POJ 1286 Necklaces of Beads （Burnside定理，有限制型）

题目链接：http://vjudge.net/problem/viewProblem.action?id=11117

就是利用每种等价情形算出置换节之后算组合数

#include <stdio.h>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
           
using namespace std;
#define lson o<<1
#define rson o<<1|1
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)
#define INF 200000000
           
typedef long long ll;
ll qpow(ll a,int k){
   ll ans=1;
   while(k>0){
     if(k&1)ans*=a;
     a*=a;
     k>>=1;
   }
   return ans;
}
int gcd(int a,int b){
   if(a==0)return b;
   return gcd(b%a,a);
}
ll rotation(int n){
   ll rot=3;
   for(int i=2;i<=n;i++){
      rot+=qpow(3LL,gcd(i,n));
   }
   return rot;
}
ll reflection(int n){
   ll ref=0;
   if(n&1){
      int cir=n/2+1;
      ref=qpow(3LL,cir)*n;
   }else{
      ref=(ll)(n/2)*qpow(3LL,n/2);
      ref+=(ll)(n/2)*qpow(3LL,1+n/2);
   }
   return ref;
}           
int main(){
   int n;
   while(~scanf("%d",&n) && n+1){
      ll ans;
      if(!n)ans=0;
      else ans=(rotation(n)+reflection(n))/(2*n);
      printf("%I64d\n",ans);
   }
   return 0;
}