A Simple Problem with Integers

Time Limit: 5000MS  Memory Limit: 131072K
Total Submissions: 108903   Accepted: 33919
Case Time Limit: 2000MS

Description

You have N integers, A1A2, ... , 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 A1A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of AaAa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of AaAa+1, ... , Ab.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
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

4
55
9
15

Hint

The sums may exceed the range of 32-bit integers.

Source

 
带lazy的线段树
 //2017-05-17
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long
#define mid ((st[id].l+st[id].r)>>1)
#define lson (id<<1)
#define rson ((id<<1)|1) using namespace std; const ll N = ;
ll arr[N];
struct Node{
ll l, r, sum, lazy;
}st[N<<]; void build(int id, int l, int r)
{
st[id].l = l; st[id].r = r; st[id].lazy = ;
if(l == r){
st[id].sum = arr[l];
return;
}
build(lson, l, mid);
build(rson, mid+, r);
st[id].sum = st[lson].sum+st[rson].sum;
} void push_down(int id)
{
if(st[id].lazy != ){
st[lson].lazy += st[id].lazy;
st[rson].lazy += st[id].lazy;
st[id].sum += (st[id].r-st[id].l+)*st[id].lazy;
st[id].lazy = ;
}
return;
} ll query(int id, int l, int r)
{
if(st[id].l == l && st[id].r == r)return st[id].sum+(r-l+)*st[id].lazy;
push_down(id);
if(r <= mid)return query(lson, l, r);
else if(l > mid)return query(rson, l, r);
else return query(lson, l, mid)+query(rson, mid+, r);
} void update(int id, int l, int r, int w)
{
if(st[id].l == l && st[id].r == r){
st[id].lazy += w;
return;
}
st[id].sum += (r-l+)*w;
if(r <= mid)update(lson, l, r, w);
else if(l > mid)update(rson, l, r, w);
else{
update(lson, l, mid, w);
update(rson, mid+, r, w);
}
} int main()
{
ll n, q;
while(scanf("%lld%lld", &n, &q)!=EOF){
for(int i = ; i <= n; i++)
scanf("%lld", &arr[i]);
build(, , n);
char op[];
ll a, b, c;
while(q--){
scanf("%s", op);
if(op[] == 'Q'){
scanf("%lld%lld", &a, &b);
printf("%lld\n", query(, a, b));
}else if(op[] == 'C'){
scanf("%lld%lld%lld", &a, &b, &c);
update(, a, b, c);
}
}
} return ;
}
 //2018-03-28

 import java.util.*;

 public class Main {

     public static void main(String[] args) {
Scanner cin = new Scanner(System.in); int n, q;
while(cin.hasNext()) {
n = cin.nextInt();
q = cin.nextInt();
SegmentTree st = new SegmentTree(n);
for(int i = 1; i <= n; i++)
st.arr[i] = cin.nextLong();
st.build(1, 1, n);
char op;
int a, b;
long c;
while(q-- > 0) {
op = cin.next().charAt(0);
if(op == 'Q') {
a = cin.nextInt();
b = cin.nextInt();
System.out.println(st.query(1, a, b));
}else if(op == 'C') {
a = cin.nextInt();
b = cin.nextInt();
c = cin.nextLong();
st.updata(1, a, b, c);
}
}
}
}
} class SegmentTree{
static class Node{
public int l, r;
long sum, lazy;
Node(int _l, int _r, long _sum, long _lazy){
this.l = _l;
this.r = _r;
this.sum = _sum;
this.lazy = _lazy;
}
} public int n;
public long [] arr;
public Node [] nodes; SegmentTree(int _n){
this.n = _n;
arr = new long[n+1];
nodes = new Node[n<<2];
} int lson(int id) {
return (id<<1);
} int rson(int id) {
return ((id<<1)|1);
} int mid(int id) {
return (nodes[id].l + nodes[id].r)>>1;
}
void build(int id, int l, int r) {
nodes[id] = new Node(l, r, 0, 0);
if(l == r) {
nodes[id].sum = arr[l];
return;
}
build(lson(id), l, mid(id));
build(rson(id), mid(id)+1, r);
nodes[id].sum = nodes[lson(id)].sum + nodes[rson(id)].sum;
} void pushDown(int id) {
if(nodes[id].lazy != 0) {
nodes[lson(id)].lazy += nodes[id].lazy;
nodes[rson(id)].lazy += nodes[id].lazy;
nodes[id].sum += (nodes[id].r-nodes[id].l+1)*nodes[id].lazy;
nodes[id].lazy = 0;
}
} long query(int id, int l, int r) {
if(nodes[id].l == l && nodes[id].r == r)
return nodes[id].sum+(r-l+1)*nodes[id].lazy;
pushDown(id);
if(r <= mid(id))return query(lson(id), l, r);
else if(l > mid(id))return query(rson(id), l, r);
else return query(lson(id), l, mid(id))+query(rson(id), mid(id)+1, r);
} void updata(int id, int l, int r, long w) {
if(nodes[id] == null)return;
if(nodes[id].l == l && nodes[id].r == r) {
nodes[id].lazy += w;
return;
}
nodes[id].sum += (r-l+1)*w;
if(r <= mid(id))updata(lson(id), l, r, w);
else if(l > mid(id))updata(rson(id), l, r, w);
else {
updata(lson(id), l, mid(id), w);
updata(rson(id), mid(id)+1, r, w);
}
}
}

