Polya 计数

View Code
/*==================================================*\
| Polya计数
| c种颜色的珠子, 组成长为s的项链, 项链没有方向和起始位置;
\*==================================================*/
// 注意超long long
int Polya(int c,int s) {
    int i,j,k,p[64], count; p[0]=1; // power of c
    for(k=1; k<=s ;k++) p[k]=p[k-1]*c;
    //reflection part
    count=(s&1)? (s*p[s/2+1]) : ((s/2)*(p[s/2]+p[s/2+1]));
    //rotation part
    for(k=1; k<=s ;k++) count+=p[gcd(k,s)];
    cout/=2*s;
    return count;
}
原文地址:https://www.cnblogs.com/zhang1107/p/3032286.html