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标 题Re: [问题] 10157
发信站批踢踢兔 (Fri Aug 5 01:21:02 2005)
转信站ptt!Group.NCTU!grouppost!Group.NCTU!ptt2
※ 引述《[email protected] (...)》之铭言:
: ※ 引述《windows2k (KERORO军曹)》之铭言:
: : 少打了一些东西
: : 要求的解 为 (深度 <=d 的 f[n][0]) - (深度<=d-1 的f[n][0])
: 下面这段是别人给我的提示:
: At each position, you have two choices: opening bracket and closing bracket.
: So dp is table [depth][position] and stores how many ways there are to reach
: that. And since you only want to allow maximum depth of d, don't allow bigger
: depths when doing dp.
是我看错还是如何,这题的关键是大数吗?:D
DP式应该是不难...
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