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DP方法应该是DP[I][J]表示长度为I的串加入第J个字符成为TIGHT WORD的概率,那么DP[I][J]=(DP[I-1][J-1]+DP[I-1][J]+DP[I-1][J+1])/nIn Reply To:同问...为什么?楼上 Posted by:Speakless at 2006-07-28 10:01:43 这样写完程序....就会出现顶楼出现的问题...无语啊..哪位大牛说一下为什么?是我想错在什么地方了? Followed by: Post your reply here: |
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