数学计数原理（Pólya）：POJ 1286 Necklace of Beads

Necklace of Beads

Description

Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 24 ). If the repetitions that are produced by rotation around the center of the circular necklace or reflection to the axis of symmetry are all neglected, how many different forms of the necklace are there?

Input

The input has several lines, and each line contains the input data n.

-1 denotes the end of the input file.

Output

The output should contain the output data: Number of different forms, in each line correspondent to the input data.

Sample Input

4
5
-1

Sample Output

21
39　　　　公式是这样子的：

　　p是颜色数，这里等于3，可以发现这2*n个置换形成了置换群，满足了群的封闭性。

　　那么只要对于每个置换找不动点就好了…… http://www.cnblogs.com/TenderRun/p/5656038.html 循环的部分和这题类似

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5 long long pow[30],phi[30],n,ans;
 6 long long Gcd(long long a,long long b){
 7     return b?Gcd(b,a%b):a;
 8 }
 9 int main(){
10     pow[0]=1;
11     for(int i=1;i<=24;i++)
12         pow[i]=pow[i-1]*3;
13     for(int i=1;i<=24;i++)
14         for(int j=i;j>=1;j--)
15             if(Gcd(i,j)==1)phi[i]+=1;
16     while(scanf("%lld",&n)!=EOF&&n!=-1){
17         if(n==0){printf("0\n");continue;}
18         for(int d=1;d<=n;d++)
19             if(n%d==0)ans+=phi[n/d]*pow[d];
20         if(n%2)ans=(ans+n*pow[(n+1)/2])/2/n;
21         else ans=(ans+n/2*(pow[n/2+1]+pow[n/2]))/2/n;
22         printf("%lld\n",ans);ans=0;
23     }
24 }
25