POJ 1286 Necklace of Beads Polya .

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long LL;
int gcd(int a,int b)
{
if(b==0) return a;
return gcd(b,a%b);
}
int euler_phi(int n)
{
int res=1;
for(int i=2;i*i<=n;i++)
if(n%i==0) {         //说明i|n
n/=i,res*=i-1;
while(n%i==0) n/=i,res*=i;  //说明i^2|n
}
if(n>1) res*=n-1;
return res;
}
LL polya(int n)
{
LL tot=0;  //方案数
for(int i=1;i<=n;i++)
tot+=pow(3,gcd(n,i));
tot/=n;
if(n%2!=0) tot+=pow(3,(n+1)/2); //odd
else tot+=(pow(3,n/2)+pow(3,n/2+1))/2;
return tot/2;
}
int main()
{
int n;

while(cin>>n&&n!=-1)
if(n==0) cout<<0<<endl;
else cout<<polya(n)<<endl;

return 0;
}