A Simple Problem with Integers-POJ3468

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 abc" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q ab" 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.

这是线段树模板题，段更新，有一个向下压的过程。

#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdlib.h>
using namespace std;
#define N 100100
#define Lson r<<1
#define Rson r<<1|1

struct node
{
    int L,R;
    long long sum,e;
    int mid()
    {
        return (L+R)/2;

    }
    int len()
    {
        return R-L+1;
    }
}a[N<<2];

void BuildTree(int r,int L,int R)
{
    a[r].L=L;
    a[r].R=R;
    a[r].e=0;
    if(L==R)
    {
        scanf("%lld",&a[r].sum);
        return;
    }
    BuildTree(Lson,L,a[r].mid());
    BuildTree(Rson,a[r].mid()+1,R);
    a[r].sum=a[Lson].sum+a[Rson].sum;
}
void Down(int r)
{
    a[Lson].sum+=a[Lson].len()*a[r].e;
    a[Lson].e+=a[r].e;
    a[Rson].sum+=a[Rson].len()*a[r].e;
    a[Rson].e+=a[r].e;
    a[r].e=0;
}
void Add(int r,int L,int R,int e)
{
    a[r].sum+=(R-L+1)*e;
    if(a[r].L==L && a[r].R==R)
    {
        a[r].e+=e;
        return;
    }
    Down(r);
    if(R<=a[r].mid())
        Add(Lson,L,R,e);
    else if(L>a[r].mid())
        Add(Rson,L,R,e);
    else
    {
        Add(Lson,L,a[r].mid(),e);
        Add(Rson,a[r].mid()+1,R,e);
    }
}
long long Qurry(int r,int L,int R)
{
    if(a[r].L==L && a[r].R==R)
    {
        return a[r].sum;
    }
    Down(r);
    if(R<=a[r].mid())
        return Qurry(Lson,L,R);
    else if(L>a[r].mid())
        return Qurry(Rson,L,R);
    else
    {
        long long int a1=Qurry(Lson,L,a[r].mid());
        long long int a2=Qurry(Rson,a[r].mid()+1,R);
        return a1+a2;
    }
}

int main()
{
    int n,m,d,b,c;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        BuildTree(1,1,n);
        char s[10];
        while(m--)
        {
            scanf("%s",s);
            if(s[0]=='Q')
            {
                scanf("%d %d",&d,&b);
                printf("%lld
",Qurry(1,d,b));
            }
            else
            {
                scanf("%d %d %d",&d,&b,&c);
                Add(1,d,b,c);
            }
        }
    }
    return 0;
}