2025年3月11日
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(30) 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(31) 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(56) 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(45) 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(20) 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(17) 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(48) 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(38) 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(68) 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(22) 评论(0) 推荐(0)