POJ2409 - Click Here

# 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.

# Solutions

Polya入门题。

考虑两种置换，翻转和旋转，显然这些置换**成群**。

翻转：

当$s$为奇数时，可以以每个点作轴构造置换，此时循环数为$\frac{s + 1}{2}$

当$s$为偶数时，可以以每个点作轴构造置换，此时循环数为$\frac{s}{2}+1$；亦可以以相邻两个点的中点作轴构造置换，此时循环数为$\frac{s}{2}$

旋转：

枚举$1$与哪个点换位，然后构造出置换。对于每个置换，都暴力分解循环。

由于对于相同的$s$，旋转的置换是固定的，因此旋转的置换可以预处理。剩下的直接套用Polya就可以求解了。

时间复杂度$O(s^3+Ts)$

# CODE

CODE - Click Here