2025 年 3月 11 日随笔档案 - sevensnight

2025年3月11日

摘要：题目: from sympy import * from Crypto.Util.number import * from secret import flag def gen(): p = getPrime(512) q = nextprime(int(str(p)[::-1])) a = p&( 阅读全文

posted @ 2025-03-11 22:12 sevensnight 阅读(38) 评论(0) 推荐(0)

特殊剪枝

摘要：题目: import sympy from Crypto.Util.number import * from secret import flag e = 65537 p1 = sympy.randprime(2 ** 1023,2 ** 1024) q1 = sympy.randprime(2 * 阅读全文

posted @ 2025-03-11 22:12 sevensnight 阅读(50) 评论(0) 推荐(0)

(p+q)>>nbits

摘要：题目: from Crypto.Util.number import * import uuid flag = "flag{" + str(uuid.uuid4()) + "}" print(flag) m = bytes_to_long(flag.encode()) p = getPrime(10 阅读全文

posted @ 2025-03-11 22:12 sevensnight 阅读(79) 评论(0) 推荐(0)

(p^q)>>nbits

摘要：题目1: from Crypto.Util.number import * from secret import flag nbits=512 p=getPrime(nbits) q=getPrime(nbits) leakBits = 262 leak = (p ^ q) >> (nbits - 阅读全文

posted @ 2025-03-11 22:11 sevensnight 阅读(89) 评论(0) 推荐(0)

p^(q>>nbits)

摘要：题目: from Crypto.Util.number import getPrime P = getPrime(512) Q = getPrime(512) print(f'P=', P) print(f'Q=', Q) N = P * Q gift = P ^ (Q >> 16) print(f 阅读全文

posted @ 2025-03-11 22:11 sevensnight 阅读(35) 评论(0) 推荐(0)

p^q

摘要：题目: from Crypto.Util.number import * p = getPrime(256) #print(f"p = {p}") q = getPrime(256) #print(f"q = {q}") n = p*q xor = p^q print(f"n = {n}") pri 阅读全文

posted @ 2025-03-11 22:11 sevensnight 阅读(29) 评论(0) 推荐(0)

首尾剪枝(p^q_rev)

摘要：题目1: from Crypto.Util.number import * from secret import flag m = bytes_to_long(flag) p = getPrime(256) q = getPrime(256) n = p * q e = 65537 _q = int 阅读全文

posted @ 2025-03-11 22:11 sevensnight 阅读(59) 评论(0) 推荐(0)

剪枝问题

摘要：剪枝:剪枝是对于树算法类的题目而言的,通过减去已经计算后的树的枝点来缩短算法的时间复杂度在密码学中,剪枝技术主要用于优化复杂的搜索算法,尤其是在格密码学和一些基于搜索的密码学问题中,以下是剪枝在密码学中的主要应用和相关算法的介绍: 1. 格密码学中的剪枝格密码学中的核心问题是格中最短向量问题(S 阅读全文

posted @ 2025-03-11 22:11 sevensnight 阅读(96) 评论(0) 推荐(0)

Boneh和Durfee攻击

摘要：题目: from Crypto.Util.number import * p = getPrime(512) q = getPrime(512) flag = b'NSSCTF{******}' n = p*q m = bytes_to_long(flag) d = getPrime(int(102 阅读全文

posted @ 2025-03-11 22:06 sevensnight 阅读(268) 评论(0) 推荐(0)

已知d部分高位部分低位攻击

摘要：题目: n = 1491726986872473433074847744274639470404353859395383179955778029337083566597447813088496581491994632704029460549590262470114966436097223810368 阅读全文

posted @ 2025-03-11 22:06 sevensnight 阅读(23) 评论(0) 推荐(0)

已知d高位攻击(位数不变)

摘要：题目: from secret import m1 def task1(): e = 149 p = getPrime(512) q = getPrime(512) n = p * q d = inverse(e,(p-1)*(q-1)) return (pow(m1, e, n), d >> 22 阅读全文

posted @ 2025-03-11 22:06 sevensnight 阅读(25) 评论(0) 推荐(0)

已知d低位攻击

摘要：题目: n=928965239796164317835697626459459187511623211851597903020857680957632483571461988826411606786230698570118329291799876234922678523041788944614862 阅读全文

posted @ 2025-03-11 22:06 sevensnight 阅读(76) 评论(0) 推荐(0)

