ACM: polya定理 数论题 poj 2409

                                                                  Let it
Bead

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

题意: 一个环上有s个珠,有c种颜色,染色通过翻转,旋转相同的方案算是同一种方案.计算有多少种不同的方案.

解题思路:

            
1.polya定理题.

              
Polya定理:设G是n个对象的一个置换群,用m种颜色涂染这n个对象,则不同染色的方案数

              
L=1/|G|*[m^p(a1)+m^p(a2)+....+m^p(an)].其中p(ai)是某个置换的循环数.

             
（1).旋转置换.

                      
我们假设依次顺时针旋转1~n个,则循环个数=gcd(i,n);

               
(2).翻转置换

                      
当n为偶数时,分两种情况,一种是中心轴在两个对称对象上,则循环个数为n/2+1,

                                         
另一种是对称轴两边分别有n/2个对象,则循环个数为n/2;

                      
当n为奇数时,对称轴就只能在一个对象上,则循环个数为n/2+1;
代码:

#include
<cstdio>
#include
<iostream>
#include
<cmath>
using namespace std;

int s, c;

int gcd(int a,int b)
{
  
 return b == 0 ? a : gcd(b,a%b);
}

int polya()
{
  
 double sum = 0;

    
  
 for(int i = 1; i <= s;
++i)
  
   
 sum +=
pow(c,gcd(s,i));  
 
  
 if(s % 2 == 0)
  
   
 sum += pow(c,s/2+1)*(s/2) +
pow(c,s/2)*(s/2);
  
 else
  
   
 sum += pow(c,s/2+1)*s;
  
 return sum / 2 / s;
}

int main()
{
//  
 freopen("input.txt","r",stdin);
  
 while(scanf("%d
%d",&c,&s) != EOF
&& (s||c))
  
 {
  
   
 printf("%d\n",polay());
  
 }
  
 return 0;
}