uva-317-找规律

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<iostream>
#include<stdio.h>
#include<memory.h>
#include<stdlib.h>
using namespace std;
int t = ;
const int N = ;
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 = -;
int mark1[] = { , ,  };
int mark2[] = { , ,  };
int main()
{
    cin >> t;
    int k = ;
    while (k <= t)
    {
        MAX = -;
        memset(num, , sizeof(num));
        for(int i = ; i < ; i++)
        {
            for(int j = ; j < ; j++)
                cin >> num[i][j];
            qsort(num[i], N, sizeof(int), f);
        }
        for(int i = -; i < N; i++)
        {
            int MAX2 = ;
            for(int j = ; j < N; j++)
            {
                if(i == j)
                {
                    for(int k = ; k < N; k++)
                        MAX2 += num[j][k] * mark1[k];
                }
                else
                {
                    for(int k = ; 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 ;
}