已知p+q高位攻击

摘要：题目1: from Crypto.Util.number import * p = getPrime(512) q = getPrime(512) n = p*q e = 65537 m = bytes_to_long(b"nbctf{[REDACTED]}") ct = pow(m,e,n) pr 阅读全文

posted @ 2025-03-11 22:06 sevensnight 阅读(117) 评论(0) 推荐(0)

已知p部分高位部分低位攻击

摘要：题目: from Crypto.Util.number import getPrime, bytes_to_long from secret import flag p = getPrime(1024) q = getPrime(1024) n = p * q e = 65537 hint1 = p 阅读全文

posted @ 2025-03-11 22:05 sevensnight 阅读(98) 评论(0) 推荐(0)

已知p高位攻击(在模p下的类型)(位数不变)

摘要：题目: from Crypto.Util.number import * from gmpy2 import * from tqdm import * # flag = b'ZLCTF{记得那年的雨季，回忆里特安静，哭过后的决定，是否还能进行}' m = bytes_to_long(flag) p 阅读全文

posted @ 2025-03-11 22:05 sevensnight 阅读(18) 评论(0) 推荐(0)

已知p高位攻击

摘要：题目: from Crypto.Util.number import * p = getPrime(512) q = getPrime(512) flag = b'NSSCTF{******}' n = p*q m = bytes_to_long(flag) e = 65537 c = pow(m, 阅读全文

posted @ 2025-03-11 22:05 sevensnight 阅读(25) 评论(0) 推荐(0)

已知m高位攻击(位数不变)

摘要：题目: from Crypto.Util.number import getPrime,bytes_to_long,long_to_bytes from random import randint from secret import flag p = getPrime(1024) q = getP 阅读全文

posted @ 2025-03-11 22:05 sevensnight 阅读(26) 评论(0) 推荐(0)

已知m高位攻击

摘要：题目: from Crypto.Util.number import * p = getPrime(512) q = getPrime(512) flag = b'NSSCTF{******}' n = p*q m = bytes_to_long(flag) e = 3 c = pow(m, e, 阅读全文

posted @ 2025-03-11 22:05 sevensnight 阅读(70) 评论(0) 推荐(0)

已知m低位(可以看作短填充)

摘要：题目: from Crypto.Util.number import * from secret import flag p = getPrime(512) q = getPrime(512) assert GCD(3, (p-1)*(q-1)) != 1 assert len(flag) == 4 阅读全文

posted @ 2025-03-11 22:05 sevensnight 阅读(23) 评论(0) 推荐(0)

SMUPE扩展进阶

摘要：题目: import random from Crypto.Util.number import * m = bytes_to_long(b'NSSCTF{******}') e = [3, 3, 5, 5, 5] cnt = 5 A = [random.randint(1, 128) for i 阅读全文

posted @ 2025-03-11 22:04 sevensnight 阅读(38) 评论(0) 推荐(0)

SMUPE扩展

摘要：题目: import random from Crypto.Util.number import * m = bytes_to_long(b'NSSCTF{******}') e = 3 cnt = 5 A = [random.randint(1, 128) for i in range(cnt)] 阅读全文

posted @ 2025-03-11 22:04 sevensnight 阅读(18) 评论(0) 推荐(0)

线性填充广播攻击

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' p = getPrime(700) q = getPrime(700) n = p*q e = 5 m1 = bytes_to_long(flag) a = getPrime( 阅读全文

posted @ 2025-03-11 22:04 sevensnight 阅读(19) 评论(0) 推荐(0)

4p-1method

摘要：题目: from Crypto.Util.number import isPrime from secret import flag import random m = int.from_bytes(flag) def getMyPrime(nbits, d): print(d) s = rando 阅读全文

posted @ 2025-03-11 22:04 sevensnight 阅读(81) 评论(0) 推荐(0)

Stereotyped messages attack(刻板消息攻击)

摘要：攻击阐述我们用b'\x00'替换消息中的x这样就有了(m+x)^e mod n=c m知道一部分 x是b'\x00\x00******'未知的 (e,n)是公钥,c是密文问题变为如何找到x Coppersmith可以解决了这个问题 (这种题本质上就是已知m高位Coppersmith求小根的变体) 阅读全文

posted @ 2025-03-11 21:54 sevensnight 阅读(86) 评论(0) 推荐(2)

短填充攻击

摘要：题目: from Crypto.Util.number import * m = bytes_to_long(b'NSSCTF{******}') p = getPrime(512) q = getPrime(512) n = p*q e = 9 r = getPrime(512) c1 = pow 阅读全文

posted @ 2025-03-11 21:53 sevensnight 阅读(54) 评论(0) 推荐(0)

