A Simple Problem with Integers

最新推荐文章于 2022-12-01 00:10:51 发布

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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

Hint

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

题解：线段树延迟做法，就是延迟对孩子节点的操作，把前面要操作的数据记录下来，等下次必须访问孩子的时候才去做，减少递归次数。

#include <iostream>
#include <cstdio>
#include <cstring>
#define mem(a) memset(a,0,1izeof(a));

using namespace std;

long long arr[400005];
long long add[400005];
long long a[100005];

void pushUp(int k)
{
	arr[k] = arr[k << 1] + arr[(k << 1) | 1];
}

void pushDown(int k,int x)    //分给左右孩子。
{
	if(add[k] != 0)
	{
		add[k << 1] += add[k];
		add[(k << 1) | 1] += add[k];
		arr[k << 1] += add[k] * (x - (x >> 1));
		arr[(k << 1) | 1] += add[k] * (x >> 1);
		add[k] = 0; 
	}
}

void segTree(int k,int l,int r)
{
	add[k] = 0;
	if(l == r)
	{
		arr[k] = a[l];
		return;
	}
	int mid = ( l + r) >> 1;
	segTree(k << 1,l,mid);
	segTree((k << 1) | 1,mid + 1,r);
	pushUp(k);
}

void update(int k,int l,int r,int x,int y,int c)
{
	if(l == x && r == y)
	{
		add[k] += c;
		arr[k] += (r - l + 1) * c;
		return;
	}
	int mid = (l + r) >> 1;
	pushDown(k,r - l + 1);
	if(x > mid)
	{
		update((k << 1) | 1,mid + 1,r,x,y,c);
	}
	else if(y <= mid)
	{
		update(k << 1,l,mid,x,y,c);
	}
	else
	{
		update(k << 1,l,mid,x,mid,c);
		update((k << 1) | 1,mid + 1,r,mid + 1,y,c);
	}
	pushUp(k);
}

long long query(int k,int l,int r,int x,int y)
{
	if(l == x && r == y)
	{
		return arr[k];
	}
	pushDown(k,r - l + 1);
	int mid = (l + r) >> 1;
	if(x > mid)
	{
		return query((k << 1) | 1,mid + 1,r,x,y);
	}
	if(y <= mid)
	{
		return query(k << 1,l,mid,x,y);
	}
	return query(k << 1,l,mid,x,mid) + query((k << 1) | 1,mid + 1,r,mid + 1,y);
}

int main()
{
	int n,m;
	while(scanf("%d%d",&n,&m) != EOF)
	{
		for(int i = 1;i <= n;i++)
		{
			scanf("%lld",a + i);
		}
		
		segTree(1,1,n);
		char s[3];
		int l,r;
		for(int i = 0;i < m;i++)
		{
			scanf("%s%d%d",s,&l,&r);
			if('Q' == s[0])
			{
				printf("%lld\n",query(1,1,n,l,r));
			}
			else
			{
				long long c;
				scanf("%lld",&c);
				update(1,1,n,l,r,c);
			}
		}
	}
	
	
	
	return 0;
}