MOBIUS

题目描述

Description

Stancu likes space travels but he is a poor software developer and will never be able to buy his own spacecraft. That is why he is preparing to steal the spacecraft of Petru. There is only one problem – Petru has locked the spacecraft with a sophisticated cryptosystem based on the ID numbers of the stars from the Milky Way Galaxy. For breaking the system Stancu has to check each subset of four stars such that the only common divisor of their numbers is 1. Nasty, isn’t it? Fortunately, Stancu has succeeded to limit the number of the interesting stars to N but, any way, the possible subsets of four stars can be too many. Help him to find their number and to decide if there is a chance to break the system.
Input

In the input file several test cases are given. For each test case on the first line the number N of interesting stars is given (1 ≤ N ≤ 10000). The second line of the test case contains the list of ID numbers of the interesting stars, separated by spaces. Each ID is a positive integer which is no greater than 10000. The input data terminate with the end of file.
Output

For each test case the program should print one line with the number of subsets with the asked property.

从n个数中选择4个数使他们的GCD = 1，求总共有多少种方法

Sample Input

4

2 3 4 5

4

2 4 6 8

7

2 3 4 5 7 6 8

Sample Output

1
0
34

#include<bits/stdc++.h>
#define reg register
#define inc(i,l,r) for(register int i=l;i<=r;++i)
using namespace std;
const int maxn=1e4+5;
template<typename T>inline void rd(T&x)
{
	char c;bool f=0;
	while((c=getchar())<'0'||c>'9')if(c=='-')f=1;
	x=c^48;
	while((c=getchar())>='0'&&c<='9')x=x*10+(c^48);
	if(f)x=-x;
}

int noprime[maxn],num[maxn],prime[maxn],tot,mob[maxn];
int n,maxx;
long long ans;

void MOBIUS()
{
	noprime[1]=1;mob[1]=1;
	inc(i,2,maxn)
	{
		if(!noprime[i])prime[++tot]=i,mob[i]=-1;
		
		for(int j=1;j<=tot,i*prime[j]<=maxn;++j)
		{
			noprime[i*prime[j]]=1;
			if(i%prime[j]==0){
				mob[i*prime[j]]=0;
				break;
			}
			mob[i*prime[j]]=-mob[i];	
		}
	}
}

long long vvi()
{
	inc(i,1,maxx)for(reg int j=i+i;j<=maxx;j+=i)
	num[i]+=num[j];
	
	inc(i,1,maxx)
	if(num[i]>3)ans+=(long long)mob[i]*num[i]*(num[i]-1)*(num[i]-2)*(num[i]-3)/24;
	re ans;
}

int main()
{
	int x;
	MOBIUS();//线性筛求μ(i)
	freopen("in.txt","r",stdin);
	while(~scanf("%d",&n))
	{
		maxx=ans=0;
		memset(num,0,sizeof(num));
		inc(i,1,n){
			rd(x);
			++num[x];
			maxx=(x>maxx?x:maxx);
		}
		if(n<4)printf("0\n");
		else printf("%lld\n",vvi());
	}
	re 0;
}

设f(i)为gcd(x,y)=1;
F(i)为f(i)倍数
详见PPT