Crypto - 随笔分类(第3页) - sevensnight

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(49) 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(64) 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)

SevensNight

随笔分类 - Crypto

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