POJ 3468 A Simple Problem with Integers (线段树区间更新)

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本文介绍了一个涉及区间操作的问题，包括区间累加值和查询区间和的算法实现。使用线段树进行高效处理，通过示例输入输出展示了算法的有效性和正确性。

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 75143		Accepted: 23146
Case Time Limit: 2000MS

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 andQ. 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

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

题目链接：http://poj.org/problem?id=3468

题目大意：给一串数，C操作对区间累加值，Q操作查询区间和

题目分析：裸的线段树区间更新问题

#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
int const MAX = 1e5 + 5;
ll sum[MAX << 2], add[MAX << 2];

void PushUp(int rt)
{
	sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}

void PushDown(int ln, int rn, int rt)
{
	if(add[rt])
	{
		sum[rt << 1] += (ll)ln * add[rt];
		sum[rt << 1 | 1] += (ll)rn * add[rt];
		add[rt << 1] += add[rt];
		add[rt << 1 | 1] += add[rt];
		add[rt] = 0;
	}
	return;
}

void Build(int l, int r, int rt)
{
	add[rt] = 0;
	if(l == r)
	{
		scanf("%lld", &sum[rt]);
		return;
	}
	int mid = (l + r) >> 1;
	Build(lson);
	Build(rson);
	PushUp(rt);
	return;
}

void Update(int L, int R, int c, int l, int r, int rt)
{
	if(L <= l && r <= R)
	{
		sum[rt] += (r - l + 1) * c;
		add[rt] += c;
		return;
	}
	int mid = (l + r) >> 1;
	PushDown(mid - l + 1, r - mid, rt);
	if(L <= mid)
		Update(L, R, c, lson);
	if(R > mid)
		Update(L, R, c, rson);
	PushUp(rt);
	return;
}

ll Query(int L, int R, int l, int r, int rt)
{
	if(L <= l && r <= R)
		return sum[rt];
	int mid = (l + r) >> 1;
	PushDown(mid - l + 1, r - mid, rt);
	ll ans = 0;
	if(L <= mid)
		ans += Query(L, R, lson);
	if(R > mid)
		ans += Query(L, R, rson);
	return ans;
}

int main()
{
	int n, q;
	scanf("%d %d", &n, &q);
	Build(1, n, 1);
	while(q --)
	{
		char s[2];
		scanf("%s", s);
		if(s[0] == 'Q')
		{
			int l, r;
			scanf("%d %d", &l, &r);
			printf("%lld\n", Query(l, r, 1, n, 1));
		}
		else 
		{
			int l, r, c;
			scanf("%d %d %d", &l, &r, &c);
			Update(l, r, c, 1, n, 1);
		}
	}
}