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;

}