POJ 3468 A Simple Problem with Integers（线段树功能：区间加减区间求和）

Posted 2020-09-12 blfbuaa

题目链接：http://poj.org/problem?id=3468

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 56005		Accepted: 16903
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 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

代码例如以下：

#include <cstdio>
#include <algorithm>
using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
//lson和rson分辨表示结点的左儿子和右儿子
//rt表示当前子树的根(root),也就是当前所在的结点
#define LL long long
const int maxn = 111111;
//maxn是题目给的最大区间,而节点数要开4倍,确切的来说节点数要开大于maxn的最小2x的两倍
LL add[maxn<<2];
LL sum[maxn<<2];
void PushUp(int rt) //把当前结点的信息更新到父结点
{
	sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void PushDown(int rt,int m)//把当前结点的信息更新给儿子结点
 {
	if (add[rt]) 
	{
		add[rt<<1] += add[rt];
		add[rt<<1|1] += add[rt];
		sum[rt<<1] += add[rt] * (m - (m >> 1));
		sum[rt<<1|1] += add[rt] * (m >> 1);
		add[rt] = 0;
	}
}
void build(int l,int r,int rt)
 {
	add[rt] = 0;
	if (l == r)
	{
		scanf("%lld",&sum[rt]);
		return ;
	}
	int m = (l + r) >> 1;
	build(lson);
	build(rson);
	PushUp(rt);
}
void update(int L,int R,int c,int l,int r,int rt) 
{
	if (L <= l && r <= R)
	{
		add[rt] += c;
		sum[rt] += (LL)c * (r - l + 1);
		return ;
	}
	PushDown(rt , r - l + 1);
	int m = (l + r) >> 1;
	if (L <= m)
		update(L , R , c , lson);
	if (m < R) 
		update(L , R , c , rson);
	PushUp(rt);
}
LL query(int L,int R,int l,int r,int rt)
 {
	if (L <= l && r <= R)
	{
		return sum[rt];
	}
	PushDown(rt , r - l + 1);
	int m = (l + r) >> 1;
	LL ret = 0;
	if (L <= m) 
		ret += query(L , R , lson);
	if (m < R)
		ret += query(L , R , rson);
	return ret;
}
int main() 
{
	int N , Q;
	scanf("%d%d",&N,&Q);//N为节点数
	build(1 , N , 1);
	while (Q--)//Q为询问次数
	{
		char op[2];
		int a , b , c;
		scanf("%s",op);
		if (op[0] == ‘Q‘) 
		{
			scanf("%d%d",&a,&b);
			printf("%lld\n",query(a , b , 1 , N , 1));
		}
		else
		{
			scanf("%d%d%d",&a,&b,&c);//c为区间a到b添加的值
			update(a , b , c , 1 , N , 1);
		}
	}
	return 0;
}