POJ 1286 Necklace of Beads Ploya定理

来源：http://poj.org/problem?id=1286

题意：用三种颜色着色一个长度为n的项链，问有多少种方法。。。

思路：继续裸的Ploya，水水更健康。。。被n=0坑了一次re，被long long坑了一次wa，，，

代码：

#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;

typedef long long ll;
int gcd(int a,int b){
	if(b == 0)
		return a;
	return gcd(b,a%b);
}
ll mi(int x){
	ll s = 1;
	for(int i = 1;i <= x;++i)
		s *= 3;
	return s;
}
int main(){
	int n;
	while(scanf("%d",&n) != EOF){
	  if(n == -1)
		  break;
	  if(n == 0){
	    printf("0\n");
		continue;
	  }
	  ll sum = 0;
	  for(int i = 1;i <= n;++i){
	    int x = gcd(n,i);
		sum += mi(x);
	  }
	  if(n % 2)
		  sum += ( n * mi( (n-1)/2 + 1 ) );
	  else
		  sum += ( n/2 * mi( n/2 ) + n/2 * mi( n/2 + 1 ));
	  printf("%lld\n",sum / (2*n));
	}
	return 0;
}