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这样做有运气成分In Reply To:快排7032MS水过 Posted by:donkeyinacm at 2010-03-08 20:27:02 if(a<=N[p].pos&&N[p].pos<=b) 换成 if(b>=N[p].pos&&N[p].pos>=a) 就超时了 这算是擦边球过的办法，我还是去学划分树吧 Followed by: Post your reply here: |
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