bzoj2818Gcd

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
不用筛$\varphi$强行莫比乌斯也是很资瓷
$\sum\limits_{i=1}^{n}\sum\limits_{i=1}^{n}isprime(gcd(i,j))$
$\sum\limits_{g=1}^{n}isprime(g)\sum\limits_{g|i}\sum\limits_{g|j}[gcd(i,j)==g]$
$\sum\limits_{g=1}^{n}isprime(g)\sum\limits_{t=1}^{\lfloor\frac{n}{g}\rfloor}\mu(t)\sum\limits_{i=1}^{\frac{n}{gt}}\sum\limits_{j=1}^{\frac{n}{gt}}1$
$\sum\limits_{p=1}^{n}\lfloor\frac{n}{p}\rfloor^{2}*\sum\limits_{g|p} isprime(g)*\mu(\frac{p}{g})$

#include<cstdio>
#include<algorithm>
#define N 10001000
#define rg register
#define ll long long
using namespace std;
int n,mu[N],prime[N/10],tot,s[N];ll ans;
bool not_prime[N];
int main(){
//  freopen("bzoj2818.in","r",stdin);
    scanf("%d",&n);mu[1]=1;
    for (rg int i=2;i<=n;++i){
        if (!not_prime[i]) prime[++tot]=i,mu[i]=-1;
        for (int j=1;prime[j]*i<=n;++j){
            not_prime[i*prime[j]]=1;
            if (i%prime[j]==0){
                mu[i*prime[j]]=0;break;
            }else mu[i*prime[j]]=-mu[i];
        }
    }
    for (rg int i=1;i<=tot;++i){
        const int p=prime[i];
        for (rg int i=1;p*i<=n;++i) s[p*i]+=mu[i];
    }int last=1;
    for (rg int i=1;i<=n;++i) s[i]+=s[i-1];
    for (rg int i=1;i<=n;i=last+1){
        last=n/(n/i);ans+=(ll)(n/i)*(n/i)*(s[last]-s[i-1]);
    }printf("%lld\n",ans);
    return 0;
}