poj-3468-A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 47659 Accepted: 14013Case 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 poj_3468_线段树 {                     //Accepted 13612K 10938MS Java 2657B 2013-08-19 20:13:19  static int []arr= new int[100010]; static int N,Q; static Tree []tree = new Tree[300000]; private static void build(int p, int left, int right ) {tree[p]=new Tree();///////////////////////////////int v,mid;tree[p].l = left;tree[p].r = right;tree[p].add = 0;if(left==right){tree[p].value = arr[left];return ;}mid = (left+right)>>1;//找分界点，进行分界v = p<<1;build( v,left,mid );build( v+1,mid+1,right );tree[p].value = tree[v].value + tree[v+1].value;}private static void update(int p, int left, int right, long add) {int v,mid;if(tree[p].l == left && tree[p].r == right){tree[p].add += add;return ;}tree[p].value += (right-left+1)*add;v = p<<1;mid = ( tree[p].l + tree[p].r )>>1;if( right <= mid )update(v,left,right,add);else if(left >= mid+1 )update( v+1,left,right,add );else{update(v,left,mid,add);update(v+1,mid+1,right,add);}}private static long query(int p, int left, int right) {int v,mid;v = p<<1;mid =(tree[p].l + tree[p].r)>>1;if (tree[p].l == left && tree[p].r ==right)return (tree[p].value + tree[p].add * ( tree[p].r-tree[p].l+1) );else{tree[v].add += tree[p].add;tree[v+1].add += tree[p].add;//增量下移 tree[p].value += (tree[p].r-tree[p].l+1)*tree[p].add;        tree[p].add=0;}if(right<=mid)        return query(v,left,right);    else if(left>=mid+1)        return query(v+1,left,right);    else        return query(v,left,mid)+query(v+1,mid+1,right);}public static void main(String[] args) {              Scanner sc =new Scanner (System.in);            while(sc.hasNext()){             N = sc.nextInt();                 Q = sc.nextInt();                            for(int i=1; i<=N; i++)            arr[i] = sc.nextInt();            build(1,1,N);            for(int i=1; i<=Q; i++){            String s = sc.next();            if(s.equalsIgnoreCase("C")){            int a = sc.nextInt();            int b = sc.nextInt();            int c = sc.nextInt();            update(1,a,b,c);            }else{            int a = sc.nextInt();            int b = sc.nextInt();            System.out.println(query(1,a,b));            }            }            }              }}class Tree{int l,r;long  value,add;Tree(int l, int r,long value,long add){this.l = l;this.r = r;this.value = value;this.add = add;}Tree(){} }

C语言版：

#include<stdio.h>#define N 100010int w[N];struct node  {    int l,r;    long long value, add;}tree[N*3];void build(int p, int left, int right) {    int v,mid;    tree[p].l=left;    tree[p].r=right;    tree[p].add=0;    if(left==right) {        tree[p].value=w[left];        return;    }    mid=(left+right)>>1;    v=p<<1;    build(v,left,mid);    build(v+1,mid+1,right);    tree[p].value=tree[v].value+tree[v+1].value;}void updata(int p, int left, int right, long long add) {    int v,mid;    if(tree[p].l==left&&tree[p].r==right){        tree[p].add+=add;        return;    }    tree[p].value+=(right-left+1)*add;    v=p<<1;    mid=(tree[p].l+tree[p].r)>>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);    }}long long query(int p, int left, int right) {    int mid,v;    v=p<<1;    mid=(tree[p].l+tree[p].r)>>1;        if(tree[p].l==left&&tree[p].r==right) {        return (tree[p].value+tree[p].add*(tree[p].r-tree[p].l+1));    }    else{//可能区间有增量，需要计算并下移    tree[v].add+=tree[p].add;        tree[v+1].add+=tree[p].add;//增量下移         tree[p].value+=(tree[p].r-tree[p].l+1)*tree[p].add;        tree[p].add=0;    }    if(right<=mid)        return query(v,left,right);    else if(left>=mid+1)        return query(v+1,left,right);    else        return query(v,left,mid)+query(v+1,mid+1,right);}int main() {    int n,q,i,a,b,c;    char s[3];    while(scanf("%d %d",&n,&q)!=EOF) {        for(i=1;i<=n;i++)            scanf("%d",&w[i]);        build(1,1,n);        for(i=0;i<q;i++) {            scanf("%s",s);            if(s[0]=='C') {                scanf("%d %d %d",&a, &b, &c);                updata(1,a,b,c);            }            else {                scanf("%d %d",&a,&b);                printf("%I64d\n",query(1,a,b));            }        }    }    return 0;}