Detail of message

Online Judge

Problem Set

Authors

Online Contests

User

Web Board
Home Page
F.A.Qs
Statistical Charts

Current Contest
Past Contests
Scheduled Contests
Award Contest

Re:for p prime, B^(P-1)==1 (mod P), B^(P-2)==B^(-1) (mod P)

Posted by uni at 2005-04-24 20:22:08 on Problem 2417
In Reply To:Re:for p prime, B^(P-1)==1 (mod P), B^(P-2)==B^(-1) (mod P) Posted by:frkstyc at 2005-04-24 20:18:48

恩，分解因数毕竟比离散对数好一些吧，呵呵，分解因数的拨拉德算法，很强的
那个是基于随机的算法，在绝大多数条件下都很快（不排除有慢的情况）

一个哥们说LOGN5可以做到确定，不过那些论文我没一个懂的

Followed by:

Re:for p prime, B^(P-1)==1 (mod P), B^(P-2)==B^(-1) (mod P) (79) frkstyc 2005-04-24 20:26:32
- Re:for p prime, B^(P-1)==1 (mod P), B^(P-2)==B^(-1) (mod P) (36) uni 2005-04-24 20:31:43
  - Re:for p prime, B^(P-1)==1 (mod P), B^(P-2)==B^(-1) (mod P) (49) frkstyc 2005-04-24 20:34:29
    - Re:for p prime, B^(P-1)==1 (mod P), B^(P-2)==B^(-1) (mod P) (13) Sinya 2010-04-20 08:51:51

Post your reply here:

Home Page Go Back To top

All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator