2025 年 3月 12 日随笔档案 - sevensnight

2025年3月12日

摘要：题目: from secret import flag assert flag[:6] == 'TPCTF{' and flag[-1] == '}' flag = flag[6:-1] assert len(set(flag)) == len(flag) xs = [] for i, c in e 阅读全文

posted @ 2025-03-12 22:36 sevensnight 阅读(82) 评论(0) 推荐(0)

Pell's Equation(佩尔方程)

摘要：题目1: from Crypto.Util.number import * flag = ******* p = getPrime(1024) q = getPrime(1024) n = p * q hb = len(flag)//2 hb1 = bytes_to_long(flag[:hb]) 阅读全文

posted @ 2025-03-12 22:26 sevensnight 阅读(177) 评论(0) 推荐(0)

MITM(中间相遇攻击)

摘要：题目: from secret import flag from Crypto.Util.number import * from Crypto.Cipher import AES from hashlib import sha256 from random import * p = getPrim 阅读全文

posted @ 2025-03-12 22:13 sevensnight 阅读(43) 评论(0) 推荐(0)

线性k

摘要：题目1: import hashlib from Crypto.Util.number import * from gmpy2 import * FLAG = b'******************' assert FLAG.startswith(b'NSSCTF{') and FLAG.ends 阅读全文

posted @ 2025-03-12 10:49 sevensnight 阅读(79) 评论(0) 推荐(0)

共享(复用)k

摘要：题目: 解题思路: 解答: 阅读全文

posted @ 2025-03-12 10:48 sevensnight 阅读(12) 评论(0) 推荐(0)

对私钥x的攻击

摘要：对私钥x的攻击本质上是解决一个**DLP**问题让G为一个阿贝尔群(交换群),我们把G中的二元操作写成乘法* 1.给定G,g和h=g<sup>a</sup>,计算a是困难的 2.这里a就叫做h的以g为底的离散对数题目: 利用如下的参数,恢复DSA的秘密钥x p = 800000000000000 阅读全文

posted @ 2025-03-12 10:48 sevensnight 阅读(19) 评论(0) 推荐(0)

NTRU格密码

摘要：题目: from Crypto.Util.number import getPrime, inverse, bytes_to_long, long_to_bytes import random import math FLAG = b'crypto{?????????????????????}' d 阅读全文

posted @ 2025-03-12 10:48 sevensnight 阅读(50) 评论(0) 推荐(0)

隐蔽的背包问题

摘要：题目: from random import shuffle from Crypto.Util.number import getPrime from random import choice, randint # flag来源 flag = b"VNCTF{xxxxxxx}" assert len 阅读全文

posted @ 2025-03-12 00:08 sevensnight 阅读(85) 评论(0) 推荐(0)

密度大于0.9408

摘要：题目: from Crypto.Util.number import * import random from Crypto.Cipher import AES import hashlib from Crypto.Util.Padding import pad from secrets impor 阅读全文

posted @ 2025-03-12 00:08 sevensnight 阅读(47) 评论(0) 推荐(0)

密度小于0.9408

摘要：题目: 解题思路: 解答: 阅读全文

posted @ 2025-03-12 00:07 sevensnight 阅读(17) 评论(0) 推荐(0)

背包密码密度的作用(LLL算法的实现)

摘要：背包问题通解: 当密度d满足以下条件时,LLL算法能有效地解决这类背包问题 ![](https://cdn.nlark.com/yuque/0/2025/png/49294098/1739349426771-401f2fa7-aefe-422b-b642-5d8f664c415d.png) 其中: 阅读全文

posted @ 2025-03-12 00:07 sevensnight 阅读(97) 评论(0) 推荐(0)

超递增背包问题(较清晰)

摘要： from Crypto.Util.number import * from secret import flag def encrypt(m): m = str(bin(m))[2:][::-1] enc = 0 for i in range(len(m)): enc += init[i] * in 阅读全文

posted @ 2025-03-12 00:07 sevensnight 阅读(95) 评论(0) 推荐(0)

