【poj2409】Let it Bead

2015年4月24日1,3200

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

转自http://www.cnblogs.com/DrunBee/archive/2012/09/10/2678378.html

题意:用k种颜色对n个珠子构成的环上色,旋转翻转后相同的只算一种,求不等价的着色方案数。

Burnside定理的应用:

当n为奇数时,有n种翻转,每种翻转都是以一个顶点和该顶点对边的中点对称。有k^(n/2+1)*n种。

当n为偶数时,有n种翻转,其中一半是以两个对应顶点,另一半是以两条对边对称。有k^(n/2+1)*n/2+k^(n/2)*n/2种。

考虑旋转:枚举旋转角度360/n*i,(0<i<=n),也就是一个置换。经过该置换,颜色仍保持不变的着色方案有k^GCD(n,i)种。

一个长度为n的环,每i个上同一种颜色,可以上多少种颜色。

假设起点在x,则x,x+i,x+2*i,……,x+k*i,……

假设在第t次,第一次回到起点,则x=(x+t*i)%n => t*i%n=0 => t=LCM(i,n)/i=n*i/GCD(n,i)/i=n/GCD(n,i)。

那么可以上n/t种颜色,即n/(n/GCD(n,i))种,所以旋转的着色方案有k^GCD(n,i)种。