【欧拉计划】11. Largest product in a grid

（本题取 $n=20$，$m=4$，$d=8$）

【思路】先 $O(n^2)$ 枚举起始数字，然后 $\mathcal O(md)$ 枚举所有方向的所有数字。时间复杂度为 $\mathcal O(n^2md)$：

#include<stdio.h>
const int dx[]={-1,-1,-1,0,0,1,1,1},dy[]={-1,0,1,-1,1,-1,0,1};
int ans;
const int matrix[20][20]=
{
    {8,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91,8},
    {49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,04,56,62,00},
    {81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,03,49,13,36,65},
    {52,70,95,23,04,60,11,42,69,24,68,56,01,32,56,71,37,02,36,91},
    {22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80},
    {24,47,32,60,99,03,45,02,44,75,33,53,78,36,84,20,35,17,12,50},
    {32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70},
    {67,26,20,68,02,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21},
    {24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},
    {21,36,23,9,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95},
    {78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14,9,53,56,92},
    {16,39,05,42,96,35,31,47,55,58,88,24,00,17,54,24,36,29,85,57},
    {86,56,00,48,35,71,89,07,05,44,44,37,44,60,21,58,51,54,17,58},
    {19,80,81,68,05,94,47,69,28,73,92,13,86,52,17,77,04,89,55,40},
    {04,52,8,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66},
    {88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69},
    {04,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36},
    {20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16},
    {20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,05,54},
    {01,70,54,71,83,51,54,69,16,92,33,48,61,43,52,01,89,19,67,48}
};
int main()
{
    for(int i=0;i<20;++i)
    {
        for(int j=0;j<20;++j)
        {
            for(int k=0;k<8;++k)
            {
                int mx=i+dx[k]*3,my=j+dy[k]*3,fac=matrix[i][j];
                if(mx<0||my<0||mx>=20||my>=20)continue;
                for(int l=0,lx=i,ly=j;l<3;++l)
                {
                    int nx=lx+dx[k],ny=ly+dy[k];
                    fac*=matrix[nx][ny];
                    lx=nx,ly=ny;
                }
                if(fac>ans)ans=fac;
            }
        }
    }
    printf("%d",ans);
    return 0;
}