https://blog.csdn.net/gungnir0711zh-cnhttps://blog.csdn.net/https://blog.csdn.net/gungnir0711/article/details/50939296https://blog.csdn.net/gungnir0711/article/details/50939296gungnir0711Sun, 20 Mar 2016 22:15:51 +0800https://blog.csdn.net/gungnir0711/article/details/50936204https://blog.csdn.net/gungnir0711/article/details/50936204gungnir0711Sun, 20 Mar 2016 13:22:04 +0800 #includ]]>https://blog.csdn.net/gungnir0711/article/details/50742138https://blog.csdn.net/gungnir0711/article/details/50742138gungnir0711Thu, 25 Feb 2016 22:58:21 +0800https://blog.csdn.net/gungnir0711/article/details/50738336https://blog.csdn.net/gungnir0711/article/details/50738336gungnir0711Thu, 25 Feb 2016 15:50:22 +0800https://blog.csdn.net/gungnir0711/article/details/50738275https://blog.csdn.net/gungnir0711/article/details/50738275gungnir0711Thu, 25 Feb 2016 15:43:32 +0800https://blog.csdn.net/gungnir0711/article/details/50737676https://blog.csdn.net/gungnir0711/article/details/50737676gungnir0711Thu, 25 Feb 2016 14:50:46 +0800https://blog.csdn.net/gungnir0711/article/details/50712607https://blog.csdn.net/gungnir0711/article/details/50712607gungnir0711Mon, 22 Feb 2016 01:27:27 +0800=右括号数。那么我们设dp[i][j]表示长度为i的串，(比)多j个的方案数]]>https://blog.csdn.net/gungnir0711/article/details/50705666https://blog.csdn.net/gungnir0711/article/details/50705666gungnir0711Sat, 20 Feb 2016 18:34:07 +0800https://blog.csdn.net/gungnir0711/article/details/50654209https://blog.csdn.net/gungnir0711/article/details/50654209gungnir0711Fri, 12 Feb 2016 05:20:58 +0800https://blog.csdn.net/gungnir0711/article/details/50615996https://blog.csdn.net/gungnir0711/article/details/50615996gungnir0711Fri, 12 Feb 2016 04:24:08 +0800https://blog.csdn.net/gungnir0711/article/details/50446324https://blog.csdn.net/gungnir0711/article/details/50446324gungnir0711Fri, 01 Jan 2016 23:19:30 +0800 #in]]>https://blog.csdn.net/gungnir0711/article/details/50443767https://blog.csdn.net/gungnir0711/article/details/50443767gungnir0711Thu, 31 Dec 2015 21:21:18 +0800https://blog.csdn.net/gungnir0711/article/details/50436639https://blog.csdn.net/gungnir0711/article/details/50436639gungnir0711Wed, 30 Dec 2015 18:24:53 +0800https://blog.csdn.net/gungnir0711/article/details/50417353https://blog.csdn.net/gungnir0711/article/details/50417353gungnir0711Mon, 28 Dec 2015 10:28:57 +0800>2))。这说明，j]]>https://blog.csdn.net/gungnir0711/article/details/50411292https://blog.csdn.net/gungnir0711/article/details/50411292gungnir0711Sun, 27 Dec 2015 07:13:28 +0800=L的子串中，含有1的个数比子串长度最大的子串的左右端点。 简单来说，就是假如我维护一个前缀和sum[i]表示前i个数中1的个数，那么求最大的( sum[i]-sum[]]>https://blog.csdn.net/gungnir0711/article/details/50411275https://blog.csdn.net/gungnir0711/article/details/50411275gungnir0711Sun, 27 Dec 2015 05:31:23 +0800https://blog.csdn.net/gungnir0711/article/details/50407272https://blog.csdn.net/gungnir0711/article/details/50407272gungnir0711Sat, 26 Dec 2015 05:38:38 +0800https://blog.csdn.net/gungnir0711/article/details/50391716https://blog.csdn.net/gungnir0711/article/details/50391716gungnir0711Thu, 24 Dec 2015 04:40:50 +0800https://blog.csdn.net/gungnir0711/article/details/50384151https://blog.csdn.net/gungnir0711/article/details/50384151gungnir0711Wed, 23 Dec 2015 06:25:11 +0800https://blog.csdn.net/gungnir0711/article/details/50376059https://blog.csdn.net/gungnir0711/article/details/50376059gungnir0711Tue, 22 Dec 2015 01:57:22 +0800