PE530 : GCD of Divisors

[egin{eqnarray*}
ans&=&sum_{i=1}^nf(i)\
&=&sum_{i=1}^nsum_{d|i}gcd(d,frac{i}{d})\
&=&sum_{i=1}^nsum_{d|i}sum_{k|d,k|frac{i}{d}}varphi(k)\
&=&sum_{k=1}^nvarphi(k)sum_{k^2|i}sigma_0(frac{i}{k^2})\
&=&sum_{k=1}^nvarphi(k)sum_{i=1}^{lfloorfrac{n}{k^2} floor}lfloorfrac{n}{k^2i} floor\
&=&sum_{k=1}^{sqrt{n}}varphi(k)S(lfloorfrac{n}{k^2} floor)
end{eqnarray*}]

其中

[S(n)=sum_{i=1}^nlfloorfrac{n}{i} floor]

枚举所有$k$，然后分段计算$S$即可。

时间复杂度

[egin{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;
}