反素数(emirp数)

摘要：反素数,英文称作emirp(prime(素数)的左右颠倒拼写),是素数的一种,把一个素数的阿拉伯字数字序列(十进制)变成由低位向高位反写出来,得到的另一个数还是素数,例如素数13,反写就是<font style= 阅读全文

posted @ 2025-03-11 21:53 sevensnight 阅读(129) 评论(0) 推荐(0)

关联信息攻击

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' p = getPrime(512) q = getPrime(512) n = p*q e = 3 m1 = bytes_to_long(flag) a = getPrime( 阅读全文

posted @ 2025-03-11 21:53 sevensnight 阅读(31) 评论(0) 推荐(0)

ROCA攻击

摘要：题目1: from Crypto.Util.number import * import random from sympy import prime FLAG=b'hgame{xxxxxxxxxxxxxxxxxx}' e=0x10001 def primorial(num): result = 1 阅读全文

posted @ 2025-03-11 21:52 sevensnight 阅读(200) 评论(0) 推荐(0)

模多项式求根-Coppersmith方法

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' m = bytes_to_long(flag) a = getPrime(512) b = getPrime(512) c = getPrime(512) d = getPri 阅读全文

posted @ 2025-03-11 21:52 sevensnight 阅读(86) 评论(0) 推荐(0)

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(135) 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)

d泄露攻击

摘要：题目: from Crypto.Util.number import * from gmpy2 import * p = getPrime(512) q = getPrime(512) assert p < q n = p*q e = 65537 phi = (p-1)*(q-1) d = inve 阅读全文

posted @ 2025-03-11 14:36 sevensnight 阅读(22) 评论(0) 推荐(1)

e很大的dp泄露攻击

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' + b'1'*80 p = getPrime(512) q = getPrime(512) n = p*q e = getPrime(128) d = inverse(e, ( 阅读全文

posted @ 2025-03-11 14:36 sevensnight 阅读(75) 评论(0) 推荐(0)

dp&dq泄露攻击-没e?

摘要：题目: from Crypto.Util.number import * from gmpy2 import * from secret import flag p = getPrime(1024) q = getPrime(1024) d = inverse(65537,(p-1)*(q-1)) 阅读全文

posted @ 2025-03-11 14:36 sevensnight 阅读(101) 评论(0) 推荐(0)

dp泄露攻击

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' + b'1'*100 p = getPrime(512) q = getPrime(512) n = p*q e = 65537 d = inverse(e, (p-1)*(q 阅读全文

posted @ 2025-03-11 14:35 sevensnight 阅读(62) 评论(0) 推荐(0)

p+1光滑攻击

摘要：题目: from Crypto.Util.number import * from random import choice flag = b'NSSCTF{******}' def getMyPrime(nbits): while True: p = 1 while p.bit_length() 阅读全文

posted @ 2025-03-11 14:35 sevensnight 阅读(142) 评论(0) 推荐(0)

p-1光滑攻击

摘要：题目: from Crypto.Util.number import * from random import choice flag = b'NSSCTF{******}' def getMyPrime(nbits): while True: p = 1 while p.bit_length() 阅读全文

posted @ 2025-03-11 14:35 sevensnight 阅读(91) 评论(0) 推荐(0)

低加密指数广播攻击CRT

摘要：题目: ('n=', '0x683fe30746a91545a45225e063e8dc64d26dbf98c75658a38a7c9dfd16dd38236c7aae7de5cbbf67056c9c57817fd3da79dc4955217f43caefde3b56a46acf5dL', 'e=' 阅读全文

posted @ 2025-03-11 14:35 sevensnight 阅读(219) 评论(0) 推荐(0)

离散对数的运用

摘要：题目: from Crypto.Util.number import * import random import sympy import gmpy2 m = bytes_to_long(b'flag{*****}') p = getPrime(512) q = getPrime(512) r = 阅读全文

posted @ 2025-03-11 14:35 sevensnight 阅读(60) 评论(0) 推荐(0)

Wiener攻击-连分数

摘要：题目: from Crypto.Util.number import * from gmpy2 import * flag = b'NSSCTF{******}' p = getPrime(256) q = getPrime(256) n = p*q d = getPrime(128) e = in 阅读全文

posted @ 2025-03-11 14:34 sevensnight 阅读(93) 评论(0) 推荐(0)

