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

Posted by frkstyc at 2005-04-24 11:42:59 on Problem 2417
In Reply To:Fermat's Little Theory,不理解，看不懂（汗）~~~~ Posted by:jane332470 at 2005-04-24 10:08:04

这个题有空间换时间的算法，可以做到O(N^(1/2)logN)，假如用hash的话可以做到O(N^(1/2))

Followed by:

Re:for p prime, B^(P-1)==1 (mod P), B^(P-2)==B^(-1) (mod P) (12) uni 2005-04-24 20:01:21
- Re:for p prime, B^(P-1)==1 (mod P), B^(P-2)==B^(-1) (mod P) (92) frkstyc 2005-04-24 20:05:35
  - Re:for p prime, B^(P-1)==1 (mod P), B^(P-2)==B^(-1) (mod P) (35) uni 2005-04-24 20:11:52
    - 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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