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.

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.

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

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

4
55
9
15

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

POJ Monthly--2007.11.25, Yang Yi


From：http://www.cnblogs.com/littlex/archive/2012/01/30/2331670.html

這題很好的體現了lazy_tag的思想，當要增加的區間覆蓋當前區間，則直接打上標記返回，當下次查詢這個區間兒子區間的時候，直接標記往下傳。可以想象，這個標記表示的是這個區間表示的整個區間的標記是整個子樹的性質，當你不需要用到這個區間的兒子區間的時候，你可以不把操作都做到底，而是要用到的時候再往下傳！

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define MAX 500000
struct node
{
    int l,r;
    long long val;//增量
    long long num;//總和
} tree[4*MAX];
long long num[MAX];
void build(int l,int r,int index)//建樹
{
    tree[index].l=l;
    tree[index].r=r;
    if(tree[index].l==tree[index].r)
    {
        tree[index].num=num[l-1];
        return ;
    }
    int mid=(l+r)/2;
    build(l,mid,2*index);
    build(mid+1,r,2*index+1);
    tree[index].num=tree[index*2].num+tree[index*2+1].num;//每個節點和
}
void update(int l,int r,int c,int index)//更新
{
    if(tree[index].l>=l&&tree[index].r<=r)
    {
        tree[index].num+=(tree[index].r-tree[index].l+1)*c;
        tree[index].val+=c;
        return ;
    }
    else
    {
        if(tree[index].val)//最關鍵部分就是這個標記下傳，將增加值val傳給他的左右孩子
        {
            tree[index*2].num+=tree[index].val*(tree[index*2].r-tree[index*2].l+1);
            tree[index*2].val+=tree[index].val;
            tree[index*2+1].num+=tree[index].val*(tree[index*2+1].r-tree[index*2+1].l+1);
            tree[index*2+1].val+=tree[index].val;
            tree[index].val=0;
        }
        if(r>=tree[index*2+1].l)
            update(l,r,c,2*index+1);
        if(l<=tree[index*2].r)
            update(l,r,c,2*index);
        tree[index].num=tree[index*2].num+tree[index*2+1].num;//每個節點和
    }
}
long long query(int a,int b,int index)//查詢
{

    if(tree[index].l>=a&&tree[index].r<=b)
        return tree[index].num;
    if(tree[index].val)         //求和同樣需要標記下傳
    {
        tree[index*2].num+=tree[index].val*(tree[index*2].r-tree[index*2].l+1);
        tree[index*2].val+=tree[index].val;
        tree[index*2+1].num+=tree[index].val*(tree[index*2+1].r-tree[index*2+1].l+1);
        tree[index*2+1].val+=tree[index].val;
        tree[index].val=0;
    }
    int mid=(tree[index].l+tree[index].r)/2;
    long long max1=0,max2=0;
    if(mid<a)
        max1= query(a,b,2*index+1);
    else if(b<=mid)
        max1=query(a,b,2*index);
    else
    {
        max1=query(a,mid,2*index);
        max2=query(mid+1,b,2*index+1);
    }
    return max1+max2;
}
int main()
{
    int n,i,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(tree,0,sizeof(tree));
        for(i=0; i<n; i++)
            scanf("%I64d",&num[i]);
        build(1,n,1);
        for(i=0; i<m; i++)
        {
            char st;
            cin>>st;
            if(st=='C')
            {
                int a,b,c;
                scanf("%d%d%d",&a,&b,&c);
                update(a,b,c,1);
            }
            else
            {
                int a,b;
                long long sum=0;
                scanf("%d%d",&a,&b);
                sum=query(a,b,1);
                printf("%I64d\n",sum);
            }
        }

    }
    return 0;
}