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题目 from Crypto.Util.number import * flag = b'qqqy{*****}' p = getPrime(512) q = getPrime(512) n = p * q p0 = p % 2**400 m = bytes_to_long(flag) c = po 阅读全文
posted @ 2025-07-25 19:42
sevensnight
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题目 ! SUCTF2025-SU_rsa from Crypto.Util.number import * from hashlib import sha256 flag = open("flag.txt").read() p = getPrime(512) q = getPrime(512) e 阅读全文
posted @ 2025-07-25 19:42
sevensnight
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题目 ! Moectf2022-Weird_E_Revenge from Crypto.Util.number import * import random from secret import flag table='abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMN 阅读全文
posted @ 2025-07-25 19:32
sevensnight
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题目 ! [SWPUCTF 2021 新生赛]crypto1 from gmpy2 import * from Crypto.Util.number import * flag = '****************************' flag = {"asfajgfbiagbwe"} p 阅读全文
posted @ 2025-07-25 19:28
sevensnight
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题目 ! CISCN2024初赛-OVO from Crypto.Util.number import * from secret import flag nbits = 512 p = getPrime(nbits) q = getPrime(nbits) n = p * q phi = (p-1 阅读全文
posted @ 2025-07-25 19:22
sevensnight
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故障注入攻击(RSA-CRT Fault Attack) 假设在计算 $S_q$ 时发生了错误, $S_q=m_q^{d_q} \bmod q$ , $m_q=m \bmod q$ , $d_q=d \bmod (q-1)$ 正确签名: $S=S_p \cdot(q^{-1} \bmod p) \c 阅读全文
posted @ 2025-07-25 19:22
sevensnight
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题目 ! SCTF2020-RSA from Crypto.Util.number import * from random import randint import codecs flag = int(codecs.encode('SCTF{*******************}'.encod 阅读全文
posted @ 2025-07-25 19:21
sevensnight
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题目 from Crypto.Util.number import * flag = b"qqqy{******}" p = getPrime(1024) q = getPrime(1024) n = p**3*q e = 65537 d = inverse(e, (p-1)*(q-1)) dp = 阅读全文
posted @ 2025-07-25 19:08
sevensnight
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题目 ! GHCTF2025-Sin from Crypto.Util.number import bytes_to_long; print((2 * sin((m := bytes_to_long(b'NSSCTF{test_flag}'))) - 2 * sin(m) * cos(2 * m)) 阅读全文
posted @ 2025-07-25 19:08
sevensnight
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题目 ! LitCTF2025-leak from Crypto.Util.number import * from enc import flag m = bytes_to_long(flag) p,q,e = getPrime(1024),getPrime(1024),getPrime(101) 阅读全文
posted @ 2025-07-25 19:08
sevensnight
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题目 ! 祥云杯2020-Exposure from Crypto.Util.number import * import gmpy2 p = getStrongPrime(512) q = getStrongPrime(512) n = p * q phi = (p - 1) * (q - 1) 阅读全文
posted @ 2025-07-25 19:07
sevensnight
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