POJ3468(KB7-C 线段树)的更多相关文章

  1. poj3468 线段树的懒惰标记

    题目链接:poj3468 题意:给定一段数组,有两种操作,一种是给某段区间加c,另一种是查询一段区间的和 思路:暴力的方法是每次都给这段区间的点加c,查询也遍历一遍区间,复杂度是n*n,肯定过不去,另 ...

  2. poj3468 A Simple Problem with Integers(线段树区间更新)

    https://vjudge.net/problem/POJ-3468 线段树区间更新(lazy数组)模板题 #include<iostream> #include<cstdio&g ...

  3. 线段树---poj3468 A Simple Problem with Integers:成段增减:区间求和

    poj3468 A Simple Problem with Integers 题意:O(-1) 思路:O(-1) 线段树功能:update:成段增减 query:区间求和 Sample Input 1 ...

  4. 洛谷P3372/poj3468(线段树lazy_tag)(询问区间和,支持区间修改)

    洛谷P3372 //线段树 询问区间和,支持区间修改 #include <cstdio> using namespace std; struct treetype { int l,r; l ...

  5. POJ3468 线段树(区间更新,区间求和,延迟标记)

    A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 97196   ...

  6. 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   ...

  7. POJ3468 A Simple Problem with Integers(线段树延时标记)

    题目地址http://poj.org/problem?id=3468 题目大意很简单,有两个操作,一个 Q a, b 查询区间[a, b]的和 C a, b, c让区间[a, b] 的每一个数+c 第 ...

  8. 【POJ3468】【zkw线段树】A Simple Problem with Integers

    Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. On ...

  9. poj3468 线段树+lazy标记

    A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 92921   ...

  10. poj3468 A Simple Problem with Integers(线段树模板 功能:区间增减,区间求和)

    转载请注明出处:http://blog.csdn.net/u012860063 Description You have N integers, A1, A2, ... , AN. You need ...

随机推荐

  1. python -m SimpleHTTPServer 8080

    启动一个简单的 web 服务器 python -m SimpleHTTPServer 8080

  2. 2018 Multi-University Training Contest 6

    A.oval-and-rectangle 题目传送门:http://acm.hdu.edu.cn/showproblem.php?pid=6362 题意:在长半轴为a,短半轴为b的椭圆内部,以y=c( ...

  3. centos7上mysql5.6版本主从复制

    做主从复制实验: 第一步:主服务器上操作 1.修改主服务器master: [root@localhost ~]# vim /etc/my.cnf server_id = 1  //[必须]服务器唯一I ...

  4. CSP攻略

    看完三篇文章应该就懂了csp是干嘛的. https://www.cnblogs.com/Wayou/p/intro_to_content_security_policy.html https://ww ...

  5. MapReducer

    MapReducer    概述        是一个分布式的计算框架(编程模型),最初由由谷歌的工程师开发,基于GFS的分布式计算框架.后来Cutting根据<Google Mapreduce ...

  6. 遗忘Windows Server 2008R2密码的处理方法

    遗忘Windows Server 2008R2的处理方法 有的时候,我们会由于各种原因忘掉了服务器密码,比如服务器太多,太杂什么的,或直接是被人黑掉了,这个时候我们想要登录,发现我们已经没有办法了,其 ...

  7. Window 上安装Node.js

    Window 上安装Node.js 1.Windows安装包下载地址 https://nodejs.org/en/ 2.下载好后,按照默认设置即可,安装路径可以更改 3.安装好后,检测PATH环境变量 ...

  8. Mysql 和 Postgresql 抛开性能的对比

    MySQL/MariaDB的当前版本是5.7.6(MariaDB为MySQL创建者Monty Widenius创建的一个MySQL分支),PostgreSQL的版本是9.4.1. 以下几个方面对比两者 ...

  9. Linux下dmesg命令处理故障和收集系统信息的7种用法

    目录: <syslog之一:Linux syslog日志系统详解> <syslog之二:syslog协议及rsyslog服务全解析> <syslog之三:建立Window ...

  10. spring mvc 数据格式化

    web.xml <web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www. ...