区间树 [成段更新] POJ 3468 A Simple Problem with Integers

可参考这个题

http://blog.youkuaiyun.com/gaotong2055/article/details/9300141

Description

You haveNintegers,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 numbersNandQ. 1 ≤N,Q≤ 100000.
The second line containsNnumbers, the initial values ofA₁,A₂, ... ,A_N. -1000000000 ≤A_i≤ 1000000000.
Each of the nextQlines represents an operation.
"Cabc" means addingcto each ofA_a,A_a+1, ... ,A_b. -10000 ≤c≤ 10000.
"Qab" means querying the sum ofA_a,A_a+1, ... ,A_b.

Output

You need to answer allQcommands 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

#include <iostream>
#include <stdio.h>
using namespace std;
int n,q;
long long tree[400000];
long long add[400000];

int a,b,c;
void build(int l, int r, int k){
	add[k] = 0;
	if(l==r){
		scanf("%lld", &tree[k]);
		return;
	}
	int m = (l+r)/2;
	build( l, m, 2*k);
	build( m+1, r, 2*k + 1);
	tree[k] = tree[2*k] + tree[2*k+1];
}

void down(int k,int m){
	if(add[k]){
		add[k*2+1] += add[k];
		add[k*2] += add[k];
		tree[k*2] += (m-m/2) * add[k];
		tree[k*2+1] += (m/2) * add[k];
		add[k] = 0;
	}
}

void update(int l,int r, int k){
	if( a <= l && b >= r){ //找到合适的区间
		add[k] += c;
		tree[k] += (long long)(r - l + 1) * c;
		return;
	}
	down(k, r-l+1); //更新子树
	int m = (l+r)/2;
	if(a <= m) update(l, m, 2*k);
	if(b > m) update(m+1, r, 2*k+1);
	tree[k] = tree[2*k] + tree[2*k+1];
}

long long query(int l, int r, int k){
	if(a <= l &&  b >= r)
	{
		return tree[k];
	}
	down(k, r-l+1); //查询的时候需要更新子树
	long long ans = 0;
	int m = (l+r)/2;
	if(a <= m)
		ans += query(l , m, k*2);
	if(b > m)
		ans += query(m+1, r, k*2+1);
	return ans;
}
int main() {
	//freopen("in.txt", "r", stdin);
	char cmd[5];
	while(scanf("%d %d", &n, &q) != EOF){
		build(1, n ,1);
		while(q--){
			scanf("%s", cmd);
			if(cmd[0] == 'Q'){
				scanf("%d %d",&a,&b);
				printf("%lld\n", query(1,n,1));
			}else{
				scanf("%d %d %d",&a, &b, &c);
				update(1, n, 1);
			}
		}
	}
	return 0;
}