置换群Polya定理（poj 2409: Let it Bead）

最新推荐文章于 2021-10-17 21:55:16 发布

Jaihk662

最新推荐文章于 2021-10-17 21:55:16 发布

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本文介绍如何利用Polya定理解决手链计数问题，详细解释了置换群burnside引理的应用，并提供了具体的算法实现。适用于m种颜色、n长度的手链制作方案计算。

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置换群burnside引理（bzoj 1004: [HNOI2008]Cards）

Polya定理公式（必须在没有限制下才能使用此公式）：

其中|G|为总置换数，m表示可用的颜色数，c(gi)为第i种置换的循环节个数

Let it Bead

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6079		Accepted: 4072

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

题意：

你有m种颜色的珠子无数个，问能用这些珠子做出多少种不同的周长为n的珍珠项链

如果两个项链能通过翻转和旋转变成一样，那么这两种项链就当做是同一种

Polya模板题

求出所有置换循环节，假设有n颗珠子，m种颜色

①顺时针循环i颗珠子，循环节个数为Gcd(n, i)

②n为奇数时：以第i颗珠子为中心翻转，循环节个数为(n-1)/2+1

n为偶数时：以第i颗珠子为中心翻转，对面的珠子也不会动，循环节个数为(n-2)/2+2

以两颗珠子中间为中心翻转，循环节个数为n/2

置换总数|G| = 2*N

#include<stdio.h>
int Gcd(int x, int y)
{
	if(y==0)
		return x;
	return Gcd(y, x%y);
}
int Pow(int x, int y)
{
	int ans = 1;
	while(y)
	{
		if(y%2)
			ans = ans*x;
		x = x*x;
		y /= 2;
	}
	return ans;
}
int main(void)
{
	int m, n, i, ans;
	while(scanf("%d%d", &m, &n), m!=0 || n!=0)
	{
		ans = 0;
		for(i=1;i<=n;i++)
			ans += Pow(m, Gcd(n, i));
		if(n%2)
			ans += n*Pow(m, n/2+1);
		else
			ans += n*Pow(m, n/2+1)/2+n*Pow(m, n/2)/2;
		printf("%d\n", ans/(2*n));
	}
	return 0;
}