用n个颜色的珠子编项链,求有多少种情况
由N(G,C) = 所有f的稳定核的和/|G|
m边形有m种旋转m种翻转
首先说旋转,有模线性方程可知每种旋转都有gcd(m,i)个循环节且每个循环节长度为n/gcd(m,i)
所以每个旋转的稳定核 = pow(n,gcd(m,i))
翻转的循环节数可有观察得知
#include<algorithm> #include<string.h> #include<stdio.h> #include<cmath> #include<iostream> using namespace std; #define ll long long int main(){ //freopen("in.cpp", "r", stdin); ll ans; int n, m; while(scanf("%d%d", &n, &m), n+m){ ans = 0; ll sum = 1; for(int i =1; i <= m; i++){ sum += pow(n, __gcd(i, m)); } ans = sum;//cout<<"*"<<ans<<endl; if(m & 1){ sum = 1; for(int i = 0; i < (m+1)/2; i++){ sum *= n; } sum *= m; ans += sum; }else{ sum = 1; for(int i = 0; i < m/2; i++){ sum *= n; } ans += m/2*(sum + sum*n); }//cout<<ans<<endl; ans/= m*2; cout<<ans<<endl; } }