POJ2409Let it Bead(polya定理)

Let it Bead

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5495		Accepted: 3678

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

//用c种颜色不同的珠子，可以造多少种不同的长度为s的手链
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int gcd(int a,int b)
{
    b=b%a;
    while(b)
    {
        a=a%b;
        swap(a,b);
    }
    return a;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m),n|m)
    {
        int sum=0;
        for(int i=1;i<=m;i++)//旋转所得方法
            sum+=pow(n,gcd(i,m));
        if(m&1)
            sum+=m*pow(n,m/2+1);//翻转所得：奇数群，对称轴在两颗珠子之间，循环个数为(n+1)/2,n个循环群
        else
            sum+=m/2*pow(n,m/2)+m/2*pow(n,m/2+1);//偶数群，对称轴过两颗珠子，循环个数（n+2）/2,n/2个循环群
        sum/=m*2;//对称轴过两颗相邻珠子之间，循环个数为(n)/2，n/2个循环群
        printf("%d\n",sum);
    }
    return 0;
}