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Re:其实很简单的

Posted by adam123 at 2008-01-18 21:52:06 on Problem 2769
In Reply To:其实很简单的 Posted by:Shinjikun at 2007-08-27 10:17:31
> 余数的差=差的余数 只要找一个k,使得k不能整除任何的差
> 把所有的差写出来,然后因式分解,从而得到所有因数,
> 筛去所有因数,得到最小的k
> 似乎是o(n^2log(n))吧
顶一下,确实强悍!

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