无耻的抄袭了结果,三组数,从每一组数中选取一个数组成正六边形的对边,总共会有27个正六边形,从27个小六边形中选取19个组成大六边形,求大六边形的最大值

#include
        
         #include
         
          #include
          
           #include
           
            using namespace std;int t = 0;const int N = 3;int num[N][N];int f(const void* i, const void* j){    int* ii = (int*) i;    int* jj = (int*) j;    return *ii > *jj;}int MAX = -1;int mark1[] = { 5, 6, 8 };int mark2[] = { 5, 7, 7 };int main(){    cin >> t;    int k = 1;    while (k <= t)    {        MAX = -1;        memset(num, 0, sizeof(num));        for(int i = 0; i < 3; i++)        {            for(int j = 0; j < 3; j++)                cin >> num[i][j];            qsort(num[i], N, sizeof(int), f);        }        for(int i = -1; i < N; i++)        {            int MAX2 = 0;            for(int j = 0; j < N; j++)            {                if(i == j)                {                    for(int k = 0; k < N; k++)                        MAX2 += num[j][k] * mark1[k];                }                else                {                    for(int k = 0; k < N; k++)                        MAX2 += num[j][k] * mark2[k];                }            }            if(MAX2 > MAX)                MAX = MAX2;        }        printf("Test #%d\n%d\n\n", k, MAX);        k++;    }    return 0;}