区间更新求和 主要用来练习splay树区间更新问题 //splay树的题解 // File Name: 3468-splay.cpp // Author: Zlbing // Created Time: 2013年08月09日 星期五 16时30分32秒 #include<iostream> #include<string> #include<algorithm> #include<cstdlib> #include<cstdio> #inclu…

题目链接 题意:两种操作:一是指定区间的数全都加上一个数,二是统计指定区间的和 参考斌神的代码 #include <iostream> #include <cstring> #include <algorithm> #include <cstdio> using namespace std; ; const int INF = 0x3f3f3f3f; typedef long long LL; struct node { int l,r; LL inc; /…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 139191   Accepted: 43086 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type o…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K                Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some gi…

A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 60745   Accepted: 18522 Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of…

A Simple Problem with Integers Time Limit: 10000MS Memory Limit: 65536K Description You have N integers, A1, A2, - , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interv…

An easy problem C Time Limit: 4000/2000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others) Problem Description N个数排成一列,有三种操作.1.给一段区间内的每个数乘上一个非负整数.2.给一段区间内的每个数加上一个非负整数.3.询问一段区间的和模上P的值. Input 第一行两个整数N(1≤N≤100000)表示数的个数,P(1≤P≤1000000000)表示模的值.接下来…

http://poj.org/problem?id=3468 真心觉得这题坑死我了,一直错,怎么改也没戏,最后tjj把q[rt].lz改成了long long 就对了,真心坑啊. 线段树的区域更新. 线段树功能:update:成段增减 query:区间求和 #include <iostream> #include <stdio.h> #include <string.h> #include <stdlib.h> using namespace std; #d…

http://poj.org/problem?id=3468 _(:зゝ∠)_我又活着回来啦,前段时间太忙了写的题没时间扔上来,以后再说. [问题描述] 成段加某一个值,然后询问区间和. [思路] 讲一下pushdown和pushup出现的几个位置. pushup: (1)build的结尾,当叶子节点分别有对应的值后,它的父亲们就等于它们求和. (2)update的结尾,因为此时当前根的左右孩子已经更新了,故它也需要更新. pushdown(延迟标记): *pushdown会出现在一切要从当前结…

Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval. In…