一开始把题目看错了,弄得很复杂,后来发现就是简单的求体积并问题。#include <cstdio>#include <cstring>#include <iostream>#include <map>#include <se
https://www.u72.net/daima/08nm.html - 2024-07-18 12:57:14 - 代码库又做了一题扫描线以后对节点的覆盖标记理解的更加深刻了。 代码如下: 1 #include <stdio.h> 2 #include <algorithm> 3 #include <strin
https://www.u72.net/daima/2rn9.html - 2024-09-01 04:33:16 - 代码库题意:给出n个矩形,求矩形中被覆盖K次以上的面积的和。解法:整体与求矩形面积并差不多,不过在更新pushup改变len的时候,要有一层循环,来更新tree[rt].len[i],其
https://www.u72.net/daima/2dvn.html - 2024-07-19 22:47:44 - 代码库CoderTime Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3183 Accepted Submission(
https://www.u72.net/daima/3k2b.html - 2024-07-20 23:21:34 - 代码库看看type = 2的操作,对于区间[l,r]内的元素对x取模,由于取模肯定不能和取模,所以只能每个元素取模,看上去不是区间更新,但是仔细一看,若区间[l,r]内所有的元
https://www.u72.net/daima/3zhx.html - 2024-07-20 20:56:20 - 代码库3295: [Cqoi2011]动态逆序对Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 861 Solved: 287[Submit][Status]Description对于序列A,它的逆序对数
https://www.u72.net/daima/3nu4.html - 2024-07-20 20:12:55 - 代码库【题目链接】 http://poj.org/problem?id=1769 【题目大意】 给出一些排序器,能够将区间li到ri进行排序,排序器按一定顺序摆放 问在排序器顺
https://www.u72.net/daima/3nek.html - 2024-09-02 10:26:45 - 代码库题目大意:有n个区间,求k个区间,使得这k个区间相交的区间内数字之和最大。数列的数字均>=0优先队列思路:按照左端点sort,然后枚举左端点,假设他被覆盖过k次,
https://www.u72.net/daima/3bk6.html - 2024-09-02 16:54:40 - 代码库Problem DescriptionThe God of sheep decides to pixelate some pictures (i.e., change them into pictures with mosaic). Here‘s how he is gonna
https://www.u72.net/daima/c2uc.html - 2024-07-11 06:26:31 - 代码库题目大意:1 l r x操作 讲 [l,r]上的节点涂成x颜色,并且每个节点的&#20540;都加上 |y-x| y为涂之前的颜色2 l r 操作,求出[l,r]上的和。思路分析:如果一个
https://www.u72.net/daima/c26u.html - 2024-07-11 06:49:36 - 代码库聪哥推荐的题目区间修改和区间查询,但是此题新颖之处就在于他的区间修改不是个定值,而是从L 到 R 分别加 F1、F2、。。。Fr-l+1 (F为斐波那契数列)想了一下
https://www.u72.net/daima/rx47.html - 2024-07-12 04:05:18 - 代码库posters时间限制:1000 ms | 内存限制:65535 KB难度:6 描述The citizens of Bytetown, AB, could not stand that the candidates in the mayoral electi
https://www.u72.net/daima/u50h.html - 2024-07-14 11:39:56 - 代码库1 /* 2 之前的思想是用回溯的方式进行颜色的更新的!如果用回溯的方法的话,就是将每一个节点的颜色都要更新 3 通过子节点的颜色情况来判断父节点的颜
https://www.u72.net/daima/u3db.html - 2024-07-14 09:40:54 - 代码库Problem DescriptionThere are n types of cells in the lab, numbered from 1 to n. These cells are put in a queue, the i-th cell belongs to typ
https://www.u72.net/daima/1069.html - 2024-07-19 07:08:46 - 代码库A Simple Problem with IntegersTime Limit: 5000MS Memory Limit: 131072KTotal Submissions: 62431 Accepted: 19141Case Time Limit: 2000MSDescrip
https://www.u72.net/daima/11wc.html - 2024-07-19 07:49:05 - 代码库http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1859Matrix SearchingTime Limit: 10 Seconds Memory Limit: 32768 KBGiven an n*
https://www.u72.net/daima/1w30.html - 2024-07-19 05:06:17 - 代码库把区间和改成,第一层|,第二层 ^。每次给出一个x,y把 第x个变成y ,输出 sum[1];#include <cstdio>#include <cstring>#include <algorithm>#include <clim
https://www.u72.net/daima/xu2w.html - 2024-07-17 03:37:04 - 代码库题目大意: 给你10000以内的星星的坐标和亮度,让你用一个W × H 的矩形去围住一个区域,使得区域内星星的亮度最大,矩形边缘上的星星不算。解题思路: 对于每
https://www.u72.net/daima/183s.html - 2024-07-19 14:36:30 - 代码库Problem DescriptionThere are n types of cells in the lab, numbered from 1 to n. These cells are put in a queue, the i-th cell belongs to typ
https://www.u72.net/daima/185d.html - 2024-07-19 14:40:35 - 代码库http://poj.org/problem?id=2482Stars in Your WindowTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9909 Accepted: 2743DescriptionFl
https://www.u72.net/daima/1c34.html - 2024-07-19 00:24:39 - 代码库