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Re:其实很简单的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! Followed by:
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