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

Posted by uni at 2005-04-24 20:11:52 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 20:05:35

恩，呵呵，不过似乎也不尽然，比如分解因数，就是有O（LOGN3）的算法

Followed by:

Re:for p prime, B^(P-1)==1 (mod P), B^(P-2)==B^(-1) (mod P) (193) frkstyc 2005-04-24 20:18:48
- Re:for p prime, B^(P-1)==1 (mod P), B^(P-2)==B^(-1) (mod P) (102) uni 2005-04-24 20:22:08
  - 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

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