Poj 2409 Let it Bead Polya count

Poj 2409 Let it Bead Polya count

Rotation can be divided into n replicas. The corresponding equivalence classes are gcd (n, I) I = 0, and n replicas exist.

The flip is divided into parity discussions. There are n replicas in an odd number, each of which has n/2 + 1

There are n replicas in an even number, with n/2 + 1 in half and n/2 in half.

I have been reading Polya theorem for a long time in my thesis, PPT, and various books. Recently I am seriously skeptical about my IQ when I am doing a math problem.

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      using namespace std;typedef long long ll;ll gcd(ll a,ll b) {return a%b==0?b:gcd(b,a%b);}ll quickpow(ll m,ll n){    ll ans=1;    while(n)    {        if(n&1) ans=ans*m;        n=(n>>1);        m=m*m;    }    return ans;}int main(){    ll c,n;    while(~scanf("%lld%lld",&c,&n))    {        if(c+n==0) break;        ll ans=quickpow(c,n);        for(int i=1;i