poj2409Let it Bead

标签：acm   algorithm   算法   poj   

polya计数法

简单模板

void solve(int n,int c)
{
    int ans = 0;
    for (int i = 1; i <= n; i++) if (n % i == 0)
        {
            ans += fun(c, i) * euler(n / i);
        }
    if (n & 1) ans += n * fun(c, n / 2 + 1);
    else ans += n / 2 * (fun(c, n / 2) + fun(c, n / 2 + 1));
    cout << ans / (2 * n) << endl;
}

#include<cstdio>
#include<iostream>
using namespace std;

int fun(int a,int x)
{
    int ans=1;
    while(x)
    {
        if(x&1)
            ans=ans*a;
        a=a*a;
        x>>=1;
    }
    return ans;
}
int euler(int n){
     int res=n,a=n;
     for(int i=2;i*i<=a;i++){
         if(a%i==0){
             res=res/i*(i-1);
             while(a%i==0) a/=i;
         }
     }
     if(a>1) res=res/a*(a-1);
     return res;
}
void solve(int n,int c)
{
    int ans = 0;
    for (int i = 1; i <= n; i++) if (n % i == 0)
        {
            ans += fun(c, i) * euler(n / i);
        }
    if (n & 1) ans += n * fun(c, n / 2 + 1);
    else ans += n / 2 * (fun(c, n / 2) + fun(c, n / 2 + 1));
    cout << ans / (2 * n) << endl;
}
int main()
{
    int n,c;
    while(~scanf("%d%d",&c,&n)&&n&&c)
    {
        solve(n,c);
    }
    return 0;
}

poj2409Let it Bead

标签：acm   algorithm   算法   poj   

原文地址：http://blog.csdn.net/zh9406/article/details/45361275