poj 3468 线段树模板题

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 89835		Accepted: 27972
Case 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 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

Hint

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

Source

POJ Monthly–2007.11.25, Yang Yi

题目大意：

就是给你一个数列，让你进行区间操作，c表示把a到b的数都加上c；q表示输出a到b的所有的数的总和。

#include<iostream>
#include<cstdio>
using namespace std;
const long long maxn=233333;
int xl[maxn];
struct love{
    int l,r;
    long long sum,add;
}liuwen[maxn*4];
void build(int p,int l,int r)
{
    liuwen[p].l=l;
    liuwen[p].r=r;
    if(l==r)
    {
        liuwen[p].sum=xl[l];
        return ;
    }
    int mid=(l+r)/2;
    build(p*2,l,mid);
    build(p*2+1,mid+1,r);
    liuwen[p].sum=liuwen[p*2].sum+liuwen[p*2+1].sum;
}
void spread(int p)
{
    if(liuwen[p].add){
        liuwen[p*2].sum+=(liuwen[p*2].r-liuwen[p*2].l+1)*(long long)liuwen[p].add;
        liuwen[p*2+1].sum+=(liuwen[p*2+1].r-liuwen[p*2+1].l+1)*(long long)liuwen[p].add;
        liuwen[p*2].add+=liuwen[p].add;
        liuwen[p*2+1].add+=liuwen[p].add;
        liuwen[p].add=0;
    }
}
void change(int p,int l,int r,long long x)
{
    if(l<=liuwen[p].l&&liuwen[p].r<=r)
    {
        liuwen[p].sum+=(long long)(liuwen[p].r-liuwen[p].l+1)*x;
        liuwen[p].add+=x;
        return ;
    }
    spread(p);
    int mid=(liuwen[p].l+liuwen[p].r)/2;
    if(l<=mid)
        change(p*2,l,r,x);
    if(mid< r)
        change(p*2+1,l,r,x);
    liuwen[p].sum=liuwen[p*2].sum+liuwen[p*2+1].sum;
}
long long ask(int p,int l,int r)
{
    if(l<=liuwen[p].l&&liuwen[p].r<=r)
    {
        return liuwen[p].sum;
    }
    spread(p);
    long long ans=0;
    int mid=(liuwen[p].l+liuwen[p].r)/2;
    if(l<=mid) ans+=ask(p*2,l,r);
    if(mid<r) ans+=ask(p*2+1,l,r);
    return ans;
}
int main()
{
    long long n,q;
    cin>>n>>q;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&xl[i]);
    }
    build(1,1,n);
    for(int i=1;i<=q;i++){
        char ss = getchar();
        while(ss != 'Q' && ss != 'C')
            ss = getchar();
        if(ss=='Q')
        {
            int a,b;
            scanf("%d%d",&a,&b);
            printf("%lld\n",ask(1,a,b));
        }
        if(ss=='C')
        {
            int a,b;
            long long c;
            scanf("%d%d%lld",&a,&b,&c);
            change(1,a,b,c);
        }
    }
    return 0;
}