【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<cstdio>#include<algorithm>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;}