C语言线段树(3)___A Simple Problem with Integers(hdu 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 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

4
55
9
15

题意：给出n个数的初始值和m条命令, 每条命令由 一个字符和 若干个数字组成. 若字符为C 那么后面跟着三个数字分别为 l,r,step,表示在区间 [ l , r ] 的每个数加上step,如果是Q，那么后面跟着两个数字分别为 l , r .输出区间 [ l , r ]的数字和.

代码“：

#include <cstdio>
#include <cstring>
#define maxn 100010
using namespace std;
struct node
{
	int left,right;
    long long lazy,sum;
    int mid(){
        return (left+right)>>1;
    }
}tree[maxn*4]; 
void CreatBtree(int l,int r,int rt){
    tree[rt].left=l;
    tree[rt].right=r;
    tree[rt].lazy=0; 
    //cout<<l<<" "<<r<<endl;
    if(l==r){
        scanf("%lld",&tree[rt].sum);
        return;
    }
    int mid=tree[rt].mid();
    CreatBtree(l,mid,rt<<1);
    CreatBtree(mid+1,r,rt<<1|1);
    tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;
}
void change(int rt,int l,int r,int step){
    if(l==tree[rt].left && r==tree[rt].right){
    	tree[rt].lazy+=step;
		return; 
	}
	tree[rt].sum+=(long long)step*(r-l+1);
	int mid=tree[rt].mid();
	if(r<=mid) change(rt<<1,l,r,step);
	else if(l>mid)change(rt<<1|1,l,r,step);
	else{
		change(rt<<1,l,mid,step);
		change(rt<<1|1,mid+1,r,step);
	}
}
long long Query(int rt,int l,int r){
    if(tree[rt].left==l && tree[rt].right==r)return tree[rt].sum+(r-l+1)*tree[rt].lazy;
    if(tree[rt].lazy!=0){
    	tree[rt<<1].lazy+=tree[rt].lazy;
		tree[rt<<1|1].lazy+=tree[rt].lazy;
		tree[rt].sum+=(long long)(tree[rt].right-tree[rt].left+1)*tree[rt].lazy;  
		tree[rt].lazy=0; 
	} 
    if(r<=tree[rt].mid()) return Query(rt<<1,l,r);
    else if(l>tree[rt].mid())return Query(rt<<1|1,l,r);
    else
        return Query(rt<<1,l,tree[rt].mid())+Query(rt<<1|1,tree[rt].mid()+1,r);
}
int main()
{
    int i,j,n,m,k,l,r,step;
    char ch;
    while(scanf("%d%d",&n,&m)!=EOF){
        CreatBtree(1,n,1);
        //printf("%d %d %d\n",tree[1].max,tree[1].left,tree[1].right); 
        for(i=0;i<m;i++){
        	scanf("%s%d%d",&ch,&l,&r);
        	if(ch=='Q')printf("%lld\n",Query(1,l,r));
        	else{
        		scanf("%d",&step);
        		change(1,l,r,step);
			}
		}
    }
    return 0;
}