HDU 4633 Who's Aunt Zhang （2013多校4 1002 polya计数）

Who's Aunt Zhang

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19    Accepted Submission(s): 16

Problem Description

Aunt Zhang, well known as 张阿姨, is a fan of Rubik’s cube. One day she buys a new one and would like to color it as a gift to send to Teacher Liu, well known as 刘老师. As Aunt Zhang is so ingenuity, she can color all the cube’s points, edges and faces with K different color. Now Aunt Zhang wants to know how many different cubes she can get. Two cubes are considered as the same if and only if one can change to another ONLY by rotating the WHOLE cube. Note that every face of Rubik’s cube is consists of nine small faces. Aunt Zhang can color arbitrary color as she like which means that she doesn’t need to color the nine small faces with same color in a big face. You can assume that Aunt Zhang has 74 different elements to color. (8 points + 12 edges + 9*6=54 small faces)

 

Input

The first line of the date is an integer T, which is the number of the text cases.
Then T cases follow, each case contains one integer K, which is the number of colors. T<=100, K<=100.

 

Output

For each case, you should output the number of different cubes.
Give your answer modulo 10007.

 

Sample Input

3 1 2 3

 

Sample Output

Case 1: 1 Case 2: 1330 Case 3: 9505

 

Source

2013 Multi-University Training Contest 4

 

Recommend

zhuyuanchen520

 

明显是polya计数的题目。

但是本题有74个元素，

总共的变换数数24种。

数起来很麻烦，数据很大。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。

于是我用代码怒搞之。

写代码找不动点。。。。共24种变换，24种都可以通过多次右旋和上旋得到。。。

300多行代码，，，，，终于找到不动点个数了。。。。。。

附上源代码，。包括找不动点。。

找出来可以直接用公式写的，就没有优化了

/*
 *  Author:kuangbin
 *  1002.cpp
 */

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>
#include <map>
#include <vector>
#include <queue>
#include <set>
#include <string>
#include <math.h>
using namespace std;
const int MOD = 10007;
long long pow_m(long long a,long long n)
{
    long long ret = 1;
    long long tmp = a%MOD;
    while(n)
    {
        if(n&1)
        {
            ret *= tmp;
            ret %=MOD;
        }
        tmp *= tmp;
        tmp %=MOD;
        n >>= 1;
    }
    return ret;
}
//求ax = 1( mod m) 的x值，就是逆元(0<a<m)
long long inv(long long a,long long m)
{
    if(a == 1)return 1;
    return inv(m%a,m)*(m-m/a)%m;
}


int b[100];
int c[100];

void rr()
{
    for(int i = 1;i <= 54;i++)
        c[i] = b[i];
    for(int i = 1;i <= 27;i++)
        b[i+9] = c[i];
    for(int i = 1;i <= 9;i++)
        b[i] = c[27+i];
    b[39] = c[37];
    b[42] = c[38];
    b[45] = c[39];
    b[44] = c[42];
    b[43] = c[45];
    b[40] = c[44];
    b[37] = c[43];
    b[38] = c[40];
    b[41] = c[41];


    b[54] = c[52];
    b[51] = c[53];
    b[48] = c[54];
    b[47] = c[51];
    b[46] = c[48];
    b[49] = c[47];
    b[52] = c[46];
    b[53] = c[49];
    b[50] = c[50];

}
void up()
{
    for(int i = 1;i <= 54;i++)
        c[i] = b[i];
    for(int i = 1;i <= 9;i++)
        b[i] = c[45+i];
    for(int i = 37;i <= 45;i++)
        b[i] = c[i-36];
    for(int i = 19;i <= 27;i++)
        b[i] = c[64-i];
    for(int i = 46;i <= 54;i++)
        b[i] = c[73-i];

    b[10] = c[12];
    b[13] = c[11];
    b[16] = c[10];
    b[17] = c[13];
    b[18] = c[16];
    b[15] = c[17];
    b[12] = c[18];
    b[11] = c[15];
    b[14] = c[14];

    b[28] = c[34];
    b[29] = c[31];
    b[30] = c[28];
    b[33] = c[29];
    b[36] = c[30];
    b[35] = c[33];
    b[34] = c[36];
    b[31] = c[35];
    b[32] = c[32];

}

bool used[1000];

int calc()
{
    memset(used,false,sizeof(used));

    int ret = 0;
    for(int i = 1;i <= 54;i++)
        if(!used[i])
    {
        ret++;
        int tmp = i;
        while(!used[tmp])
        {
            used[tmp] = true;
            tmp = b[tmp];
        }
    }
    return ret;

