线段树 poj 3468

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

 

#include<iostream>
#include<stdio.h>
using namespace std;
struct node
{
    long long l_,r_,number,mark;
};
class segment_tree //这个代码是A不了的 应为用类 重复申请内存勒  把类取消了就可以了   思路最重要嘛
{
    private:
        node data[*];//储存 需要开四倍  怎么算的我也不知道 二叉树叶子节点 为n  总结点一定小于4n？
    public:
        void built(long long l,long long r,long long i)//建立线段树
        {
            data[i].l_=l;data[i].r_=r;data[i].mark=;data[i].number=;
            if(l==r){   //如果是叶子节点
            cin>>data[i].number; return ;}
            else
            {
                int mid=(l+r)/;
                built(l,mid,i*); //不是就往下建立左右孩子
                built(mid+,r,i*+);
            }
            data[i].number=data[i*].number+data[i*+].number;
        }
        void down_mark(long long i)//更新缓存
        {
            data[i*].number+=data[i].mark*(data[i*].r_-data[i*].l_+);
            data[i*+].number+=data[i].mark*(data[i*+].r_-data[i*+].l_+);
            data[i*].mark+=data[i].mark;  data[i*+].mark+=data[i].mark;
            data[i].mark=;
        }
        void add(long long l,long long r,long long x,long long i)//刷新区域；
        {
            if(data[i].l_==l&&data[i].r_==r)
            {
                data[i].number+=(data[i].r_-data[i].l_+)*x;
                data[i].mark+=x; return ;
            }
            if(data[i].mark) down_mark(i);
            int mid=(data[i].l_+data[i].r_)/;
            if(l>=mid+) add(l,r,x,i*+);//区间全在右孩子
            else if(r<=mid) add(l,r,x,i*); //区间全在左孩子
            else{                              //左右各一部分
                add(l,mid,x,i*);
                add(mid+,r,x,i*+);
            }
            data[i].number=data[i*].number+data[i*+].number;
        }
        long long query(long long l,long long r,long long i)//查询
        {
            if(l==data[i].l_&&data[i].r_==r)
            return data[i].number;
            if(data[i].mark) down_mark(i);
            int mid=(data[i].l_+data[i].r_)/;
            if(l>=mid+) return query(l,r,i*+);
            else if(r<=mid) return query(l,r,i*);
            else{
                return query(l,mid,i*)+query(mid+,r,i*+);
            }
        }
};
long long m,n,a,b,c;
char ch[];
int main()//这里吐槽一句 真的不想用scanf。。。 太不习惯了
{
    while(scanf("%lld%lld",&m,&n)!=EOF)
    {
        segment_tree q;
        q.built(,m,);
        while(n--)
        {
            scanf("%s,",&ch);//cin>>ch;
            if(ch[]=='Q'){
                scanf("%lld%lld",&a,&b);//cin>>a>>b;
                 cout<<q.query(a,b,)<<endl;
            }
            else{
                scanf("%lld%lld%lld",&a,&b,&c);//cin>>a>>b>>c;
                q.add(a,b,c,);
            }
        }
    }
    return ;
}