poj 3468 线段树成段更新

http://poj.org/problem?id=3468

A Simple Problem with Integers

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

4
55
9
15

Hint

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

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Life is so hard! many courage and not care the result!

成段更新不容易啊，看了两三天了，仍像云里雾里，不懂啊！

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>

#define Maxx 100010

int n,m;
int str[Maxx];
long long ans;

struct Node
{
    int left,right;
    long long add,sum;
}tree[Maxx*4];

void build(int l,int r,int rt)
{
    tree[rt].left=l;
    tree[rt].right=r;
    tree[rt].add=0;

    if(l == r)
    {
        tree[rt].sum=str[l];
        return ;
       }
    int mid=(l+r)/2;
    build(l,mid,2*rt);
    build(mid+1,r,2*rt+1);
    tree[rt].sum=tree[rt*2].sum+tree[rt*2+1].sum;
}

void update(int l,int r,int add,int rt)
{
    if(tree[rt].left>=l && tree[rt].right<=r)
    {
        tree[rt].sum+=(tree[rt].right-tree[rt].left+1)*add;
        tree[rt].add+=add;
        return ;
       }

    if(tree[rt].left>r || tree[rt].right<l)
    {
        return ;
       }

    if(tree[rt].add)
    {
        tree[2*rt].sum += (tree[2*rt].right-tree[2*rt].left+1)*tree[rt].add;
        tree[2*rt].add += tree[rt].add;
        tree[2*rt+1].sum += (tree[2*rt+1].right-tree[2*rt+1].left+1)*tree[rt].add;
        tree[2*rt+1].add += tree[rt].add;
        tree[rt].add = 0;
       }

    update(l,r,add,rt*2);
    update(l,r,add,rt*2+1);
    tree[rt].sum =tree[rt*2].sum+tree[rt*2+1].sum;
}

void query(int l,int r,int rt)
{
    if(tree[rt].left>r || tree[rt].right<l)
    {
            return ;
       }

    if(tree[rt].left>=l && tree[rt].right<=r)
    {
        ans+=tree[rt].sum;
        return ;
       }

    if(tree[rt].add)
    {
        tree[2*rt].sum += (tree[2*rt].right-tree[2*rt].left+1)*tree[rt].add;
        tree[2*rt].add += tree[rt].add;
        tree[2*rt+1].sum += (tree[2*rt+1].right-tree[2*rt+1].left+1)*tree[rt].add;
        tree[2*rt+1].add += tree[rt].add;
        tree[rt].add = 0;
       }

    query(l,r,rt*2);
    query(l,r,rt*2+1);
    tree[rt].sum=tree[rt*2].sum+tree[rt*2+1].sum;
}

int main()
{
    int i,j,k,t,a,b,c;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&str[i]);
           }
        build(1,n,1);
        char st[2];
        for(i=1;i<=m;i++)
        {
                   scanf("%s",st);
                 if(st[0] == ‘Q‘)
                {
                    scanf("%d%d",&a,&b);
                    ans=0;
                    query(a,b,1);
                    printf("%lld\n",ans);
                   }
                else
                {
                    scanf("%d%d%d",&a,&b,&c);
                    update(a,b,c,1);
                   }
           }
       }
    return 0;
}

poj 3468 线段树成段更新