BZOJ3212: Pku3468 A Simple Problem with Integers(线段树)
3212: Pku3468 A Simple Problem with Integers
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 2530 Solved: 1096
[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 number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4
Sample Output
55
9
15
HINT
The sums may exceed the range of 32-bit integers.
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define ll long long
using namespace std;
const int maxn=;
int Lazy[maxn]; ll sum[maxn];
void pushup(int Now){ sum[Now]=sum[Now<<]+sum[Now<<|];}
void pushdown(int Now,int L,int R){
if(Lazy[Now]){
int Mid=(L+R)>>;
sum[Now<<]+=(ll)Lazy[Now]*(Mid-L+);
sum[Now<<|]+=(ll)Lazy[Now]*(R-Mid);
Lazy[Now<<]+=Lazy[Now];
Lazy[Now<<|]+=Lazy[Now];
Lazy[Now]=;
}
}
void build(int Now,int L,int R)
{
if(L==R) {
scanf("%lld",&sum[Now]);
return ;
}
int Mid=(L+R)>>;
build(Now<<,L,Mid); build(Now<<|,Mid+,R);
pushup(Now);
}
ll query(int Now,int L,int R,int l,int r)
{
if(l<=L&&r>=R) return sum[Now];
int Mid=(L+R)>>; ll res=;
pushdown(Now,L,R);
if(l<=Mid) res+=query(Now<<,L,Mid,l,r);
if(r>Mid) res+=query(Now<<|,Mid+,R,l,r);
pushup(Now);
return res;
}
void update(int Now,int L,int R,int l,int r,int c)
{
if(l<=L&&r>=R){ sum[Now]+=(ll)(R-L+)*c; Lazy[Now]+=c; return ;}
int Mid=(L+R)>>; pushdown(Now,L,R);
if(l<=Mid) update(Now<<,L,Mid,l,r,c);
if(r>Mid) update(Now<<|,Mid+,R,l,r,c);
pushup(Now);
}
int main()
{
int N,M,L,R,x; char opt[];
scanf("%d%d",&N,&M);
build(,,N);
rep(i,,M){
scanf("%s%d%d",opt,&L,&R);
if(opt[]=='Q') printf("%lld\n",query(,,N,L,R));
else scanf("%d",&x),update(,,N,L,R,x);
}
return ;
}
BZOJ3212: Pku3468 A Simple Problem with Integers(线段树)的更多相关文章
- bzoj3212 Pku3468 A Simple Problem with Integers 线段树
3212: Pku3468 A Simple Problem with Integers Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 2046 So ...
- bzoj 3212 Pku3468 A Simple Problem with Integers 线段树基本操作
Pku3468 A Simple Problem with Integers Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 2173 Solved: ...
- 2018 ACMICPC上海大都会赛重现赛 H - A Simple Problem with Integers (线段树,循环节)
2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 H - A Simple Problem with Integers (线段树,循环节) 链接:https://ac.nowcoder.co ...
- 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 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 ...
- 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 ...
- poj 3468 A Simple Problem with Integers 线段树第一次 + 讲解
A Simple Problem with Integers Description You have N integers, A1, A2, ... , AN. You need to deal w ...
- Poj 3468-A Simple Problem with Integers 线段树,树状数组
题目:http://poj.org/problem?id=3468 A Simple Problem with Integers Time Limit: 5000MS Memory Limit ...
随机推荐
- 5.3 Components — Passing Properties to A Component
1. 默认情况下,一个组件在它使用的模板范围中没有访问属性. 例如,假想你有一个blog-post组件被用来展示一个blog post: app/templates/components/blog-p ...
- cocos代码研究(22)Widget子类Layout学习笔记
理论基础 一个包含控件的容器. 子节点可以根据布局类型重新排序,它还可以开启剪裁,设置背景图像和颜色.继承自Widget,以及LayoutProtocol. 被 HBox, PageView, Rel ...
- sql server 将时间中的时分秒改为00:00:00
select convert(varchar(10),getdate(),120
- AtCoder Grand Contest 029 Solution
A: Solved. 签. #include <bits/stdc++.h> using namespace std; #define ll long long #define N 200 ...
- 2018 Multi-University Training Contest 6 Solution
A - oval-and-rectangle 题意:给出一个椭圆的a 和 b,在$[0, b]中随机选择c$ 使得四个顶点在椭圆上构成一个矩形,求矩形周长期望 思路:求出每种矩形的周长,除以b(积分) ...
- nodejs+express工程 在npm install之后或使用npm install bootstrap命令安装bootstrap之后
nodejs+express工程 在npm install之后或使用npm install bootstrap命令安装bootstrap之后引入bootstrap文件 如果你的静态资源存放在多个目录下 ...
- Python笔记 #16# Pandas: Operations
10 Minutes to pandas #Stats # shift 这玩意儿有啥用??? s = pd.Series([1,5,np.nan], index=dates).shift(0) # s ...
- Python3:爬取新浪、网易、今日头条、UC四大网站新闻标题及内容
Python3:爬取新浪.网易.今日头条.UC四大网站新闻标题及内容 以爬取相应网站的社会新闻内容为例: 一.新浪: 新浪网的新闻比较好爬取,我是用BeautifulSoup直接解析的,它并没有使用J ...
- Java实现文件上传到服务器(FTP方式)
Java实现文件上传到服务器(FTP方式) 1,jar包:commons-net-3.3.jar 2,实现代码: //FTP传输到数据库服务器 private boolean uploadServer ...
- 20145104张家明 《Java程序设计》第3周学习总结
20145104张家明 <Java程序设计>第4周学习总结 教材学习内容总结 第四章 认识对象 4.1 类与对象 4.1.1 定义类 类定义时使用class关键词,建立实例要用new关键词 ...