C - A Simple Problem with Integers
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 100010
using namespace std;
int n,m;
struct nond{
int l,r;
long long sum,flag;
}tree[MAXN*];
void up(int now){
tree[now].sum=tree[now*].sum+tree[now*+].sum;
}
void build(int now,int l,int r){
tree[now].l=l;tree[now].r=r;
if(tree[now].l==tree[now].r){
scanf("%lld",&tree[now].sum);
return ;
}
int mid=(tree[now].l+tree[now].r)/;
build(now*,l,mid);
build(now*+,mid+,r);
up(now);
}
void down(int now){
tree[now*].flag+=tree[now].flag;
tree[now*].sum+=(tree[now*].r-tree[now*].l+)*tree[now].flag;
tree[now*+].flag+=tree[now].flag;
tree[now*+].sum+=(tree[now*+].r-tree[now*+].l+)*tree[now].flag;
tree[now].flag=;
}
void change(int now,int l,int r,int k){
if(tree[now].l==l&&tree[now].r==r){
tree[now].sum+=(tree[now].r-tree[now].l+)*k;
tree[now].flag+=k;
return ;
}
if(tree[now].flag) down(now);
int mid=(tree[now].l+tree[now].r)/;
if(r<=mid) change(now*,l,r,k);
else if(l>mid) change(now*+,l,r,k);
else{ change(now*,l,mid,k);change(now*+,mid+,r,k); }
up(now);
}
long long query(int now,int l,int r){
if(tree[now].l==l&&tree[now].r==r)
return tree[now].sum;
if(tree[now].flag) down(now);
int mid=(tree[now].l+tree[now].r)/;
if(r<=mid) return query(now*,l,r);
else if(l>mid) return query(now*+,l,r);
else return query(now*,l,mid)+query(now*+,mid+,r);
}
int main(){
scanf("%d%d",&n,&m);
build(,,n);
for(int i=;i<=m;i++){
char c;int x,y,z;
scanf("\n%c%d%d",&c,&x,&y);
if(c=='Q') cout<<query(,x,y)<<endl;
else if(c=='C'){ scanf("%d",&z);change(,x,y,z); }
}
}
C - A Simple Problem with Integers的更多相关文章
- POJ 3468 A Simple Problem with Integers(线段树 成段增减+区间求和)
A Simple Problem with Integers [题目链接]A Simple Problem with Integers [题目类型]线段树 成段增减+区间求和 &题解: 线段树 ...
- POJ 3468 A Simple Problem with Integers(线段树/区间更新)
题目链接: 传送门 A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Description Yo ...
- poj 3468:A Simple Problem with Integers(线段树,区间修改求和)
A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 58269 ...
- ACM: A Simple Problem with Integers 解题报告-线段树
A Simple Problem with Integers Time Limit:5000MS Memory Limit:131072KB 64bit IO Format:%lld & %l ...
- poj3468 A Simple Problem with Integers (线段树区间最大值)
A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 92127 ...
- POJ3648 A Simple Problem with Integers(线段树之成段更新。入门题)
A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 53169 Acc ...
- BZOJ-3212 Pku3468 A Simple Problem with Integers 裸线段树区间维护查询
3212: Pku3468 A Simple Problem with Integers Time Limit: 1 Sec Memory Limit: 128 MB Submit: 1278 Sol ...
- POJ 3468 A Simple Problem with Integers(线段树区间更新区间查询)
A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 92632 ...
- A Simple Problem with Integers(树状数组HDU4267)
A Simple Problem with Integers Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (J ...
- A Simple Problem with Integers
A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 77964 Acc ...
随机推荐
- ElasticSearch源码解析(五):排序(评分公式)
ElasticSearch源码解析(五):排序(评分公式) 转载自:http://blog.csdn.net/molong1208/article/details/50623948 一.目的 一个 ...
- 2017 Multi-University Training Contest - Team 2 &hdu 6055 Regular polygon
Regular polygon Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)T ...
- CodeForces--626C--Block Towers (二分)
Block Towers Time Limit: 2000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submit ...
- sql将一个表中的数据插入到另一个表中
sql将一个表中的数据插入到另一个表中 列名不一定要相同,只要你在HH中列出要插入列的列表跟select from mm表中的选择的列的列表一一对应就可以了,当然两边的数据类型应该是兼容的. ...
- bzoj1588: [HNOI2002]营业额统计(权值线段树)
1588: [HNOI2002]营业额统计 Time Limit: 5 Sec Memory Limit: 162 MBSubmit: 16863 Solved: 6789[Submit][Sta ...
- grunt的学习和使用
目前正在编写公司的部分组件,可能一个组件会包含很多js和css,为了项目上使用方便,应该压缩成一个js库,以供开发者使用,同时也可以减少很多http请求,提高页面访问速度.基于此,学习了grunt自动 ...
- SfMLearner论文笔记——Unsupervised Learning of Depth and Ego-Motion from Video
1. Abstract 提出了一种无监督单目深度估计和相机运动估计的框架 利用视觉合成作为监督信息,使用端到端的方式学习 网络分为两部分(严格意义上是三个) 单目深度估计 多视图姿态估计 解释性网络( ...
- HTML 打印 换页
打印 HTML 无法强制换页其实是一件很令人困扰的事,要达到这个功能其实可以透过 CSS 的 Pagebreak 来处理. 强制分页有大概只有二种用的到: { page-break-after: al ...
- Django分页器及自定义分页器
Django的分页器 view from django.shortcuts import render,HttpResponse # Create your views here. from app0 ...
- C#之考勤系统
闲来无聊,搞搞C#,下面就是我写的一个Demo 员工类 using System; using System.Collections.Generic; using System.Linq; using ...