BZOJ2818:Gcd(莫比乌斯反演)

Description

给定整数N，求1<=x,y<=N且Gcd(x,y)为素数的
数对(x,y)有多少对.

Input

一个整数N

Output

如题

Sample Input

4

Sample Output

4

HINT

hint

对于样例(2,2),(2,4),(3,3),(4,2)
1<=N<=10^7

Solution 

同BZOJ2820……

Code

#include<iostream>
#include<cstdio>
#define N (10000000)
using namespace std;

int n,m,vis[N+5],prime[N+5],mu[N+5],cnt;
long long sum[N+5];

void Get_mu()
{
    mu[1]=1;
    for (int i=2; i<=n; ++i)
    {
        if (!vis[i]){prime[++cnt]=i; mu[i]=-1;}
        for (int j=1; j<=cnt && prime[j]*i<=n; ++j)
        {
            vis[prime[j]*i]=true;
            if (i%prime[j]==0) break;
            mu[prime[j]*i]=-mu[i];
        }
    }
    for (int i=1; i<=cnt; ++i)
        for (int j=1; j*prime[i]<=n; ++j)
            sum[j*prime[i]]+=mu[j];
    for (int i=1; i<=n; ++i) sum[i]+=sum[i-1];
}

long long Calc(int n,int m)
{
    long long ans=0; if (n>m) swap(n,m);
    for (int l=1,r; l<=n; l=r+1)
    {
        r=min(n/(n/l),m/(m/l));
        ans+=(sum[r]-sum[l-1])*(n/l)*(m/l);
    }
    return ans;
}

int main()
{
    scanf("%d",&n);
    Get_mu();
    printf("%lld
",Calc(n,n)); 
}