题目链接：

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3736

Pocket Cube is a 3-D combination puzzle. It is a 2 × 2 × 2 cube, which means it is constructed by 8 mini-cubes. For a combination of 2 × 2 mini-cubes which sharing a whole cube face, you can twist it 90 degrees in clockwise or counterclockwise direction, this twist operation is called one twist step.

Considering all faces of mini-cubes, there will be totally 24 faces painted in 6 different colors (Indexed from 0), and there will be exactly 4 faces painted in each kind of color. If 4 mini-cubes‘ faces of same color rely on same large cube face, we can call the large cube face as a completed face.

  

Now giving you an color arrangement of all 24 faces from a scrambled Pocket Cube, please tell us the maximum possible number of completed faces in no more than N twist steps.

Index of each face is shown as below:

Input

There will be several test cases. In each test case, there will be 2 lines. One integer N (1 ≤ N ≤ 7) in the first line, then 24 integers C_i seperated by a sinle space in the second line. For index 0 ≤ i < 24, C_i is color of the corresponding face. We guarantee that the color arrangement is a valid state which can be achieved by doing a finite number of twist steps from an initial cube whose all 6 large cube faces are completed faces.

Output

For each test case, please output the maximum number of completed faces during no more than N twist step(s).

Sample Input

1
0 0 0 0 1 1 2 2 3 3 1 1 2 2 3 3 4 4 4 4 5 5 5 5
1
0 4 0 4 1 1 2 5 3 3 1 1 2 5 3 3 4 0 4 0 5 2 5 2

Sample Output

6
2

题目意思：

给2*2*2的魔方，求在给定的n步内最多可以凑成多少完整的面。

解题思路：

由于n<=7,所以暴搜即可。

每种状态下有6种选择，（不是12种，因为转动右边可以转化为转动左边，转动下面可以转化为转动上面，转动后面可以转化为转动前面）

所以只用考虑左边、下面、后面的各两种情况。

找出各位置新的状态下在原来的位置标号。

Tips: 当数组指针作为参数传给函数时，不能在里面用memcpy,因为此时并不知道该数组的大小

当全局数组指针作为参数传给函数时，每个递归函数里面对指针所指内容所做的修改，都会生效，并且有可能访问的都是同一个内存区域。

当传数组指针时，最好使用局部数组这样才不会有问题。

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 30

int cur[Maxn],vv[Maxn],n,ans;

int next[][24]=
{
    {6,1,12,3,5,11,16,7,8,9,4,10,18,13,14,15,20,17,22,19,0,21,2,23},//左边往内
    {20,1,22,3,10,4,0,7,8,9,11,5,2,13,14,15,6,17,12,19,16,21,18,23},//左边往外
    {0,1,11,5,4,16,12,6,2,9,10,17,13,7,3,15,14,8,18,19,20,21,22,23},//下面往内
    {0,1,8,14,4,3,7,13,17,9,10,2,6,12,16,15,5,11,18,19,20,21,22,23},//下面往外
    {2,0,3,1,6,7,8,9,23,22,10,11,12,13,14,15,16,17,18,19,20,21,5,4},//后面往下
    {1,3,0,2,23,22,4,5,6,7,10,11,12,13,14,15,16,17,18,19,20,21,9,8} //后面忘上
};
int cal(int * cur) //统计完整面的个数
{
    int res=0;

    if(cur[0]==cur[1]&&cur[0]==cur[2]&&cur[0]==cur[3])
        res++;
    if(cur[4]==cur[5]&&cur[4]==cur[10]&&cur[4]==cur[11])
        res++;
    if(cur[6]==cur[7]&&cur[6]==cur[12]&&cur[6]==cur[13])
        res++;
    if(cur[8]==cur[9]&&cur[8]==cur[14]&&cur[8]==cur[15])
        res++;
    if(cur[16]==cur[17]&&cur[16]==cur[18]&&cur[16]==cur[19])
        res++;
    if(cur[20]==cur[21]&&cur[20]==cur[22]&&cur[20]==cur[23])
        res++;
    return res;
}
void dfs(int num,int * cur)
{
    ans=max(ans,cal(cur));
    if(num>n)
        return ;
    int vv[Maxn]; //该数组定义成全局变量会有问题
    for(int i=0;i<6;i++)
    {
        for(int j=0;j<24;j++)
            vv[j]=cur[next[i][j]]; //如果vv是全局的，vv和cur会指向同一块内存，访问会出现混乱
        dfs(num+1,vv);
    }
}

int main()
{
   //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   while(~scanf("%d",&n))
   {
       for(int i=0;i<24;i++)
            scanf("%d",&cur[i]);
       ans=0;
       dfs(1,cur);
       printf("%d\n",ans);
   }

   return 0;
}