2409 Let it Bead //polya 定理

Let it Bead

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2209		Accepted: 1363

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

1 1
2 1
2 2
5 1
2 5
2 6
6 2
0 0

Sample Output

1
2
3
5
8
13
21

Source

Ulm Local 2000

/* 问题：给你m种不同颜色的珠子，你能穿出多少长度为n的“本质不同”首饰来。这里的“本质不同”指首饰经过翻转或旋转，不会与其他方案相同。我们称之为着色方案。例如图中给出的2种颜色，穿出5个珠子的首饰的各种方案示意图。 思路： 置换：【1,2,…,n | a1, a2,…an】 表示1被a1取代（a1为1到n的某数），2被a2取代……， a1，a2…，an不相同 置换群：置换群的元素是置换，运算时置换的连接。例如：【1,2,3,4 | 3,1,2,4】【1,2,3,4 | 4,3,2,1】 = 【1,2,3,4 | 2,4,3,1】 轮换：记(a1a2…an) = 【a1,a2,a3,…an | a2,a3,…,an,a1】 方案与方案之间其实是可以看做一种置换，即由其中一种方案通过置换可以变成另一种方案。例如有4个珠子，2种颜色，我们看一下前两个珠子白色，后两个黑色的情况。这种方案可以通过置换【2，3，4，1】可以变成第一个和第四个是黑色，第二个和第三个为白色的方案； Burnside引理： 置换会使一个着色方案变成另一个方案。但是有些方案比较特殊，他在某些置换下会置换到他自身。我们称之在置换f下不变。记C(f)为置换f下保持不变的着色方案个数。那么就有一个结论：本质不同的着色方案数为所有置换f的C(f)值的平均数。 我们设f的置换群为G，则Burnside引理：等价类数目 N = sum{C(f)} / |G|， f属于G。 这个公式虽然简单，但是对于C(f)的求解还是个问题，于是介绍下面的定理。 Polya定理： 设G是N个对象上的置换群，用M种颜色涂染这N个对象，则不同的染色方案数为： LG =( Mλ(g1)+Mλ(g2)+…+Mλ(gp))/|G| 其中G=（g1,g2, …gp）, λ(gk)为置换gk的轮换的个数。 利用置换论中轮换的概念（轮换概念=循环概念），我们将置换f分解为若干循环。记f的循环节数位m(f)。 则有定理:C(f) = k ^ m(f) ，其中看为着色数。 于是有Polya定理：不同的着色方案数 l = sum(k ^ m(f)) / |G| ，其中f属于G。 对于本题还有如下结论： 在此种模型下，所有的置换可以通过一个置换旋转和翻转全部得来。 （1）旋转 将置换顺时针旋转i格，其循环节数的为gcd(n, i)； （2）翻转 当n为奇数：共有n个循环节数为(n+1)/2的循环群 当 n为偶数：共有n/2个循环节数(n+2)/2的循环群，和n/2个循环节数n/2的循环群。 Polya定理是个非常实用的定理，记p为格子数，s为着色数，则它可以在O(ps)内结果着色问题。 */ #include <iostream> #include <math.h> using namespace std; typedef long long int64; int64 gcd(int64 a, int64 b) { if (b) return gcd(b, a % b); else return a; } int main() { int64 n, i; double m; while(cin>>m>>n, n || m)//n=length ,m is the number of colors { int64 ans = 0; for (i = 1; i <= n; i++) ans += (int64)pow( m, (int)gcd(n, i) ); //m^gcd(n,i) if (n & 1) ans += (int64)(pow(m, (int)(n + 1) / 2) * n); else { ans += (int64)(pow(m, (int)n / 2) * (n / 2)); ans += (int64)(pow(m, (int)(n + 2) / 2) * (n / 2)); } ans /= (2 * n); cout<<ans<<endl; } return 0; }