PE530 : GCD of Divisors

\[\begin{eqnarray*}
ans&=&\sum_{i=1}^nf(i)\\
&=&\sum_{i=1}^n\sum_{d|i}\gcd(d,\frac{i}{d})\\
&=&\sum_{i=1}^n\sum_{d|i}\sum_{k|d,k|\frac{i}{d}}\varphi(k)\\
&=&\sum_{k=1}^n\varphi(k)\sum_{k^2|i}\sigma_0(\frac{i}{k^2})\\
&=&\sum_{k=1}^n\varphi(k)\sum_{i=1}^{\lfloor\frac{n}{k^2}\rfloor}\lfloor\frac{n}{i}\rfloor\\
&=&\sum_{k=1}^{\sqrt{n}}\varphi(k)S(\lfloor\frac{n}{k^2}\rfloor)
\end{eqnarray*}\]

其中

\[S(n)=\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor\]

枚举所有$k$，然后分段计算$S$即可。

时间复杂度

\[\begin{eqnarray*}
T(n)&=&O(\sqrt{n}+\sum_{i=1}^{\sqrt{n}}\sqrt{\frac{n}{i^2}})\\
&=&O(\sqrt{n}\sum_{i=1}^{\sqrt{n}}\frac{1}{i})\\
&=&O(\sqrt{n}\log n)
\end{eqnarray*}\]

#include<cstdio>
typedef long long ll;
const int N=31622800;
const ll n=1000000000000000LL;
int i,j,k,tot,p[N/10],phi[N];bool v[N];ll ans;
inline ll F(ll n){
  ll ret=0;
  for(ll i=1,j;i<=n;i=j+1){
    j=n/(n/i);
    ret+=n/i*(j-i+1);
  }
  return ret;
}
int main(){
  for(phi[1]=1,i=2;i<=n/i;i++){
    if(!v[i])phi[i]=i-1,p[tot++]=i;
    for(j=0;j<tot;j++){
      k=i*p[j];
      if(k>n/k)break;
      v[k]=1;
      if(i%p[j])phi[k]=phi[i]*(p[j]-1);else{
        phi[k]=phi[i]*p[j];
        break;
      }
    }
  }
  for(i=1;i<=n/i;i++)ans+=F(n/i/i)*phi[i];
  return printf("%lld",ans),0;
}

PE530 : GCD of Divisors

原文：http://www.cnblogs.com/clrs97/p/5986601.html