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75000 is enough... simple maths, just program to calculate the minimum n that a[n]=a[n+1]=1(mod 10^5)

Posted by richardhuang at 2006-01-24 00:13:33 on Problem 2748
In Reply To:这题怎么知道750000为一个循环节??能否证明一下? Posted by:fjnu_jxd_010 at 2006-01-23 23:37:26

Followed by:

Re:75000 is enough... simple maths, just program to calculate the minimum n that a[n]=a[n+1]=1(mod 10^5) (29) fjnu_jxd_010 2006-01-24 17:45:13
- the proof is trivial... (220) frkstyc 2006-01-24 17:50:12
  - 因为循环是二阶的？ (195) Gamme 2006-01-24 19:53:43
    - 二阶的递推 (0) frkstyc 2006-01-24 20:13:51
Re:750000 is enough... simple maths, just program to calculate the minimum n that a[n]=a[n+1]=1(mod 10^5) (0) encapsulation 2006-01-26 17:23:46
Re (0) wesley007 2006-02-05 22:39:41

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