}


void rr2()
{
    for(int i = 1;i <= 12;i++)
        c[i] = b[i];
    b[1] = c[4];
    for(int i = 2;i <= 4;i++)
        b[i] = c[i-1];
    b[5] = c[8];
    for(int i = 6;i <= 8;i++)
        b[i] = c[i-1];
    b[9] = c[12];
    for(int i = 10;i <= 12;i++)
        b[i] = c[i-1];
}
void up2()
{
        for(int i = 1;i <= 12;i++)
        c[i] = b[i];
    b[1] = c[3];
    b[2] = c[7];
    b[3] = c[11];
    b[4] = c[8];
    b[5] = c[4];
    b[6] = c[2];
    b[7] = c[10];
    b[8] = c[12];
    b[9] = c[1];
    b[10] = c[6];
    b[11] = c[9];
    b[12] = c[5];
}
int calc2()
{
    memset(used,false,sizeof(used));

    int ret = 0;
    for(int i = 1;i <= 12;i++)
        if(!used[i])
    {
        ret++;
        int tmp = i;
        while(!used[tmp])
        {
            used[tmp] = true;
            tmp = b[tmp];
        }
    }
    return ret;

}

void rr3()
{
    for(int i = 1;i <= 8;i++)
        c[i] = b[i];
        b[1] = c[4];
    for(int i = 2;i <= 4;i++)
        b[i] = c[i-1];
    b[5] = c[8];
    for(int i = 6;i <= 8;i++)
        b[i] = c[i-1];
}
void up3()
{
    for(int i = 1;i <= 8;i++)
        c[i] = b[i];
    b[1]=c[4];
    b[5]=c[1];
    b[8]=c[5];
    b[4]=c[8];
    b[2]=c[3];
    b[3]=c[7];
    b[7]=c[6];
    b[6]=c[2];
}
int calc3()
{
    memset(used,false,sizeof(used));

    int ret = 0;
    for(int i = 1;i <= 8;i++)
        if(!used[i])
    {
        ret++;
        int tmp = i;
        while(!used[tmp])
        {
            used[tmp] = true;
            tmp = b[tmp];
        }
    }
    return ret;

}
int num1[40];
int num2[40];
int num3[40];
int num[40];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int cnt = 0;
    for(int i = 0;i < 4;i++)
        for(int j = 0;j < 4;j++)
    {
        for(int x = 1;x <= 54;x++)
            b[x] = x;
        for(int x = 0;x < i;x++)
            up();
        for(int x = 0;x < j;x++)
            rr();
        num1[cnt++]=calc();
    }
    for(int x = 1;x <= 54;x++)
        b[x] = x;
    rr();
    for(int i = 0;i < 4;i++)
    {
        num1[cnt++]=calc();
        up();
    }
    for(int x = 1;x <= 54;x++)
        b[x] = x;
    rr();
    rr();
    rr();
    for(int i = 0;i < 4;i++)
    {
        num1[cnt++]=calc();
        up();
    }
    cnt = 0;
    for(int i = 0;i < 4;i++)
        for(int j = 0;j < 4;j++)
    {
        for(int x = 1;x <= 12;x++)
            b[x] = x;
        for(int x = 0;x < i;x++)
            up2();
        for(int x = 0;x < j;x++)
            rr2();
        num2[cnt++]=calc2();
        //printf("%d
",calc2());
    }

    for(int x = 1;x <= 12;x++)
        b[x] = x;
    rr2();
    for(int i = 0;i < 4;i++)
    {
        num2[cnt++]=calc2();
        up2();
    }
    for(int x = 1;x <= 12;x++)
            b[x] = x;
    rr2();
    rr2();
    rr2();
    for(int i = 0;i < 4;i++)
    {
        num2[cnt++]=calc2();
        up2();
    }
    cnt = 0;
    for(int i = 0;i < 4;i++)
        for(int j = 0;j < 4;j++)
    {
        for(int x = 1;x <= 8;x++)
            b[x] = x;
        for(int x = 0;x < i;x++)
            up3();
        for(int x = 0;x < j;x++)
            rr3();
        num3[cnt++]=calc3();
        //printf("%d
",calc3());
    }
    for(int x = 1;x <= 8;x++)
        b[x] = x;
    rr3();
    for(int i = 0;i < 4;i++)
    {
        num3[cnt++]=calc3();
        up3();
    }
    for(int x = 1;x <= 8;x++)
        b[x] = x;
    rr3();
    rr3();
    rr3();
    for(int i = 0;i < 4;i++)
    {
        num3[cnt++]=calc3();
        up3();
    }
    for(int i = 0;i < 24;i++)
        num[i] = num1[i]+num2[i]+num3[i];
    //for(int i = 0;i <24;i++)
   //     printf("%d
",num[i]);
    int T;
    scanf("%d",&T);
    int n ;
    int iCase = 0;
    while(T--)
    {
        iCase++;
        scanf("%d",&n);
        int ans = 0;

        for(int i = 0;i < 24;i++)
        {
            ans += pow_m(n,num[i]);
            ans %=MOD;
        }
        ans *= inv(24,MOD);
        ans %=MOD;
        printf("Case %d: %d
",iCase,ans);
    }
    return 0;
}