(Description)
求$$sum_{i=1}^ngcd(i,n)$$
(Solution)
[ egin{aligned}
sum_{i=1}^ngcd(i,n)
&=sum_{d=1}^ndsum_{i=1}^n[gcd(i,n)=d]\
&=sum_{d=1}^ndsum_{i=1}^{lfloorfrac{n}{d}
floor}[gcd(i,lfloorfrac{n}{d}
floor)=1]
end{aligned}
]
后一项不需要再化了，因为就是(phi(lfloorfrac{n}{d}
floor))。
所以
[sum_{i=1}^ngcd(i,n)=sum_{d=1}^nd*phi(lfloorfrac{n}{d}
floor)
]
因为(gcd(i,n)mid n)，所以
[ egin{aligned}
sum_{i=1}^ngcd(i,n)
&=sum_{d=1}^nd*phi(lfloorfrac{n}{d}
floor)\
&=sum_{dmid n}d*phi(lfloorfrac{n}{d}
floor)
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;
}