Crypto - 随笔分类(第5页) - sevensnight

已知d部分高位部分低位攻击

摘要：题目: n = 1491726986872473433074847744274639470404353859395383179955778029337083566597447813088496581491994632704029460549590262470114966436097223810368 阅读全文

posted @ 2025-03-11 22:06 sevensnight 阅读(23) 评论(0) 推荐(0)

已知d高位攻击(位数不变)

摘要：题目: from secret import m1 def task1(): e = 149 p = getPrime(512) q = getPrime(512) n = p * q d = inverse(e,(p-1)*(q-1)) return (pow(m1, e, n), d >> 22 阅读全文

posted @ 2025-03-11 22:06 sevensnight 阅读(25) 评论(0) 推荐(0)

已知d低位攻击

摘要：题目: n=928965239796164317835697626459459187511623211851597903020857680957632483571461988826411606786230698570118329291799876234922678523041788944614862 阅读全文

posted @ 2025-03-11 22:06 sevensnight 阅读(76) 评论(0) 推荐(0)

已知p+q高位攻击

摘要：题目1: from Crypto.Util.number import * p = getPrime(512) q = getPrime(512) n = p*q e = 65537 m = bytes_to_long(b"nbctf{[REDACTED]}") ct = pow(m,e,n) pr 阅读全文

posted @ 2025-03-11 22:06 sevensnight 阅读(117) 评论(0) 推荐(0)

已知p部分高位部分低位攻击

摘要：题目: from Crypto.Util.number import getPrime, bytes_to_long from secret import flag p = getPrime(1024) q = getPrime(1024) n = p * q e = 65537 hint1 = p 阅读全文

posted @ 2025-03-11 22:05 sevensnight 阅读(98) 评论(0) 推荐(0)

已知p高位攻击(在模p下的类型)(位数不变)

摘要：题目: from Crypto.Util.number import * from gmpy2 import * from tqdm import * # flag = b'ZLCTF{记得那年的雨季，回忆里特安静，哭过后的决定，是否还能进行}' m = bytes_to_long(flag) p 阅读全文

posted @ 2025-03-11 22:05 sevensnight 阅读(18) 评论(0) 推荐(0)

已知p高位攻击

摘要：题目: from Crypto.Util.number import * p = getPrime(512) q = getPrime(512) flag = b'NSSCTF{******}' n = p*q m = bytes_to_long(flag) e = 65537 c = pow(m, 阅读全文

posted @ 2025-03-11 22:05 sevensnight 阅读(25) 评论(0) 推荐(0)

已知m高位攻击(位数不变)

摘要：题目: from Crypto.Util.number import getPrime,bytes_to_long,long_to_bytes from random import randint from secret import flag p = getPrime(1024) q = getP 阅读全文

posted @ 2025-03-11 22:05 sevensnight 阅读(26) 评论(0) 推荐(0)

已知m高位攻击

摘要：题目: from Crypto.Util.number import * p = getPrime(512) q = getPrime(512) flag = b'NSSCTF{******}' n = p*q m = bytes_to_long(flag) e = 3 c = pow(m, e, 阅读全文

posted @ 2025-03-11 22:05 sevensnight 阅读(70) 评论(0) 推荐(0)

已知m低位(可以看作短填充)

摘要：题目: from Crypto.Util.number import * from secret import flag p = getPrime(512) q = getPrime(512) assert GCD(3, (p-1)*(q-1)) != 1 assert len(flag) == 4 阅读全文

posted @ 2025-03-11 22:05 sevensnight 阅读(23) 评论(0) 推荐(0)

SMUPE扩展进阶

摘要：题目: import random from Crypto.Util.number import * m = bytes_to_long(b'NSSCTF{******}') e = [3, 3, 5, 5, 5] cnt = 5 A = [random.randint(1, 128) for i 阅读全文

posted @ 2025-03-11 22:04 sevensnight 阅读(38) 评论(0) 推荐(0)

SMUPE扩展

摘要：题目: import random from Crypto.Util.number import * m = bytes_to_long(b'NSSCTF{******}') e = 3 cnt = 5 A = [random.randint(1, 128) for i in range(cnt)] 阅读全文

posted @ 2025-03-11 22:04 sevensnight 阅读(18) 评论(0) 推荐(0)

线性填充广播攻击

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' p = getPrime(700) q = getPrime(700) n = p*q e = 5 m1 = bytes_to_long(flag) a = getPrime( 阅读全文

posted @ 2025-03-11 22:04 sevensnight 阅读(19) 评论(0) 推荐(0)

4p-1method

摘要：题目: from Crypto.Util.number import isPrime from secret import flag import random m = int.from_bytes(flag) def getMyPrime(nbits, d): print(d) s = rando 阅读全文

posted @ 2025-03-11 22:04 sevensnight 阅读(81) 评论(0) 推荐(0)

Stereotyped messages attack(刻板消息攻击)

摘要：攻击阐述我们用b'\x00'替换消息中的x这样就有了(m+x)^e mod n=c m知道一部分 x是b'\x00\x00******'未知的 (e,n)是公钥,c是密文问题变为如何找到x Coppersmith可以解决了这个问题 (这种题本质上就是已知m高位Coppersmith求小根的变体) 阅读全文

posted @ 2025-03-11 21:54 sevensnight 阅读(86) 评论(0) 推荐(2)

短填充攻击

摘要：题目: from Crypto.Util.number import * m = bytes_to_long(b'NSSCTF{******}') p = getPrime(512) q = getPrime(512) n = p*q e = 9 r = getPrime(512) c1 = pow 阅读全文

posted @ 2025-03-11 21:53 sevensnight 阅读(54) 评论(0) 推荐(0)

反素数(emirp数)

摘要：反素数,英文称作emirp(prime(素数)的左右颠倒拼写),是素数的一种,把一个素数的阿拉伯字数字序列(十进制)变成由低位向高位反写出来,得到的另一个数还是素数,例如素数<font style="color:rgb(32, 33, 34);">13</font>,反写就是<font style= 阅读全文

posted @ 2025-03-11 21:53 sevensnight 阅读(129) 评论(0) 推荐(0)

关联信息攻击

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' p = getPrime(512) q = getPrime(512) n = p*q e = 3 m1 = bytes_to_long(flag) a = getPrime( 阅读全文

posted @ 2025-03-11 21:53 sevensnight 阅读(31) 评论(0) 推荐(0)

ROCA攻击

摘要：题目1: from Crypto.Util.number import * import random from sympy import prime FLAG=b'hgame{xxxxxxxxxxxxxxxxxx}' e=0x10001 def primorial(num): result = 1 阅读全文

posted @ 2025-03-11 21:52 sevensnight 阅读(200) 评论(0) 推荐(0)

模多项式求根-Coppersmith方法

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' m = bytes_to_long(flag) a = getPrime(512) b = getPrime(512) c = getPrime(512) d = getPri 阅读全文

posted @ 2025-03-11 21:52 sevensnight 阅读(86) 评论(0) 推荐(0)

SevensNight

随笔分类 - Crypto

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