\(Description\)
求\[\sum_{i=1}^n\gcd(i,n)\]
\(Solution\)
\[ \begin{aligned}
\sum_{i=1}^n\gcd(i,n)
&=\sum_{d=1}^nd\sum_{i=1}^n[\gcd(i,n)=d]\ &=\sum_{d=1}^nd\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}[\gcd(i,\lfloor\frac{n}{d}\rfloor)=1]
\end{aligned}
\]
后一项不需要再化了，因为就是\(\phi(\lfloor\frac{n}{d}\rfloor)\)。
所以
\[\sum_{i=1}^n\gcd(i,n)=\sum_{d=1}^nd*\phi(\lfloor\frac{n}{d}\rfloor)\]
因为\(\gcd(i,n)\mid n\)，所以
\[ \begin{aligned}
\sum_{i=1}^n\gcd(i,n)
&=\sum_{d=1}^nd*\phi(\lfloor\frac{n}{d}\rfloor)\ &=\sum_{d\mid n}d*\phi(\lfloor\frac{n}{d}\rfloor)
\end{aligned}
\]
约数可以\(O(\sqrt{n})\)枚举，\(\phi\)可以\(O(\sqrt{n})\)求，复杂度为\(因子个数*\sqrt{n}\)。
//928kb 56ms
//注意d！
#include <cmath>
#include <cstdio>
typedef long long LL;
const int N=1<<16;
int cnt,P[N>>3];
LL n;
bool Not_p[N+3];
void Make_Table(int N)
{
for(int i=2; i<=N; ++i)
{
if(!Not_p[i]) P[++cnt]=i;
for(int j=1; j<=cnt&&i*P[j]<=N; ++j)
{
Not_p[i*P[j]]=1;
if(!(i%P[j])) break;
}
}
}
LL Phi(LL x)
{
LL res=1;
for(int i=1; i<=cnt&&1ll*P[i]*P[i]<=x; ++i)
if(!(x%P[i]))
{
x/=P[i], res*=(P[i]-1);
while(!(x%P[i])) x/=P[i], res*=P[i];
}
if(x>1) res*=x-1;
return res;
}
int main()
{
scanf("%lld",&n);
Make_Table(sqrt(n)+1);
LL res=0;
int lim=sqrt(n);
for(int i=1,lim=sqrt(n); i<=lim; ++i)
if(!(n%i)) res+=1ll*i*Phi(n/i)+1ll*(n/i)*Phi(i);//!
if(1ll*lim*lim==n) res-=lim*Phi(lim);
printf("%lld",res);
return 0;
}