树状数组关于区间修改区间求和的问题

最新推荐文章于 2024-06-17 23:09:11 发布

free-dancer

最新推荐文章于 2024-06-17 23:09:11 发布

阅读量386

点赞数

CC 4.0 BY-SA版权

分类专栏：树状数组

本文链接：https://blog.youkuaiyun.com/Joined/article/details/77512679

树状数组专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一种基础的数据结构算法——区间修改与区间求和问题的实现方法，通过使用差分数组和树状数组来高效地处理一系列数值的更新与查询操作。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Total Submission(s) : 77 Accepted Submission(s) : 19

Problem 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, ... , 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

题意：

就是给你N个数M个操作，然后输出每一个查询区间的和。

这是最基础的区间修改区间求和的实现问题。

#if 0
#include<iostream>
#include<cstring>
#include<stdio.h>

using namespace std;
const int MAXN=100000+50;
typedef long long ll;
ll  b[MAXN],c[MAXN];
ll  a[MAXN]; 

int lowbit(ll  x)
{
	return x&(-x);
}

void add_B(ll  *a,ll  x,ll  v)
{
	for( ; x>0; x-=lowbit(x))
		a[x]+=v;
}
ll sum_B(ll  *a,ll  x)
{
	ll  s=0;
	for(; x<MAXN; x+=lowbit(x))
	{
		s+=a[x];
	}
	
	return s;
}

void add_C(ll  *a,ll  x,ll v)
{
	for( ; x<MAXN; x+=lowbit(x))
	{
		a[x]+=v;
	}
	
}
ll sum_C(ll *a, ll x)
{
	ll s=0;
	for( ; x>0; x-=lowbit(x))
	{
		s+=a[x];
	}
	return s;
}

void add(ll x,ll v)
{
	add_B(b,x-1,v);
	add_C(c,x,v*x);
}
void go_add(ll a,ll b,ll v)
{
	add(b,v);
	add(a-1,-v); //
}

ll sum(ll x)
{
	return sum_C(c,x)+sum_B(b,x)*x;
}
ll go_sum(ll a,ll b)
{
	return sum(b)-sum(a-1);
}

int main()
{


	ll n,t;
	char ssc;

	scanf("%lld%lld",&n,&t);
	
	memset(a,0,sizeof(a));
	memset(b,0,sizeof(b));
	memset(c,0,sizeof(c));
	
	for(int i=3; i<=n+2; i++)
	{
		ll x;
		scanf("%lld",&x);
		go_add(i,i,x);
	}
	

	while(t--)
	{
		
		cin>>ssc;
		if(ssc=='Q')
		{
			ll ai,bi;
			
			scanf("%lld%lld",&ai,&bi);
			ai+=2;
			bi+=2;
	
			
			printf("%lld\n",go_sum(ai,bi));	
		}
		else if(ssc=='C')
		{
			ll ai,bi;
			ll v;
			scanf("%lld%lld%lld",&ai,&bi,&v);
			
			ai+=2;
			bi+=2;
			go_add(ai,bi,v);
		}

	}


}
#endif