POJ3468——A Simple Problem with Integers【线段树区间更新求和】

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本文介绍了一种使用线段树实现区间更新与查询区间和的算法，并提供了详细的代码实现。该算法适用于处理大量对区间进行加法操作及查询区间元素总和的问题。

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You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

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

Hint

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

线段树区间更新求和板子，需要用到延迟标记。

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;

struct SegmentTree{
	int l, r;
	long long sum, add;
	#define l(x) tree[x].l
	#define r(x) tree[x].r
	#define sum(x) tree[x].sum
	#define add(x) tree[x].add
}tree[100010*4];
int a[100010],n, m;

void build(int p, int l, int r){
	l(p) = l, r(p) = r;
	if(l == r){ sum(p) = a[l]; return ;}
	int mid = (l+r)/2;
	build(p*2,l,mid);
	build(p*2+1, mid+1, r);
	sum(p) = sum(p*2) + sum(p*2+1);
}

void spread(int p){
	if(add(p)){
		sum(p*2) += add(p)*(r(p*2) - l(p*2)+1);
		sum(p*2+1) += add(p)*(r(p*2+1) - l(p*2+1)+1);
		add(p*2) += add(p);
		add(p*2+1) += add(p);
		add(p) = 0;
	}
}

void change(int p, int l, int r, int d){
	if(l <= l(p) && r >= r(p)){
		sum(p) += (long long)d * (r(p)-l(p)+1);
		add(p) += d;
		return;
	}
	spread(p);
	int mid = (l(p)+r(p))/2;
	if(l <= mid) change(p*2, l, r, d);
	if(r > mid) change(p*2+1, l, r, d);
	sum(p) = sum(p*2) + sum(p*2+1);
}

long long ask(int p, int l, int r){
	if(l <=l(p) && r >= r(p)) return sum(p);
	spread(p);
	int mid = (l(p)+r(p))/2;
	long long val = 0;
	if(l <= mid) val += ask(p*2, l, r);
	if(r > mid) val += ask(p*2+1, l, r);
	return val;
}

int main(){
	scanf("%d %d",&n, &m);
	for(int i=1; i<=n; i++) scanf("%d", &a[i]);
	build(1, 1, n);
	while(m--){
		char op[2];
		int l, r, d;
		scanf("%s%d%d",op, &l, &r);
		if(op[0] == 'C'){
			scanf("%d", &d);
			change(1, l, r, d);
		}
		else printf("%lld\n", ask(1, l, r));
	}
	return 0;
}