POJ3468 线段树模板题

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 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

4 55 9 15

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

题意很明显，只有两种操作，需要使用线段树（对于区间的修改~）：update（)和query（） 用到了__int64需要注意

代码如下：

#include <stdio.h>
#include <iostream>
#include <string.h>

using namespace std;

const int MAXN = 100000 + 10;
struct Node{
	int l, r;
	__int64 k, add;
};
Node znode[4*MAXN];
__int64 zfun(__int64 a, __int64 b) { 
	return a+b;
}
void build(int l, int r, int n){
	int mid = (l+r)/2;
	znode[n].l = l;
	znode[n].r = r;
	znode[n].add = 0;
	znode[n].k = 0;
	if(l==r) return;
	else {
		build(l, mid, 2*n);
		build(mid+1, r, 2*n+1);
	}
}
void update(int l, int r, int n, __int64 val){
	znode[n].k += (r-l+1)*val;
	if(l<=znode[n].l && znode[n].r<=r) znode[n].add += val;
	else{
		int mid = (znode[n].l + znode[n].r) / 2;
		if(r<=mid) update(l, r, 2*n, val);
		else if(l>mid) update(l, r, 2*n+1, val);
		else {
			update(l, mid, 2*n, val);
			update(mid+1, r, 2*n+1, val);
		}
	}
}
__int64 query(int l, int r, int n){				//l、r是待查询的区间端点
	int mid = (znode[n].l + znode[n].r) / 2;
	if(znode[n].l==l && znode[n].r==r) return znode[n].k;
	else{
		if(znode[n].add){			//下放延迟的标记
			znode[2*n].add += znode[n].add;
			znode[2*n].k += znode[n].add*(znode[2*n].r-znode[2*n].l+1);
			znode[2*n+1].add += znode[n].add;
			znode[2*n+1].k += znode[n].add*(znode[2*n+1].r-znode[2*n+1].l+1);
			znode[n].add = 0;
		}
		if(r<=mid) return query(l, r, 2*n);
		else if(l>mid) return query(l, r, 2*n+1);
		else return zfun(query(l, mid, 2*n), query(mid+1, r, 2*n+1));
	}
}
int main(){
//	freopen("in.txt", "r", stdin);
	int id = 1, n, m;
	while(scanf("%d%d", &n, &m)!=EOF){
		build(1, n, 1);
		int i, s;
		for(i=1; i<=n; i++){
			scanf("%d", &s);
			update(i, i, 1, s);
		}
		char ch[2];
		int a, b;
		__int64 c;
		for(i=1; i<=m; i++){
			scanf("%s%d%d", ch, &a, &b);
			if(ch[0]=='Q') {
				__int64 ans = query(a, b, 1);
				printf("%I64d\n", ans);
			}
			else if(ch[0]=='C'){
				scanf("%I64d", &c);
				update(a, b, 1, c);
			}
		}
	}
	return 0;
}