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题意：就是一个集合分开，有两种区分 对子：两个相同数字，顺子：连续三个不同数字，问最多分多少个

解法：贪心，如果当前数字不构成顺子就取对子 /2，如果可以取顺子，那么先取顺子再取对子

窗口是屏幕上的一个矩形区域。窗口分为3种：重叠窗口、弹出窗口和子窗口。每个窗口都有由系统绘制的“非客户区”和应用程序绘制的“客户区”。在 MFC 中，CWnd 类为各种窗口提供了基类。1 通过 HWND 获得 CWnd 指针通过 HWND 获得 Cwnd 指针可以调用

题意：给你n个数，问你最多能组成多少顺子和对子题解：和bzoj上的一道题类似，不过bzoj上对子是3张牌，所以无脑先选对子即可，然而这道题对子是两张牌，因此之前的贪心还是有些问题的，比如： 1 2 3 3 4 5 ，答案应该是2,。但是我们仍然可以考虑先选对子，然后让每种牌最多留下一个对子（可以自己证明，留的多了没用）然后我们判断留下的对子能否组成两个顺子即可，当然遍历的过程中顺子

Monkey A lives on a tree, he alw

Little A has come to college and majoYou are giving ...

Do you like painting? Little D doesn’t like painting, especially mes

Little A is an astronomy lover, and he has found that the sky was so beautiful! So he is counting stars now! There are n stars in the sky, and little A has connected them by m non-directional edge...

A。dp但是m次数太多，开不出那么大的数组，所以要用算完的值替代新的值，这样来回循环，滚动数组，我用的方法比较麻烦，看到有%2的，也有^1的我写的是每一次从0到n循环一边，能到的话概率相加还有一种思路是当前的概率至与上一次的概率有关，每一次都是dp[i][j] = dp[i-1][ (j-x+n)%n ] + dp[i][  (j+x)%n ]I题就是硬暴，开始想用dfs，发现不对

题意：给你4*n的矩形，用1*2的方格铺满。题解：经典问题，我们可以先用公式找出前几项，然后求递推式即可，但是n很大，所以要快速幂搞搞。我们可以由前几项推得递推公式为：f[i]=f[i-1]+5*f[i-2]+f[i-3]-f[i-4];因此我们可以构造矩阵为：1 5 1 -11 0 0 00 1 0 00 0 1 0       然后矩阵快速幂求解就好啦。

Problem Description Nike likes playing cards and makes a problem of it. Now give you n integers, ai(1≤i≤n) We define two identical numbers (eg: 2,2) a

题意：删除指定数字，求剩下数字的与或非值 解法：保存一下前缀和后缀

解法：发现。。最多15个，那么。。暴力一下啦

Problem Description Bob's school has a big playground, boys and girls always play games here after school. To protect boys and girls from getting hurt

Problem Description Monkey A lives on a tree, he always plays on this tree. One day, monkey A learned about one of the bit operations, xor. He was kee

Covering Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 187 Accepted Submission(s): 107 Problem

CS Course Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0 Problem Des