POJ - 3468 树状数组区间查询区间更新模板

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数据结构----线段树/树状数组/ST表同时被 2 个专栏收录

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本文介绍了一种处理大规模数组中区间加法与求和查询的高效算法，通过使用差分数组和前缀和技巧，实现了对给定区间的快速修改与求和。适用于数据结构与算法竞赛及实际应用中对大量数据进行区间操作的需求。

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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

Hint

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

写一次以后当模板用

#include<iostream>
#include<cstdio>
//#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int M = 100000+100;
ll a[M];
ll sum[M];
ll c[2][M];
int n;
int lowbit(int x)
{
	return x&(-x);
}
void add(int x,int d,int k)
{
	while(x<=n)
	{
		c[k][x]+=d;
		x+=lowbit(x);
	}
}
ll ask(int x,int k)
{
	ll ans=0;
	while(x)
	{
		ans+=c[k][x];
		x-=lowbit(x);
	}
	return ans;
}
int main()
{
	int q;
	cin>>n>>q;
	for(int i =1 ;i <= n;i++)
	{
		scanf("%I64d",&a[i]);
		sum[i]=sum[i-1]+a[i];
	//	cout<<sum[i]<<endl;
	}
	char s[2];
	int l,r;
	ll ans=0;
	for(int i = 1; i <=q; i++)
	{
		scanf("%s%d%d",s,&l,&r);
		if(s[0]=='C')
		{
			int d;
			scanf("%d",&d);
			add(r+1,-d,0),add(r+1,-(r+1)*d,1);
			add(l,d,0),add(l,l*d,1);
		}
		else
		{
			ans=sum[r]+(r+1)*ask(r,0)-ask(r,1);
			ans-=sum[l-1]+l*ask(l-1,0)-ask(l-1,1);
			printf("%I64d\n",ans);
		}
	}
	return 0;
 }