低解密指数攻击(wiener)-e大

摘要：题目: from secret import flag from Crypto.Util.number import * m = bytes_to_long(flag) p = getPrime(512) q = getPrime(512) #取个512比特的随机质数 N = p * q phi = 阅读全文

posted @ 2025-03-11 14:34 sevensnight 阅读(89) 评论(0) 推荐(0)

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 14:34 sevensnight 阅读(49) 评论(0) 推荐(0)

低加密指数攻击-e小

摘要：题目: from Crypto.Util.number import * from gmpy2 import * flag = b'NSSCTF{******}' p = getPrime(5120) q = getPrime(5120) n = p*q e = 97 phi = (p-1)*(q- 阅读全文

posted @ 2025-03-11 14:34 sevensnight 阅读(62) 评论(0) 推荐(0)

威尔逊定理(筛法的运用)

摘要：题目: #sage from Crypto.Util.number import bytes_to_long from sympy import nextprime FLAG = b'hgame{xxxxxxxxxxxxxxxxxxxxxx}' m = bytes_to_long(FLAG) def 阅读全文

posted @ 2025-03-11 14:34 sevensnight 阅读(36) 评论(0) 推荐(0)

小明文攻击

摘要：题目 from Crypto.Util.number import * from gmpy2 import * flag = b'NSSCTF{******}' p = getPrime(5120) q = getPrime(5120) n = p*q e = 97 phi = (p-1)*(q-1 阅读全文

posted @ 2025-03-11 14:34 sevensnight 阅读(54) 评论(0) 推荐(0)

共p_已知q1,q2,e1,c1,e2,c2,求m

摘要：题目: e1 = 14606334023791426 p = 12100977273546023536494062298943380761921192601549408745367474761433129504006367972242229828654949369815069069496510610 阅读全文

posted @ 2025-03-11 14:33 sevensnight 阅读(49) 评论(0) 推荐(0)

已知n与phi分解n

摘要：题目: from Crypto.Util.number import getPrime from math import prod from sympy import nextprime from random import choices with open('flag.txt', 'rb') a 阅读全文

posted @ 2025-03-11 14:33 sevensnight 阅读(85) 评论(0) 推荐(0)

e与(p-1)或(q-1)均不互素

摘要：题目: from Crypto.Util.number import bytes_to_long from secret import flag e = 0x14 p = 7330895897249035860738209657929637460767893905398244379628076799 阅读全文

posted @ 2025-03-11 14:24 sevensnight 阅读(59) 评论(0) 推荐(0)

e与(q-1)互素,但用上题方法求不出

摘要：题目: c = 2485360255306619684345131431867350432205477625621366642887752720125176463993839766742234027524 n = 0x2CAA9C09DC1061E507E5B7F39DDE3455FCFE127A2 阅读全文

posted @ 2025-03-11 14:24 sevensnight 阅读(27) 评论(0) 推荐(0)

e较大且与(p-1)或(q-1)中任意一个互素

摘要：题目: from Crypto.Util.number import * from secret import flag m=bytes_to_long(flag) p=getPrime(512) q=getPrime(512) print('p=',p) print('q=',q) n=p*q e 阅读全文

posted @ 2025-03-11 14:24 sevensnight 阅读(24) 评论(0) 推荐(0)

e与phi不互素

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' p = getPrime(512) q = getPrime(512) e = 65537*2 n = p*q m = bytes_to_long(flag) c = pow( 阅读全文

posted @ 2025-03-11 14:24 sevensnight 阅读(80) 评论(0) 推荐(0)

CRT的进阶使用

摘要：题目: from Crypto.Util.number import * from secert import flag m = bytes_to_long(flag) e = 260792700 q,p,q_,p_ = [getPrime(512) for _ in range(4)] gift 阅读全文

posted @ 2025-03-11 14:11 sevensnight 阅读(55) 评论(0) 推荐(0)

中国剩余定理(CRT)的运用

摘要：题目: from Crypto.Util.number import * from secret import flag flag=bytes_to_long(flag) l=flag.bit_length()//3 + 1 n=[] N=1 while len(n) < 3: p = 4*getP 阅读全文

posted @ 2025-03-11 14:11 sevensnight 阅读(88) 评论(0) 推荐(0)

欧拉函数的运用

摘要：题目: from Crypto.Util.number import * flag = b'******' p = getPrime(256) q = getPrime(256) n = (p**3) * q e = 65537 phi = (p-1)*(q-1) m = bytes_to_long 阅读全文

posted @ 2025-03-11 14:10 sevensnight 阅读(38) 评论(0) 推荐(0)

