Bzoj 2818: Gcd(莫比乌斯反演)

2818: Gcd

Time Limit: 10 Sec Memory Limit: 256 MB

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

Source

湖北省队互测

/*
莫比乌斯反演.
算是模板题了吧....
*/
#include<iostream>
#include<cstdio>
#define LL long long
#define MAXN 10000010
using namespace std;
int pri[MAXN],tot,mu[MAXN];
LL n,g[MAXN],sum[MAXN],ans;
bool vis[MAXN];
void pre()
{
mu[1]=1;
for(int i=2;i<=n;i++)
{
if(!vis[i]) pri[++tot]=i,mu[i]=-1,g[i]=1;
for(int j=1;j<=tot&&i*pri[j]<=n;j++)
{
vis[i*pri[j]]=true;
if(i%pri[j])
{
mu[i*pri[j]]=-mu[i];
g[i*pri[j]]=-g[i]+mu[i];
}
else
{
mu[i*pri[j]]=0;
g[i*pri[j]]=mu[i];
break;
}
}
}
for(int i=1;i<=n;i++) sum[i]=sum[i-1]+g[i];
}
int main()
{
LL last;
cin>>n;
pre();
for(LL i=1;i<=n;i=last+1)
{
last=n/(n/i);
ans+=(LL)(n/i)*(n/i)*(sum[last]-sum[i-1]);
}
printf("%lld",ans);
return 0;
}