POJ - 3468 A Simple Problem with Integers （区间求和）

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

Hint

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

题意：区间修改，区间求和

思路：典型的线段树区间求和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define lson(x) ((x) << 1)
#define rson(x) ((x) << 1 | 1)
#define ll long long
using namespace std;
const int maxn = 100005;

struct seg {
	ll w;
	int flag;
}; 

struct segment_tree {
	seg node[maxn<<2];

	void update(int pos) {
		node[pos].w = node[lson(pos)].w + node[rson(pos)].w;
	}

	void build(int l, int r, int pos) {
		node[pos].flag = 0;
		if (l == r) {
			scanf("%lld", &node[pos].w);
			return;
		}
		int m = l + r >> 1;
		build(l, m, lson(pos));
		build(m+1, r, rson(pos));
		update(pos);
	}

	void push(int l, int r, int pos) {
		int len = r - l + 1;
		if (node[pos].flag != 0) {
			node[lson(pos)].flag += node[pos].flag;
			node[rson(pos)].flag += node[pos].flag;
			node[lson(pos)].w += (ll)node[pos].flag * (len - (len >> 1));
			node[rson(pos)].w += (ll)node[pos].flag * (len >> 1);
			node[pos].flag = 0;
		}
	}

	void modify(int l, int r, int pos, int x, int y, int z) {
		if (x <= l && y >= r) {
			node[pos].w += (ll)(r-l+1) * z;
			node[pos].flag += z;
			return;
		}
		push(l, r, pos);
		int m = l + r >> 1;
		if (x <= m)
			modify(l, m, lson(pos), x, y, z);
		if (y > m)
			modify(m+1, r, rson(pos), x, y, z);
		update(pos);
	}

	ll query(int l, int r, int pos, int x, int y) {
		if (x <= l && y >= r)
			return node[pos].w;
		push(l, r, pos);
		int m = l + r >> 1;
		ll res = 0;
		if (x <= m)
			res += query(l, m, lson(pos), x, y);
		if (y > m)
			res += query(m+1, r, rson(pos), x, y);
		return res;
	}

} tree;

int main() {
	int n, q, a, b, c;
	char str[10];
	while (scanf("%d%d", &n, &q) != EOF) {
		tree.build(1, n, 1);
		for (int i = 0; i < q; i++) {
			scanf("%s", str);
			if (str[0] == 'Q') {
				scanf("%d%d", &a, &b);
				printf("%lld\n",tree.query(1, n, 1, a, b));
			}
			else if (str[0] == 'C'){
				scanf("%d%d%d", &a, &b, &c);
				tree.modify(1, n, 1, a, b, c);
			}
		}
	}
	return 0;
}

POJ - 3468 A Simple Problem with Integers （区间求和）