POJ 2409 Let it Bead

最新推荐文章于 2018-08-10 16:11:09 发布

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本文介绍了一种基于Polya定理的独特手链计数算法，该算法用于估算使用特定数量的颜色和珠子能制作的不同手链的最大数量。文章详细解释了如何通过考虑旋转和翻转操作来确定手链的独特性，并提供了实现这一算法的C++代码。

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Description

"Let it Bead" company is located upstairs at 700 Cannery Row in Monterey, CA. As you can deduce from the company name, their business is beads. Their PR department found out that customers are interested in buying colored bracelets. However, over 90 percent of the target audience insists that the bracelets be unique. (Just imagine what happened if two women showed up at the same party wearing identical bracelets!) It's a good thing that bracelets can have different lengths and need not be made of beads of one color. Help the boss estimating maximum profit by calculating how many different bracelets can be produced.

A bracelet is a ring-like sequence of s beads each of which can have one of c distinct colors. The ring is closed, i.e. has no beginning or end, and has no direction. Assume an unlimited supply of beads of each color. For different values of s and c, calculate the number of different bracelets that can be made.

Input

Every line of the input file defines a test case and contains two integers: the number of available colors c followed by the length of the bracelets s. Input is terminated by c=s=0. Otherwise, both are positive, and, due to technical difficulties in the bracelet-fabrication-machine, cs<=32, i.e. their product does not exceed 32.

Output

For each test case output on a single line the number of unique bracelets. The figure below shows the 8 different bracelets that can be made with 2 colors and 5 beads.

Sample Input

Sample Output

Source

Ulm Local 2000

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

polya定理~

定理内容：本质不同为1/|置换总数|*（个数*所有颜色数^（循环节长度）之和）种。

所以我们考虑置换种类。

1.旋转：每次可以旋转0~n-1长度，总共循环lcm(i,n)/i次即可转到原位，经历了n/(lcm(i,n)/i)个位置，所以循环节长度为gcd(i,n)。

2.翻转：分为n%2==1和n%2==0两种。

2.1n%2==1时，对称轴为某一点及它对面的两点的中点。所以个数为n种。循环节n/2+1。

2.2n%2==0时，对称轴为某对应两点或对应四点的中点。个数分别为n/2和n/2。循环节分别为n/2+1和n/2。

然后按照公式写就可以了~

注意先输入的是m（即颜色数）！

#include<cstdio>
#include<iostream>
using namespace std;

int n,m,ans;

int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0' || ch>'9') {if(ch=='-') f=-1;ch=getchar();}
	while(ch>='0' && ch<='9') {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}

void writ(int u)
{
	if(u>9) writ(u/10);
	putchar((u%10)+'0');
}

int mi(int u,int v)
{
	int now=1;
	while(v)
	{
		if(v&1) now*=u;
		u*=u;v>>=1;
	}
	return now;
}

int gcd(int u,int v)
{
	return v ? gcd(v,u%v):u;
}

int main()
{
	while(m=read())
	{
		n=read();ans=0;
		for(int i=0;i<n;i++) ans+=mi(m,gcd(i,n));
		if(n&1)
		{
			ans+=n*mi(m,(n>>1)+1);
			ans/=n*2;writ(ans);puts("");
		}
		else
		{
			ans+=n/2*mi(m,(n>>1)+1)+n/2*mi(m,n>>1);
			ans/=n*2;writ(ans);puts("");
		}
	}
	return 0;
}