hdu 3923 polya定理

刚开始，没什么思路，后来才知道要用到polya定理，本来还以为是一道排列问题，最后却变成了一道群的轮换问题，最后由于数据非常大，又要mod除法运算，因此要求乘法逆元。

polya定理 参考：http://www.docin.com/p-43391964.html

                                    http://www.cppblog.com/sdfond/archive/2010/02/06/107403.aspx

                                    http://hi.baidu.com/lewutian/blog/item/c75dff06c4693f6a03088100.html

mod除法运算：http://wenku.baidu.com/view/6f2879cca1c7aa00b52acb5f.html

                               http://wenku.baidu.com/view/6f2879cca1c7aa00b52acb5f.html

#include<cstdio>
#include<cstring>
#include<cmath>

long long colorgcd[10005];
const int mod=1000000007;

int gcd(int a,int b){    return b?gcd(b,a%b):a;}

int Extended_Euclid(int x,int mod)
{
    int X1=1,X2=0,X3 = mod;
    int Y1=0,Y2=1,Y3 = x;
    while(true)
    {
        if (Y3 == 0)    return 0; //无逆元
        if (Y3 == 1)    return Y2; //Y2为逆元
        int Q = X3 / Y3;
        int T1 = X1 - Q*Y1, T2 = X2 - Q*Y2, T3 = X3 - Q*Y3;
        X1 = Y1; X2 = Y2; X3 = Y3;
        Y1 = T1; Y2 = T2; Y3 = T3;
    }
}

long long polya(int color,int n)
{
    int i;
    long long sum=0;
    colorgcd[0] = 1;
    for(i=1;i<=n;i++)
        colorgcd[i] = (colorgcd[i-1] * color) % mod;
    for(i=1; i<=n; i++)//旋转
    {
        sum += colorgcd[gcd(i,n)];
        sum %= mod;
    }
    if(n&1)//翻转
    {
        sum += n*colorgcd[n/2+1];
        sum %= mod;
        sum *= Extended_Euclid(2*n,mod);
        sum %= mod;
    }
    else
    {
        sum += (n/2)*colorgcd[n/2];
        sum %= mod;
        sum += (n/2)*colorgcd[n/2+1];
        sum %= mod;
        sum *= Extended_Euclid(2*n,mod);
        sum %= mod;
    }
    sum = (sum + mod) %mod;
    return sum;
}

int main()
{
    //freopen(“out.txt”,”w”,stdout);
    int cas,icas=0;
    scanf(“%d”,&cas);
    while(cas–)
    {
        int color,num;
        scanf(“%d%d”,&color,&num);
        memset(colorgcd,0,sizeof(colorgcd));
        printf(“Case #%d: %I64d\n”,++icas,polya(color,num));
    }
    return 0;
}