poj 2409 Let it Bead Polya波利亚定理

Let it Bead

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5054 Accepted: 3362

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 12 12 25 12 52 66 20 0

Sample Output

123581321

Source

Ulm Local 2000

我是小白，初识Polya定理。

不过看了各种代码，自己琢磨了半天，似乎有这么一回事：

对于一个置换

1,2,3,4,...,n

1,2,3,4,...,n

现在进行顺时针，或逆时针旋转（就是把底下一行推进一格或置后一格，再把多出的元素补到空位上）

操作k次，那么得到的置换群可以表示为循环的个数为gcd(n,k);

现在考虑翻转操作:

如果n是奇数：那么循环个数为 (n+1)/2;

如果n是偶数：那么要循环n次，其中n/2个置换群可以表示为 n/2个循环，还有n/2个可以表示为(n+2)/2个循环。

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<climits>#include<queue>#include<vector>#include<map>#include<sstream>#include<set>#include<stack>#include<cctype>#include<utility>#pragma comment(linker, "/STACK:102400000,102400000")#define PI 3.1415926535897932384626#define eps 1e-10#define sqr(x) ((x)*(x))#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)#define  lson   num<<1,le,mid#define rson    num<<1|1,mid+1,ri#define MID   int mid=(le+ri)>>1#define zero(x)((x>0? x:-x)<1e-15)#define mk    make_pair#define _f     first#define _s     secondusing namespace std;//const int INF=    ;typedef long long ll;//const ll inf =1000000000000000;//1e15;//ifstream fin("input.txt");//ofstream fout("output.txt");//fin.close();//fout.close();//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);const int INF =0x3f3f3f3f;//const int maxn=    ;//const int maxm=    ;//by yskysker123int n,m;int main(){    while(~scanf("%d%d",&m,&n)&&(m||n))    {        int ans=0;        for(int i=1;i<=n;i++)        {            ans+=pow(m,__gcd(i,n));        }        if(n&1)         ans+=n*pow(m,(n+1)/2);        else          ans+=n/2*pow(m,n/2)+n/2*pow(m,(n+2)/2);        ans/=2*n;        printf("%d\n",ans);    }    return 0;}