https://blog.youkuaiyun.com/hq572241670zh-cnhttps://blog.youkuaiyun.com/https://blog.youkuaiyun.com/hq572241670/article/details/50478418https://blog.youkuaiyun.com/hq572241670/article/details/50478418hq572241670Thu, 07 Jan 2016 19:35:10 +0800https://blog.youkuaiyun.com/hq572241670/article/details/50465966https://blog.youkuaiyun.com/hq572241670/article/details/50465966hq572241670Tue, 05 Jan 2016 23:15:36 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550853https://blog.youkuaiyun.com/hq572241670/article/details/41550853hq572241670Thu, 27 Nov 2014 18:47:19 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550851https://blog.youkuaiyun.com/hq572241670/article/details/41550851hq572241670Thu, 27 Nov 2014 18:47:17 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550849https://blog.youkuaiyun.com/hq572241670/article/details/41550849hq572241670Thu, 27 Nov 2014 18:47:14 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550847https://blog.youkuaiyun.com/hq572241670/article/details/41550847hq572241670Thu, 27 Nov 2014 18:47:11 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550845https://blog.youkuaiyun.com/hq572241670/article/details/41550845hq572241670Thu, 27 Nov 2014 18:47:09 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550843https://blog.youkuaiyun.com/hq572241670/article/details/41550843hq572241670Thu, 27 Nov 2014 18:47:06 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550841https://blog.youkuaiyun.com/hq572241670/article/details/41550841hq572241670Thu, 27 Nov 2014 18:47:04 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550839https://blog.youkuaiyun.com/hq572241670/article/details/41550839hq572241670Thu, 27 Nov 2014 18:47:01 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550835https://blog.youkuaiyun.com/hq572241670/article/details/41550835hq572241670Thu, 27 Nov 2014 18:46:56 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550833https://blog.youkuaiyun.com/hq572241670/article/details/41550833hq572241670Thu, 27 Nov 2014 18:46:54 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550831https://blog.youkuaiyun.com/hq572241670/article/details/41550831hq572241670Thu, 27 Nov 2014 18:46:51 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550829https://blog.youkuaiyun.com/hq572241670/article/details/41550829hq572241670Thu, 27 Nov 2014 18:46:48 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550825https://blog.youkuaiyun.com/hq572241670/article/details/41550825hq572241670Thu, 27 Nov 2014 18:46:46 +0800=k， 然后分成m个可空组，问有多少种情况。 解法： 先看从n个数中取r个相差>=k的数的方法数，可以发现 dp[i][j] = dp[1][j-1] + dp[2][j-1] + ... + dp[i-k][j-1]，(dp[i][1] = i)  即维护一个前缀和即可，可以在O(r*n)内得出。 然后n个不同的数分成m个可以空的组，即为第]]>https://blog.youkuaiyun.com/hq572241670/article/details/41550823https://blog.youkuaiyun.com/hq572241670/article/details/41550823hq572241670Thu, 27 Nov 2014 18:46:43 +0800name的映射，每进入一层括号，使father = now， 遇到右括号')'，则father = fa[father]，用vector存每个节点的子节点，然后最后dfs输出即可。 代码： #include #]]>https://blog.youkuaiyun.com/hq572241670/article/details/41550821https://blog.youkuaiyun.com/hq572241670/article/details/41550821hq572241670Thu, 27 Nov 2014 18:46:40 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550819https://blog.youkuaiyun.com/hq572241670/article/details/41550819hq572241670Thu, 27 Nov 2014 18:46:38 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550817https://blog.youkuaiyun.com/hq572241670/article/details/41550817hq572241670Thu, 27 Nov 2014 18:46:35 +0800https://blog.youkuaiyun.com/hq572241670/article/details/41550815https://blog.youkuaiyun.com/hq572241670/article/details/41550815hq572241670Thu, 27 Nov 2014 18:46:33 +0800