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Re:回馈社会In Reply To:回馈社会 Posted by:temp_ptr at 2011-12-07 21:13:38 这道题用不着这么复杂，先眼睛看出N==2的边界情况，然后就是一个 (N-2)*K^2的dp,方程很容易推出就不写了，由于K的数据量比较小，这个复杂度也很低。 不过楼主的方法可以算K的值很大的情况，ORZ一下 Followed by: Post your reply here: |
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