POJ 1286 波利亚原理

Necklace of Beads

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4189		Accepted: 1726

Description

Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 24 ). If the repetitions that are produced by rotation around the center of the circular necklace or reflection to the axis of symmetry are all neglected, how many different forms of the necklace are there?

Input

The input has several lines, and each line contains the input data n.
-1 denotes the end of the input file.

Output

The output should contain the output data: Number of different forms, in each line correspondent to the input data.

Sample Input

4
5
-1

Sample Output

21
39

Source

Xi'an 2002

这个题应该算是波利亚原理的入门题目了。

他只输入n，而且颜色数量只有3钟，然后又不去模。所以不用考虑逆元什么的

有两点要注意：

1、n=32的时候答案其实已经超过int了，要用int64才能装得下

2、n=0的时候答案不是1而是0

以上两个地方我害我WA了几次。。。

我的代码：

#include<stdio.h>
#include<string.h>

typedef __int64 ll;

ll eular(ll n)
{
	ll ret=1,i;
	for(i=2;i*i<=n;i++)
	{
		if(n%i==0)
		{
			ret=ret*(i-1);
			n=n/i;
			while(n%i==0)
			{
				n=n/i;
				ret=ret*i;
			}
		}
		if(n==1)
			break;
	}
	if(n>1)
		ret=ret*(n-1);
	return ret;
}

ll exmod(ll a,ll b)
{
	ll ret=1,i;
	for(i=1;i<=b;i++)
		ret=ret*a;
	return ret;
}

int main()
{
	ll n,ans,i;
	while(scanf("%I64d",&n)!=EOF)
	{
		if(n==-1)
			break;
		ans=0;
		if(n==0)
		{
			printf("0\n");
			continue;
		}
		for(i=1;i*i<n;i++)
		{
			if(n%i==0)
			{
				ans=ans+eular(n/i)*exmod(3,i);
				ans=ans+eular(i)*exmod(3,n/i);
			}
		}
		if(i*i==n)
			ans=ans+eular(n/i)*exmod(3,i);
		if(n&1)
			ans=ans+exmod(3,n/2+1)*n;
		else
			ans=ans+exmod(3,n/2)*n/2+exmod(3,n/2+1)*n/2;
		printf("%I64d\n",ans/n/2);
	}
	return 0;
}