【莫比乌斯反演】BZOJ2818 Gcd

题面在这里

反演裸题不解释

示例程序：

#include
#include
using namespace std;
typedef long long ll;

const int maxn=10000005;
int n,p[maxn],mu[maxn],N=maxn-5;
ll f[maxn];
bool vis[maxn];
void prepare(){
    mu[1]=1;N=min(N,n);
    for (int i=2;i<=N;i++){
        if (!vis[i]) p[++p[0]]=i,mu[i]=-1;
        for (int j=1;j<=p[0]&&i*p[j]<=N;j++){
            vis[i*p[j]]=1;
            if (i%p[j]==0) {mu[i*p[j]]=0;break;}
             else mu[i*p[j]]=-mu[i];
        }
    }
    for (int i=1;i<=p[0];i++)
     for (int j=1;p[i]*j<=N;j++) f[p[i]*j]+=mu[j];
    for (int i=1;i<=N;i++) f[i]+=f[i-1];
}
int main(){
    scanf("%d",&n);ll ans=0;
    prepare();
    for (int d=1,lst=0;d<=n;d=lst+1){
        lst=n/(n/d);
        ans+=(ll)(f[lst]-f[d-1])*(n/d)*(n/d);
    }
    printf("%lld",ans);
    return 0;
}