POJ 3468 A Simple Problem with Integers

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本文介绍了一种使用线段树优化区间加查询操作的算法，通过懒惰传播技术，有效地处理大规模数据集上的区间加法和区间求和操作，适用于需要频繁更新和查询的场景。

A Simple Problem with Integers

https://vjudge.net/problem/POJ-3468
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
4
55
9
15
Hint
The sums may exceed the range of 32-bit integers.
C++

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN = 100000 + 10;
typedef long long LL;

LL num[MAXN];

struct node{
	int l, r;
	LL lazy, sum;
}segTree[MAXN << 2];

void pushup(int x)
{
	segTree[x].sum = segTree[x << 1].sum + segTree[x << 1 | 1].sum;
}

void pushdown(int x)
{
	if(segTree[x].lazy)
	{
		segTree[x << 1].lazy += segTree[x].lazy; 
		segTree[x << 1 | 1].lazy += segTree[x].lazy;
		segTree[x << 1].sum += (segTree[x << 1].r - segTree[x << 1].l + 1) * segTree[x].lazy;
		segTree[x << 1 | 1].sum += (segTree[x << 1 | 1].r - segTree[x << 1 | 1].l + 1) * segTree[x].lazy;
		segTree[x].lazy = 0;
	}
}

void build(int x, int l, int r)
{
	segTree[x].l = l;
	segTree[x].r = r;
	segTree[x].lazy = 0;
	if(l == r)
	{
		segTree[x].sum = num[l];
		return;
	}
	int mid = l + r >> 1;
	build(x << 1, l, mid);
	build(x << 1 | 1, mid + 1, r);
	pushup(x);
}

void update(int x, int l, int r, LL val)
{
	if(l <= segTree[x].l && segTree[x].r <= r)
	{
		segTree[x].sum += (segTree[x].r - segTree[x].l + 1) * val;
		segTree[x].lazy += val;
		return;
	}
	pushdown(x);
	int mid = segTree[x].l + segTree[x].r >> 1;
	if(r <= mid) update(x << 1, l, r, val);
	else if(l > mid) update(x << 1 | 1, l, r, val);
	else
	{
		update(x << 1, l, mid, val);
		update(x << 1 | 1, mid + 1, r, val);
	}
	pushup(x);
}

LL query(int x, int l, int r)
{
	if(l <= segTree[x].l && segTree[x].r <= r)
		return segTree[x].sum;
	pushdown(x);
	int mid = segTree[x].l + segTree[x].r >> 1;
	if(r <= mid) return query(x << 1, l, r);
	else if(l > mid) return query(x << 1 | 1, l, r);
	else return query(x << 1, l, mid) + query(x << 1 | 1, mid + 1, r);
}

int main()
{
	int n, m;
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i ++)
		scanf("%lld", &num[i]);
	build(1, 1, n);
	char op;
	int a, b;
	LL c;
	for(int i = 0; i < m; i ++)
	{
		getchar();
		scanf("%c", &op);
		if(op == 'Q')
		{
			scanf("%d%d", &a, &b);
			printf("%lld\n", query(1, a, b));
		}
		else
		{
			scanf("%d%d%lld", &a, &b, &c);
			update(1, a, b, c);
		}
	}
	return 0;
}