Poj1468【线段树】

Description

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.
Source

POJ Monthly--2007.11.25, Yang Yi

#include<stdio.h>
#include<string.h>
#define Min -10010
__int64 segtree[400000], a[100010], mark[400000];
void build(int l , int r, int id)
{
	mark[id] = 0;
	if(l == r)
	{
		segtree[id] = a[l];
		return ;
	}
	int mid = (l+r)>>1;
	build(l, mid, id<<1);
	build(mid+1, r, id<<1|1);
	segtree[id] = segtree[id<<1] + segtree[id<<1|1];
}
 void Pushdown(int id, int l, int r)
 {
 	if(mark[id])
 	{
 		mark[id<<1]  += mark[id];
 		mark[id<<1|1] += mark[id];
 		int mid = (l+r)>>1;
		segtree[id<<1] += (mark[id] * (mid - l + 1) );
		segtree[id<<1|1] +=  ( mark[id] * (r - mid) );
		mark[id] = 0;
 	}
 }
 __int64 query(int l, int r, int id, int L, int R)
 {
 	if(L <= l && r <= R)
 	{
 		return segtree[id];
 	}
 	Pushdown(id, l, r);
	int mid = (l+r)>>1;
	__int64 sum = 0;
	if(L <= mid)
	sum += query(l, mid, id<<1, L, R);
	if(R >= mid +1)
	sum += query(mid+1, r, id<<1|1, L, R);
	return sum;
 }

 void update(int l, int r, int id, int left, int right, int cnt)
 {
 	if(left <= l && r <= right)
 	{
 		mark[id] += cnt;
 		segtree[id] += (cnt * (r-l+1) );
 		return ;
 	}
 	Pushdown(id, l, r);
 	int mid = (l+r)>>1;
 	if(left <= mid)
 	update(l, mid, id<<1, left, right, cnt);
 	if(right >= mid+1)
 	update(mid+1, r, id<<1|1, left, right, cnt);
 	segtree[id] = segtree[id<<1] + segtree[id<<1|1];
 }
int main()
{
	int i, j, k, n, m;
	scanf("%d%d", &n, &m);
	for(i = 1; i <= n; ++i)
	scanf("%I64d", &a[i]);
	build(1, n, 1);
	char s[10];
	for(i = 0; i < m; ++i)
	{
		scanf("%s", s);
		if(strcmp(s, "Q") == 0)
		{
			int x, y;
			scanf("%d%d", &x, &y);
			printf("%I64d\n", query(1, n, 1, x, y) );
		}
		else
		{
			int x, y, z;
			scanf("%d%d%d", &x, &y, &z);
			update(1, n, 1, x, y, z);
		}
	}
	return 0;
}

题意：给出一段区间，每个点都有一个值，共有2种操作，一种是将区间i~j内的值全部加上c，其次是查询区间i~j的和。

思路：标准的线段树模板题，但是这里要用上延迟更新避免超时。