Codeforces 548E Mike ans Foam (与质数相关的容斥多半会用到莫比乌斯函数)

题面

链接：CF548E

Description

Mike is a bartender at Rico's bar. At Rico's, they put beer glasses in a special shelf. There are n kinds of beer at Rico's numbered from 1 to n. i-th kind of beer has a**i milliliters of foam on it.

Maxim is Mike's boss. Today he told Mike to perform q queries. Initially the shelf is empty. In each request, Maxim gives him a number x. If beer number x is already in the shelf, then Mike should remove it from the shelf, otherwise he should put it in the shelf.

After each query, Mike should tell him the score of the shelf. Bears are geeks. So they think that the score of a shelf is the number of pairs (i, j) of glasses in the shelf such that $i< j $and where is the greatest common divisor of numbers \(a\) and \(b\).

Mike is tired. So he asked you to help him in performing these requests.

Input

The first line of input contains numbers \(n\) and \(q (1 ≤ n, q ≤ 2 × 10^5)\), the number of different kinds of beer and number of queries.

The next line contains n space separated integers, \(a_1, a_2, ... , a_n (1 ≤ a_i ≤ 5 × 10^5)\), the height of foam in top of each kind of beer.

The next q lines contain the queries. Each query consists of a single integer integer x(1 ≤ x ≤ n), the index of a beer that should be added or removed from the shelf.

Output

For each query, print the answer for that query in one line.

Examples

---

input	output
5 6 1 2 3 4 6 1 2 3 4 5 1	0 1 3 5 6 2

题目分析

如题，用\(\mu\)进行容斥即可。

代码实现

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#define MAXN 0x7fffffff
typedef long long LL;
const int N=500005;
using namespace std;
inline int Getint(){register int x=0,f=1;register char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}return x*f;}
int prime[N],mu[N];
bool vis[N],chk[200005];
int a[200005],cnt[N];

int main(){
	mu[1]=1;
	for(int i=2;i<=5e5;i++){
		if(!vis[i])prime[++prime[0]]=i,mu[i]=-1;
		for(int j=1;j<=prime[0]&&i*prime[j]<=5e5;j++){
			vis[i*prime[j]]=1;
			if(i%prime[j]==0)break;
			mu[i*prime[j]]=-mu[i];
		}
	} 

	int n=Getint(),q=Getint();
	for(int i=1;i<=n;i++)a[i]=Getint();
	LL ans=0;
	for(int j=1;j<=q;j++){
		int x=Getint(),ret=0;
		if(!chk[x]){
			chk[x]=1,x=a[x];
			for(int i=1,lim=sqrt(x);i<=lim;i++){
				if(x%i==0){
					ret+=mu[i]*cnt[i],cnt[i]++;
					if(i*i!=x)ret+=mu[x/i]*cnt[x/i],cnt[x/i]++;
				}
			}
			ans+=ret;
		}else{
			chk[x]=0,x=a[x];
			for(int i=1,lim=sqrt(x);i<=lim;i++){
				if(x%i==0){
					cnt[i]--,ret+=mu[i]*cnt[i];
					if(i*i!=x)cnt[x/i]--,ret+=mu[x/i]*cnt[x/i];
				}
			}
			ans-=ret;
		}
		cout<<ans<<'\n';
	}
	return 0;
}
/*
8 8
77770 436260 150552 164101 453308 1 1 329637
1 2 3 4 5 6 7 8
*/