poj-3468-A Simple Problem with Integers(线段树)

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 47608 Accepted: 13992Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... ,A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... ,A_b.

Output

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

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

import java.util.Scanner;public class Main {/** * @param args */static int[] w;static Node[] tree;public static void main(String[] args) {// TODO Auto-generated method stub        Scanner input=new Scanner(System.in);        while(input.hasNext()){        int N=input.nextInt();        int Q=input.nextInt();        w=new int[N+1];        tree=new Node[N*3];        for(int i=1;i<N+1;i++)        w[i]=input.nextInt();        Build(1,1,N);//先建树，第一个1是数组坐标，第二个1和N是区间范围        for(int i=0;i<Q;i++){        String s=input.next();                if(s.charAt(0)=='Q'){                int a=input.nextInt();                int b=input.nextInt();                long result=Query(1,a,b);                System.out.println(result);                }                else{                int a=input.nextInt();                int b=input.nextInt();                int c=input.nextInt();                Updata(1,a,b,c);                }        }        }}private static void Updata(int p, int left, int right, int add) {// TODO Auto-generated method stubif(tree[p].left==left && tree[p].right==right){tree[p].add+=add;return;}tree[p].value += (right-left+1)*add;int v=p<<1;int mid=(tree[p].left+tree[p].right)>>1;if(right<=mid)Updata(v,left,right,add);else if(left>=mid+1)Updata(v+1,left,right,add);else{Updata(v,left,mid,add);    Updata(v+1,mid+1,right,add);}}private static long Query(int p, int left, int right) {// TODO Auto-generated method stubint len=tree[p].right-tree[p].left+1;int v=p<<1;if(tree[p].left==left && tree[p].right==right)return tree[p].value+tree[p].add*len;else{tree[v].add+=tree[p].add;tree[v+1].add+=tree[p].add;tree[p].value+=len*tree[p].add;tree[p].add=0;}int mid=(tree[p].left+tree[p].right)>>1;if(right<=mid)return Query(v,left,right);else if(left>=mid+1)return Query(v+1,left,right);elsereturn Query(v,left,mid)+Query(v+1,mid+1,right);}private static void Build(int p, int left, int right) {// TODO Auto-generated method stubtree[p]=new Node();tree[p].left=left;tree[p].right=right;tree[p].add=0;if(left==right){tree[p].value=w[left];return;}int mid=(left+right)>>1;        int v=p<<1;        Build(v,left,mid);        Build(v+1,mid+1,right);        tree[p].value=tree[v].value+tree[v+1].value;}}class Node{int left;int right;long value;long add;//增量public int getLeft() {return left;}public void setLeft(int left) {this.left = left;}public int getRight() {return right;}public void setRight(int right) {this.right = right;}public long getValue() {return value;}public void setValue(long value) {this.value = value;}public long getAdd() {return add;}public void setAdd(long add) {this.add = add;}Node(){}}