Crypto - 随笔分类(第8页) - sevensnight

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(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)

SevensNight

随笔分类 - Crypto

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