超递增背包问题

摘要：知识拓展: 背包加密 - CTF Wiki 背包问题:假定一个背包可以承重W,现在有n个物品,其重量分别为a1,a2,a3,…,an,问装哪些物品可以恰好使得背包装满(其中每个物品只能被装一次) 背包密码的原理由背包问题引申而来,令a为明文二进制序列,x为公钥,这样便构成了一个不错的密码体系,超递增阅读全文

posted @ 2025-03-12 00:06 sevensnight 阅读(212) 评论(0) 推荐(0)

Twisted Hessian曲线

摘要：题目: from Crypto.Util.number import * from secret import flag def add_THcurve(P, Q): if P == (0, 0): return Q if Q == (0, 0): return P x1, y1 = P x2, y 阅读全文

posted @ 2025-03-12 00:05 sevensnight 阅读(52) 评论(0) 推荐(0)

Twisted Hessian曲线(求a)

摘要：题目: from Crypto.Util.number import * from Crypto.Cipher import AES from Crypto.Util.Padding import pad from random import randint import hashlib from 阅读全文

posted @ 2025-03-12 00:05 sevensnight 阅读(127) 评论(0) 推荐(0)

Twisted Edwards曲线

摘要：题目: #sagemath from Crypto.Util.number import * def add(P, Q): (x1, y1) = P (x2, y2) = Q x3 = (x1*y2 + y1*x2) * inverse(1 + d*x1*x2*y1*y2, p) % p y3 = 阅读全文

posted @ 2025-03-12 00:05 sevensnight 阅读(65) 评论(0) 推荐(0)

Smart's attack

摘要：题目1: from Crypto.Util.number import * from secret import flag assert flag[:5]=='cnss{' and flag[-1] == '}' flag_k = bytes_to_long(flag[5:-1].encode()) 阅读全文

posted @ 2025-03-12 00:05 sevensnight 阅读(107) 评论(0) 推荐(0)

Elliptic(椭圆)曲线

摘要：题目: from Crypto.Util.number import getPrime from libnum import s2n from secret import flag p = getPrime(256) a = getPrime(256) b = getPrime(256) E = E 阅读全文

posted @ 2025-03-12 00:04 sevensnight 阅读(99) 评论(0) 推荐(0)

seed>>64

摘要：题目: from Crypto.Util.number import * flag = b'Spirit{*****************}' plaintext = bytes_to_long(flag) length = plaintext.bit_length() a = getPrime( 阅读全文

posted @ 2025-03-12 00:04 sevensnight 阅读(27) 评论(0) 推荐(1)

三元LCG

摘要：题目: from Crypto.Util.number import * from secret import FLAG p = getPrime(128) step = len(FLAG) // 3 xs = [bytes_to_long(FLAG[:step]), bytes_to_long(F 阅读全文

posted @ 2025-03-12 00:03 sevensnight 阅读(41) 评论(0) 推荐(0)

二元LCG

摘要：题目: from secret import flag assert flag[:5] == b'cazy{' assert flag[-1:] == b'}' flag = flag[5:-1] assert (len(flag) == 24) class my_LCG: def __init__ 阅读全文

posted @ 2025-03-12 00:03 sevensnight 阅读(46) 评论(0) 推荐(0)

LCG隔位输出(多次连续加密)

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' class LCG: def __init__(self, seed, a, b, m): self.seed = seed # 初始种子 self.a = a # 乘数 se 阅读全文

posted @ 2025-03-12 00:03 sevensnight 阅读(57) 评论(0) 推荐(0)

LCG变体

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' class LCG: def __init__(self, seed, a, b, m): self.seed = seed # 初始种子 self.a = a # 乘数 se 阅读全文

posted @ 2025-03-12 00:03 sevensnight 阅读(58) 评论(0) 推荐(0)

已知output(被混淆)

摘要：题目: from Crypto.Util.number import getPrime from random import choice, randint # flag来源 flag = b"VNCTF{xxxxxxx}" assert len(flag)<100 FLAG1=flag[:32] 阅读全文

posted @ 2025-03-12 00:03 sevensnight 阅读(66) 评论(0) 推荐(1)

已知output(逆元不出来)

