ACM-ICPC 2018 徐州赛区网络预赛 Ryuji doesn't want to study（公式化简+前缀和）

最新推荐文章于 2018-11-12 17:47:12 发布

鸡冠花12138

最新推荐文章于 2018-11-12 17:47:12 发布

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分类专栏： # 前缀和 # 树状数组

本文链接：https://blog.youkuaiyun.com/yu121380/article/details/82590202

树状数组同时被 2 个专栏收录

9 篇文章

订阅专栏

前缀和

4 篇文章

订阅专栏

本文介绍了一种使用树状数组求解特定区间内加权和的问题，该问题涉及区间查询和区间更新操作，通过巧妙地转换公式，将原本复杂的区间求和问题转化为两个前缀和的求解。

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Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i].

Unfortunately, the longer he learns, the fewer he gets.

That means, if he reads books from l to r, he will get

a[l]×L+a[l+1]×(L−1)+⋯+a[r−1]×2+a[r] (L is the length of [ l, r ] that equals to r - l + 1).

Now Ryuji has qq questions, you should answer him:

1. If the question type is 1, you should answer how much knowledge he will get after he reads books [ l, r ].

2. If the question type is 2, Ryuji will change the ith book's knowledge to a new value.

Input

First line contains two integers nn and qq (n，q≤100000).

The next line contains n integers represent a[i](a[i]≤1e9) .

Then in next qq line each line contains three integers a, b, c, if a=1, it means question type is 1, and bb, ccrepresents [ l , r ]. if a = 2 , it means question type is 2 , and bb, c means Ryuji changes the bth book' knowledge to c

Output

For each question, output one line with one integer represent the answer.

样例输入复制

样例输出复制

10
8

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

解析：有a[i]*r-i+1=a[i]*(n-i+1)-a[i]*(n-r),即可化简公式：

$(\sum_{i=l}^{r}a[i]*(n-i+1))-(n-r)*\sum_{i=l}^{r}a[i].$

画图即可看出就是求两个前缀和。这里我用树状数组进行求前缀和。当然也可以用线段树。

#include<bits/stdc++.h>
using namespace std;

#define e exp(1)
#define pi acos(-1)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define mem(a,b) memset(a,b,sizeof(a))
int gcd(int a,int b){return b?gcd(b,a%b):a;}

const int maxn=1e6+5;
int n,q;
ll a[maxn];
ll c1[maxn],c2[maxn];
int lowbit(int x)
{
	return x&(-x);
}
void add1(int i,ll value)
{
	while(i<=n)
	{
		c1[i]+=value;
		i+=lowbit(i);
	}
}
void add2(int i,ll value)
{
	while(i<=n)
	{
		c2[i]+=value;
		i+=lowbit(i);
	}
}
ll getsum1(int i)
{
	ll ans=0;
	while(i>0)
	{
		ans+=c1[i];
		i-=lowbit(i);
	}
	return ans;
}
ll getsum2(int i)
{
	ll ans=0;
	while(i>0)
	{
		ans+=c2[i];
		i-=lowbit(i);
	}
	return ans;
}

int main()
{
	while(~scanf("%d%d",&n,&q))
	{
		mem(c1,0);mem(c2,0);
		for(int i=1; i<=n; i++)
		{
			scanf("%lld",&a[i]);
			add1(i,a[i]*(n-i+1));
			add2(i,a[i]);
		}
		while(q--)
		{
			int op,x,y;scanf("%d%d%d",&op,&x,&y);
			if(op==1)
			{
				ll ans=getsum1(y)-getsum1(x-1);
				ans-=(getsum2(y)-getsum2(x-1))*(n-y);
				printf("%lld\n",ans); 
			}
			else
			{
				add1(x,(n-x+1)*(y-a[x]));
				add2(x,y-a[x]);
				a[x]=y;
			}
		}
	}
	return 0;
}