util.number import * def generkey(k): p, q = getPrime(k), getPrime(k) pubkey = p**2 * q n = pubkey l = (p-1)*(q-1) / gcd(p-1, q-1) pri 阅读全文

posted @ 2025-03-11 21:52 sevensnight 阅读(98) 评论(0) 推荐(0)

Schemidt-Samoa密码

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' p = getPrime(512) q = getPrime(512) n = p*p*q e = n c = pow(bytes_to_long(flag), e, n) d 阅读全文

posted @ 2025-03-11 21:52 sevensnight 阅读(46) 评论(0) 推荐(0)

基于多项式上的RSA加密

摘要：环(Ring) 环是一个集合RR,配备两种二元运算(加法+和乘法×),满足: 加法构成阿贝尔群(交换律、结合律、存在零元、逆元) 乘法满足结合律,且对加法有分配律例子:整数集ZZ,实数集RR,矩阵环Mn(R) 环的定义比群(Group)更复杂,因为它涉及两种运算,但比域(Field)更一般,因为环阅读全文

posted @ 2025-03-11 21:52 sevensnight 阅读(111) 评论(0) 推荐(0)

证书修复

摘要：题目: #break.pem BEGIN BREAK PEM PRIVATE MIIEowIBAAKCAQEAw6JUixKmoIZjLyR1Qc/D/3mfTC3YvqKienLM7Nt/83UqpYeg 9rOw02xLtIqgBdVyVkI+MknQdB5tB1W/bo95M8JjmNxi5r 阅读全文

posted @ 2025-03-11 21:46 sevensnight 阅读(161) 评论(0) 推荐(0)

证书格式

摘要：RSA_PKCS#1私钥解密题目: #私钥 BEGIN RSA PRIVATE KEY MIIEvQIBADANBgkqhkiG9w0BAQEFAASCBKcwggSjAgEAAoIBAQDLzJrQLXi4OjFl Yr5EAfQHWnF0ZyZsHHGfNNRLlhL1haAbV2AzTQCC 阅读全文

posted @ 2025-03-11 21:46 sevensnight 阅读(110) 评论(0) 推荐(0)

求解多项式GCD(共e)(Half-GCD算法)

摘要：题目: from Crypto.Util.number import * from secret import flag m = bytes_to_long(flag) p1, q1 = getPrime(512), getPrime(512) n1 = p1*q1 e = 65537 p2, q2 阅读全文

posted @ 2025-03-11 21:46 sevensnight 阅读(136) 评论(0) 推荐(0)

AMM算法进阶

摘要：题目: from Crypto.Util.number import * import os from gmpy2 import * def getMyPrime1(nbits): while True: n = 2*1009*7*getPrime(nbits//2)*getPrime(nbits/ 阅读全文

posted @ 2025-03-11 21:45 sevensnight 阅读(57) 评论(0) 推荐(0)

AMM算法

摘要：题目: from Crypto.Util.number import * import os from gmpy2 import * def getMyPrime(nbits): while True: n = 2*1009*getPrime(nbits//2)*getPrime(nbits//2) 阅读全文

posted @ 2025-03-11 21:45 sevensnight 阅读(83) 评论(0) 推荐(0)

共模攻击进阶

摘要：题目: from Crypto.Util.number import * p = getPrime(700) q = getPrime(700) n = p*q e1 = 3*getPrime(16) e2 = 3*getPrime(16) flag = b'NSSCTF{******}' c1 = 阅读全文

posted @ 2025-03-11 21:45 sevensnight 阅读(51) 评论(0) 推荐(0)

限定范围的p+1

摘要：题目: from Crypto.Util.number import * import random primes = [] for i in range(1000): primes.append(getPrime(64)) def getMyPrime(nbits: int): while Tru 阅读全文

posted @ 2025-03-11 21:45 sevensnight 阅读(49) 评论(0) 推荐(0)

限定范围的p-1

摘要：题目: from Crypto.Util.number import * import random primes = [] for i in range(1000): primes.append(getPrime(64)) def getMyPrime(nbits: int): while Tru 阅读全文

posted @ 2025-03-11 21:45 sevensnight 阅读(17) 评论(0) 推荐(0)

广播攻击-CRT不互素?

摘要：题目: from Crypto.Util.number import * from gmpy2 import * import random flag = b'******' flag = bytes_to_long(flag) nl = [] cl = [] def getn(bits): n = 阅读全文

posted @ 2025-03-11 21:45 sevensnight 阅读(55) 评论(0) 推荐(0)

低加密指数广播攻击-e不小

摘要：题目: m = xxxxxxxx e = 65537 n c n = 2047491889405177853330526234560188092808828447112182375404972535407247715587377884805507384334582069788664108684261 阅读全文

posted @ 2025-03-11 21:44 sevensnight 阅读(30) 评论(0) 推荐(0)

连分数进阶

摘要：题目: from Crypto.Util.number import * from gmpy2 import * flag = b'******' flag = bytes_to_long(flag) p = getPrime(1024) r = getPrime(175) a = inverse( 阅读全文

posted @ 2025-03-11 21:44 sevensnight 阅读(19) 评论(0) 推荐(0)

连分数攻击

摘要：题目: import hashlib import sympy from Crypto.Util.number import * flag = 'GWHT{************}' flag1 = flag[:19].encode() flag2 = flag[19:].encode() ass 阅读全文

posted @ 2025-03-11 21:44 sevensnight 阅读(102) 评论(0) 推荐(0)

连分数利用

摘要：题目: from Crypto.Util.number import * from gmpy2 import * flag = b'NSSCTF{******}' m1 = bytes_to_long(flag[:len(flag)//2]) m2 = bytes_to_long(flag[len( 阅读全文

posted @ 2025-03-11 21:43 sevensnight 阅读(74) 评论(0) 推荐(0)

高次Rabin算法攻击

摘要：题目: from Crypto.Util.number import * from gmpy2 import * flag = b'NSSCTF{******}' p = getPrime(256) q = getPrime(256) assert p%4 == 3 and q%4 == 3 n = 阅读全文

posted @ 2025-03-11 21:43 sevensnight 阅读(72) 评论(0) 推荐(0)

扩展欧几里得

摘要：题目: from Crypto.Util.number import * flag = b'******' m1 = bytes_to_long(flag[:len(flag)//2]) m2 = bytes_to_long(flag[len(flag)//2:]) assert 186086294 阅读全文

posted @ 2025-03-11 21:43 sevensnight 阅读(53) 评论(0) 推荐(0)

d泄露但n未知

摘要：题目: from Crypto.Util.number import * from gmpy2 import * flag="ctfshow{***}" m=bytes_to_long(flag.encode()) e=65537 p=getPrime(128) q=getPrime(128) n= 阅读全文

posted @ 2025-03-11 14:36 sevensnight 阅读(29) 评论(0) 推荐(1)

SevensNight

随笔分类 - Crypto

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