sevensnight

2025年3月11日

摘要：题目: m = xxxxxxxx e = 65537 n c n = 2047491889405177853330526234560188092808828447112182375404972535407247715587377884805507384334582069788664108684261 阅读全文

posted @ 2025-03-11 21:44 sevensnight 阅读(27) 评论(0) 推荐(0)

连分数进阶

摘要：题目: from Crypto.Util.number import * from gmpy2 import * flag = b'******' flag = bytes_to_long(flag) p = getPrime(1024) r = getPrime(175) a = inverse( 阅读全文

posted @ 2025-03-11 21:44 sevensnight 阅读(9) 评论(0) 推荐(0)

连分数攻击

摘要：题目: import hashlib import sympy from Crypto.Util.number import * flag = 'GWHT{************}' flag1 = flag[:19].encode() flag2 = flag[19:].encode() ass 阅读全文

posted @ 2025-03-11 21:44 sevensnight 阅读(77) 评论(0) 推荐(0)

连分数利用

摘要：题目: from Crypto.Util.number import * from gmpy2 import * flag = b'NSSCTF{******}' m1 = bytes_to_long(flag[:len(flag)//2]) m2 = bytes_to_long(flag[len( 阅读全文

posted @ 2025-03-11 21:43 sevensnight 阅读(55) 评论(0) 推荐(0)

高次Rabin算法攻击

摘要：题目: from Crypto.Util.number import * from gmpy2 import * flag = b'NSSCTF{******}' p = getPrime(256) q = getPrime(256) assert p%4 == 3 and q%4 == 3 n = 阅读全文

posted @ 2025-03-11 21:43 sevensnight 阅读(32) 评论(0) 推荐(0)

扩展欧几里得

摘要：题目: from Crypto.Util.number import * flag = b'******' m1 = bytes_to_long(flag[:len(flag)//2]) m2 = bytes_to_long(flag[len(flag)//2:]) assert 186086294 阅读全文

posted @ 2025-03-11 21:43 sevensnight 阅读(33) 评论(0) 推荐(0)

d泄露但n未知

摘要：题目: from Crypto.Util.number import * from gmpy2 import * flag="ctfshow{***}" m=bytes_to_long(flag.encode()) e=65537 p=getPrime(128) q=getPrime(128) n= 阅读全文

posted @ 2025-03-11 14:36 sevensnight 阅读(18) 评论(0) 推荐(1)

d泄露攻击

摘要：题目: from Crypto.Util.number import * from gmpy2 import * p = getPrime(512) q = getPrime(512) assert p < q n = p*q e = 65537 phi = (p-1)*(q-1) d = inve 阅读全文

posted @ 2025-03-11 14:36 sevensnight 阅读(15) 评论(0) 推荐(1)

e很大的dp泄露攻击

摘要：题目: from Crypto.Util.number import * flag = b'NSSCTF{******}' + b'1'*80 p = getPrime(512) q = getPrime(512) n = p*q e = getPrime(128) d = inverse(e, ( 阅读全文

posted @ 2025-03-11 14:36 sevensnight 阅读(44) 评论(0) 推荐(0)

dp&dq泄露攻击-没e?

摘要：题目: from Crypto.Util.number import * from gmpy2 import * from secret import flag p = getPrime(1024) q = getPrime(1024) d = inverse(65537,(p-1)*(q-1)) 阅读全文

posted @ 2025-03-11 14:36 sevensnight 阅读(70) 评论(0) 推荐(0)

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