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' class LCG: def __init__(self, seed, a, b, m): self.seed = seed # 初始种子 self.a = a # 乘数 se 阅读全文

posted @ 2025-03-12 00:03 sevensnight 阅读(20) 评论(0) 推荐(0)

已知output

摘要：题目: from Crypto.Util.number import * flag = b'Spirit{****************************************}' plaintext = bytes_to_long(flag) length = plaintext.bit 阅读全文

posted @ 2025-03-12 00:03 sevensnight 阅读(34) 评论(0) 推荐(0)

已知n,output

摘要：题目: from Crypto.Util.number import * flag = b'Spirit{********************************}' plaintext = bytes_to_long(flag) length = plaintext.bit_length( 阅读全文

posted @ 2025-03-12 00:02 sevensnight 阅读(41) 评论(0) 推荐(0)

已知a,n,output[n],output[n+1]

摘要：题目: from Crypto.Util.number import * flag = b'Spirit{*********************}' plaintext = bytes_to_long(flag) length = plaintext.bit_length() a = getPr 阅读全文

posted @ 2025-03-12 00:02 sevensnight 阅读(15) 评论(0) 推荐(0)

已知a,n,output[i],求b

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' class LCG: def __init__(self, seed, a, b, n): self.seed = seed # 初始种子 self.a = a # 乘数 se 阅读全文

posted @ 2025-03-12 00:02 sevensnight 阅读(16) 评论(0) 推荐(0)

已知a,b,n,seed[i]

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' class LCG: def __init__(self, seed, a, b, n): self.seed = seed # 初始种子 self.a = a # 乘数 se 阅读全文

posted @ 2025-03-12 00:02 sevensnight 阅读(51) 评论(0) 推荐(0)

已知a,b,n,c

摘要：题目: from Crypto.Util.number import * flag = b'Spirit{*****************************}' plaintext = bytes_to_long(flag) length = plaintext.bit_length() a 阅读全文

posted @ 2025-03-12 00:02 sevensnight 阅读(18) 评论(0) 推荐(0)

已知seed,a,b,n,c

摘要：知识拓展: 1.LCG的基本原理 LCG是一种基于线性同余关系的伪随机数生成器,它通过一个简单的递推公式生成一系列伪随机数,虽然这些数看起来是随机的,但实际上它们是通过确定性的算法生成的,因此称为"伪随机数" 2. LCG的递推公式 LCG的核心是一个递推公式,用于生成下一个伪随机数,公式如下: X 阅读全文

posted @ 2025-03-12 00:02 sevensnight 阅读(43) 评论(0) 推荐(0)

[羊城杯 2023]Danger_RSA

摘要：题目: from Crypto.Util.number import * m = bytes_to_long(flag) def get_key(a, nbit): assert a >= 2 while True: X = getRandomInteger(nbit // a) s = getRa 阅读全文

posted @ 2025-03-12 00:02 sevensnight 阅读(49) 评论(0) 推荐(0)

AES加密解密

摘要：题目: from Cryptodome.Cipher import AES import os import gmpy2 from flag import FLAG from Cryptodome.Util.number import * def main(): key = os.urandom(2 阅读全文

posted @ 2025-03-12 00:02 sevensnight 阅读(89) 评论(0) 推荐(0)

RC4流密码

摘要：题目: get buf unsign s[256] get buf t[256] we have key:hello world we have flag:???????????????????????????????? for i:0 to 256 set s[i]:i for i:0 to 25 阅读全文

posted @ 2025-03-12 00:01 sevensnight 阅读(12) 评论(0) 推荐(0)

MT19937预测

摘要： Python中内置了一个random库,用来产生随机数其内置的算法为梅森算法(Mersenne Twister) 梅森旋转算法可以产生高质量的伪随机数,并且效率高效,弥补了传统伪随机数生成器的不足,梅森旋转算法的最长周期取自一个梅森素数:2<sup>19937</sup>-1 由此命名为梅森旋转算阅读全文

posted @ 2025-03-12 00:01 sevensnight 阅读(325) 评论(0) 推荐(0)

MD5爆破

该文被密码保护。阅读全文

posted @ 2025-03-12 00:01 sevensnight 阅读(42) 评论(0) 推荐(0)

SevensNight

公告