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 cdistinct 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.
给定颜色种数和环上的珠子总数，问有多少种染色方案（通过旋转和翻转相同的算同一种）。
Input
Every line of the input 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
polya置换的裸题了，考虑旋转，我们枚举所有可能的旋转方式，所以得到的循环节个数为gcd(i,n)，因此答案为(sumlimits_{i=1}^{n} c^{gcd(n,i)})
再考虑一下翻转，我们分奇数和偶数进行讨论，如果是奇数，那么不论如何找对称轴，都必定会形成(frac{n}{2}+1)个循环节，再乘上(n)即可；如果是偶数，那么就会有(frac{n}{2})和(frac{n-2}{2}+2)两种循环节情况，然后每种情况各占(frac{n}{2})条对称轴
最后把答案除一下置换总数(2*n)即可
/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline int read(){
int x=0,f=1;char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x>=10) print(x/10);
putchar(x%10+'0');
}
int gcd(int a,int b){return !b?a:gcd(b,a%b);}
int mlt(int a,int b){
int res=1;
for (;b;b>>=1,a=a*a) if (b&1) res=res*a;
return res;
}
int main(){
while (true){
int m=read(),n=read(),ans=0;
if (!n&&!m) break;
for (int i=1;i<=n;i++) ans+=mlt(m,gcd(n,i));
if (n&1) ans+=n*mlt(m,n/2+1);
else ans+=(mlt(m,n/2+1)+mlt(m,n/2))*(n>>1);
printf("%d
",ans/(2*n));
}
return 0;
}