poj 1286 Necklace of Beads（polya 定理）

最新推荐文章于 2018-07-24 15:44:13 发布

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最新推荐文章于 2018-07-24 15:44:13 发布

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本文探讨了由红、蓝、绿三种颜色珠子组成的圆形项链的不同形态计数问题，在忽略旋转和反射产生的重复形态后，如何计算不同形式的项链数量。通过具体的输入输出示例，介绍了计算方法，并给出了C++实现代码。

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Necklace of Beads

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7571		Accepted: 3152

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

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题解：与poj 2409类似。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long 
using namespace std;
LL ans,n,m;
LL gcd(LL x,LL y)
{
	LL r;
	while (y){
		r=x%y;
		x=y;
		y=r;
	}
	return x;
}
LL pow(LL num,LL x)
{
	LL base=num; LL ans=1;
	while (x){
		if (x&1)  ans=ans*base;
		x>>=1;
		base=base*base;
	}
	return ans;
}
int main()
{
	n=3;
	while (scanf("%lld",&m)!=EOF){
		if (m==0){
			printf("0\n");
			continue;
		}
		if (m==-1)  break;
		ans=0;
		for (LL i=1;i<=m;i++)
		 ans+=pow(n,gcd(m,i));
		if (m%2)  ans+=pow(n,m/2+1)*m;
		else ans+=pow(n,m/2+1)*(m/2),
		     ans+=pow(n,m/2)*(m/2);
		printf("%lld\n",ans/(m*2));
	}
}