多因子变形2

摘要：题目: from random import randint from gmpy2 import * from Crypto.Util.number import * def getprime(bits): while 1: n = 1 while n.bit_length() < bits: n 阅读全文

posted @ 2025-03-11 14:10 sevensnight 阅读(56) 评论(0) 推荐(0)

多因子变形1

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' p = getPrime(512) q = getPrime(512) e = 65537 while True: r = 2*getPrime(100)*e+1 if isP 阅读全文

posted @ 2025-03-11 14:10 sevensnight 阅读(27) 评论(0) 推荐(0)

梅森素数

摘要：梅森数是形如2n-1的数(n是正整数),记为Mn;如果梅森数是素数就称梅森素数(Mersenne prime) 截至2024年10月已知52个梅森素数,最大的是2136279841－1 梅森素数-WiKi 题目: from Crypto.Util.numb 阅读全文

posted @ 2025-03-11 14:09 sevensnight 阅读(304) 评论(0) 推荐(1)

RSA_m大于n

摘要：题目: n = 5806423918988431929295638568708977996508831527187617629322924822521525912798714215691620371904190364350417977398803895295936744855557922349009 阅读全文

posted @ 2025-03-11 14:09 sevensnight 阅读(72) 评论(0) 推荐(0)

差值分解素因数(m大于n)

摘要：题目: from Crypto.Util.number import * import gmpy2 FLAG = b'flag{ ' m = bytes_to_long(FLAG) p = # a prime q = # a prime n = p*q e = 65537 c = pow(m,e,n 阅读全文

posted @ 2025-03-11 14:09 sevensnight 阅读(17) 评论(0) 推荐(0)

通过求根公式求p和q

摘要：题目: import gmpy2 p =#被小男娘偷走了 q =#被小男娘摸走了 n = p * q phi = (p - 1) * (q - 1) m =#nian e = 0xe6b1bee47bd63f615c7d0a43c529d219 d = gmpy2.invert(e, phi) pr 阅读全文

posted @ 2025-03-11 14:09 sevensnight 阅读(31) 评论(0) 推荐(0)

共n_已知n,e1,c1,e2,c2,求m

摘要：题目: from gmpy2 import * from Crypto.Util.number import * flag = '***************' p = getPrime(512) q = getPrime(512) m1 = bytes_to_long(bytes(flag.en 阅读全文

posted @ 2025-03-11 14:06 sevensnight 阅读(82) 评论(0) 推荐(0)

多因子_已知e,p,q,r,求m

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' + b'1'*170 p = getPrime(512) q = getPrime(512) r = getPrime(512) n = p*q*r e = 65537 phi 阅读全文

posted @ 2025-03-11 14:06 sevensnight 阅读(10) 评论(0) 推荐(0)

共e_已知e,n1,c1,n2,c2,求m

摘要：题目: from Crypto.Util.number import * flag = b'G0D{******}' p1 = getPrime(512) q = getPrime(512) p2 = getPrime(512) n1 = p1*q n2 = p2*q e = 65537 m = b 阅读全文

posted @ 2025-03-11 14:04 sevensnight 阅读(141) 评论(0) 推荐(0)

已知n,e,c,但p,q相差不大，运用费马定理求m

摘要：题目: from Crypto.Util.number import * import gmpy2 flag = b'NSSCTF{******}' p = getPrime(512) q = gmpy2.next_prime(p - getPrime(256)) n = p*q e = 65537 阅读全文

posted @ 2025-03-11 14:04 sevensnight 阅读(18) 评论(0) 推荐(0)

已知n,e,c,但p,q相差不大,求m

摘要：题目: from Crypto.Util.number import * import gmpy2 flag = b'NSSCTF{******}' p = getPrime(512) q = gmpy2.next_prime(p) n = p*q e = 65537 phi = (p-1)*(q- 阅读全文

posted @ 2025-03-11 14:00 sevensnight 阅读(34) 评论(0) 推荐(0)

已知n,e,c,分解n,得p,q,求m

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' p = getPrime(256) q = getPrime(256) n = p*q e = 65537 phi = (p-1)*(q-1) m = bytes_to_lon 阅读全文

posted @ 2025-03-11 13:59 sevensnight 阅读(32) 评论(0) 推荐(0)

已知p,q,e,c,求出d,再求m

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' p = getPrime(512) q = getPrime(512) n = p*q e = 65537 phi = (p-1)*(q-1) m = bytes_to_lon 阅读全文

posted @ 2025-03-11 13:56 sevensnight 阅读(84) 评论(0) 推荐(0)

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