oval-and-rectangle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 237    Accepted Submission(s): 93 Problem Description Patrick Star find an oval. The half of longer axes is on the x-axis with…

Pinball Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 568    Accepted Submission(s): 244 Problem Description There is a slope on the 2D plane. The lowest point of the slope is at the origin.…

Problem E. Matrix from Arrays Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1384    Accepted Submission(s): 630 Problem Description Kazari has an array A length of L, she plans to generate a…

Everything Has Changed Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 292    Accepted Submission(s): 155Special Judge Problem Description Edward is a worker for Aluminum Cyclic Machinery. His…

Beautiful Now Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Anton has a positive integer n, however, it quite looks like a mess, so he want…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6667 题意: 有 n个班级,每个班级有a个人.b个奶茶,每个班的人不能喝自己的奶茶,只能喝别人班的奶茶,问你最多有多少人喝到奶茶. 思路: 明显一道贪心题: n=3 3   4  min=3 4   2  min=2 4   2  min=2 首先根据min值从大到小排序(先消除大的更优),两个min值是肯定可以消掉一个的,ans+=min*2,而且消到最后最多只有一个a.b都不为0. 然后就是要先…

Description Mr Wang wants some boys to help him with a project. Because the project is rather complex, the more boys come, the better it will be. Of course there are certain requirements. Mr Wang selected a room big enough to hold the boys. The boy w…

Description Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay…

链接:https://www.nowcoder.com/acm/contest/147/E 来源:牛客网 题目描述 Niuniu likes to play OSU! We simplify the game OSU to the following problem. Given n and m, there are n clicks. Each click may success or fail. For a continuous success sequence with length X,…

以前我们学习了线段树可以知道,线段树的每一个节点都储存的是一段区间,所以线段树可以做简单的区间查询,更改等简单的操作. 而后面再做有些题目,就可能会碰到一种回退的操作.这里的回退是指回到未做各种操作之前的状态. 回退的时候,如果暴力点,就直接将每步所操作的线段树都存下来,然后直接翻阅回去,这种方法虽然简单,但是对空间和时间的需求太大了,肯定不能过. 所以这时候我们就可以选择可持久化操作. 可持久化是数据结构里面的一种方法,其总体就是把一个数据结构的历史状态全部都保存下来,从而能够快速的查找之前出…