poj 1286&&poj2409 Polya计数 颜色匹配

#include <iostream>
#include <math.h>using namespace std;#define LL long long
LL gcd(LL a, LL b)
{
    return b ? gcd(b, a % b) : a;
}LL polya(LL n)
{
    LL ret = ;
    for(LL i = ; i < n; i++)
        ret += pow(, gcd(i, n));
    //flip them...
    if( n &  )//odd
        ret += n * pow(, n /  + );//symmetric axis's num is n, and a cycle of (n + 1) / 2, with the length of 2, and 2 cycles with length of 1...      else//even      ret += n / 2 * pow(3, n / 2) + (n / 2) * pow(3, n / 2 + 1);//symmetric axis's num is n, categoried by the beeds, for n/2 axis which through the beed, they formed (n/2-1) cycles with the length of 2, and 2 cycles with the length of 1; for the n/2 axis which not through the beed, they formed (n/2) cycles with the length of 2.
    else//even
   ret += n /  * pow(, n / ) + (n / ) * pow(, n /  + );//    return ret / n / ;//the average of them(according to Polya Theorem.)
}int main()
{
    LL n;
    while(cin>> n && n != -)
    {
        if (n <= ) cout <<  << endl;
        else cout << polya(n) << endl;
    }
    return ;
}

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>using namespace std;
#define n 3__int64 m;int gcd(int a, int b)
{
    b = b % a;
    while (b)
    {
        a = a % b;
        swap(a, b);
    }
    return a;
}int main()
{
    while (scanf("%lld", &m)!=EOF)
    {
        if(m==-)
        break;
        __int64 ans = ;
        for (int i = ; i <= m; i++)
            ans += pow(n*1.0, gcd(i, m)*1.0);
        if (m & )
            ans += m * pow(n*1.0, (m /  + )*1.0);
        else
            ans += m /  * pow(n*1.0, (m / )*1.0) + m /  * pow(n*1.0, (m /  + )*1.0);
        ans /= m * ;
        printf("%I64d\n", ans);
    }
    return ;
}