解释:此题难点主要是如何既能做到能及时更新一片区域的数值而又不会超时(因为每次更新成绩都要费好多时间的),所以这题 就要用到线段树的精华思想了:延迟标记 延迟标记主要是帮忙标记下某段数值是否曾经更新过,如果有更新过,那么只有在访问这段区间的时候,延迟标记才会把他积累的 数值传给下面的节点,起到节省时间的...
线段树的区间修改和点修改相比多了一个add数组,add数组的作用是在进行区间更新的时候不必将和该区间有关的所有线段都更新,某个线段的add数组的含义是:该线段的子线段的sum值都应该增加add[](该线段的sum值已经增加了)但现在还没增加,也就是相当于延迟的作用某个线段的add延迟信息在要去查询该线段的子区间的时候...
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 ...
A Simple Problem with Integers Description You haveNintegers,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. Input ...
POJ - 3468 A Simple Problem with Integers 线段树 A Simple Problem with Integers Description You haveNintegers,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...
给你n个数,然后有两种操作,求某段区间的和 or 将某段区间元素加若干值。 解题思路: 线段树。。。模仿别人的。1516MS。 貌似zkw式线段树1329MS,两个树状数组可以做到1047MS,等以后再学,TAT。 #include <stdio.h> #define L(x) (x<<1) #define R(x) (x<<1|1) ...
poj 3468 A Simple Problem with Integers 简介:点击打开链接poj 3468 思路:线段树成段更新 分析: 1 最基础的线段树的成段更新的题目,我们只要建好线段树然后进行更新即可 2 注意由于输入的数最大为10^9,因此我们应该使用long long,区间的和已经区间的延时标记都要......
3212: Pku3468 A Simple Problem with Integers Time Limit: 1 Sec Memory Limit: 128 MB Submit: 1053 Solved: 468 [Submit][Status][Discuss] 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 nu...
HDU4267 A Simple Problem with Integers 线段树/树状数组 2015-09-03 21:08 −HDU4267 A Simple Problem with Integers 线段树/树状数组 2012长春网络赛A题 Problem Description Let A1, A2, ... , AN be N elements. You need to deal with two kin... ...
MODandnintegers:A0, A1, A2 ... An-1, you have to writeKnested loops and calculate the summation of allAiwhereiis the value of any nested loop variable.' Input Input starts with an integerT (≤ 100), denoting the number of test cases. Each case starts with three integers:n (1 ≤...