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=;
   while(k>){
     if(k&)ans*=a;
     a*=a;
     k>>=;
   }
   return ans;
}
int gcd(int a,int b){
   if(a==)return b;
   return gcd(b%a,a);
}
ll rotation(int n){
   ll rot=;
   for(int i=;i<=n;i++){
      rot+=qpow(3LL,gcd(i,n));
   }
   return rot;
}
ll reflection(int n){
   ll ref=;
   if(n&){
      int cir=n/+;
      ref=qpow(3LL,cir)*n;
   }else{
      ref=(ll)(n/)*qpow(3LL,n/);
      ref+=(ll)(n/)*qpow(3LL,+n/);
   }
   return ref;
}
int main(){
   int n;
   while(~scanf("%d",&n) && n+){
      ll ans;
      if(!n)ans=;
      else ans=(rotation(n)+reflection(n))/(*n);
      printf("%I64d\n",ans);
   }
   return ;
}