2025年3月11日
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(40) 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(16) 评论(0) 推荐(0)
梅森素数
摘要： 梅森数是形如2<sup>n</sup>-1的数(n是正整数),记为M<sub>n</sub>;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(251) 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(51) 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(9) 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(20) 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(62) 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(8) 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(128) 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(13) 评论(0) 推荐(0)