[Project Euler] Problem 11

In the 2020 grid below, four numbers along a diagonal line have been marked in red.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
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 08 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 09 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 09 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 08 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 08 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

The product of these numbers is 26 63 78 14 = 1788696.

What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 2020 grid?

求四个方向上，连续四个数乘积的最大值

没有太多值得分析的，直接编写程序

#include < iostream >
using namespace std;

int grid[ 20 ][ 20 ] = {
8 , 2 , 22 , 97 , 38 , 15 , 0 , 40 , 0 , 75 , 4 , 5 , 7 , 78 , 52 , 12 , 50 , 77 , 91 , 8 ,
49 , 49 , 99 , 40 , 17 , 81 , 18 , 57 , 60 , 87 , 17 , 40 , 98 , 43 , 69 , 48 , 4 , 56 , 62 , 0 ,
81 , 49 , 31 , 73 , 55 , 79 , 14 , 29 , 93 , 71 , 40 , 67 , 53 , 88 , 30 , 3 , 49 , 13 , 36 , 65 ,
52 , 70 , 95 , 23 , 4 , 60 , 11 , 42 , 69 , 24 , 68 , 56 , 1 , 32 , 56 , 71 , 37 , 2 , 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 , 3 , 45 , 2 , 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 , 2 , 62 , 12 , 20 , 95 , 63 , 94 , 39 , 63 , 8 , 40 , 91 , 66 , 49 , 94 , 21 ,
24 , 55 , 58 , 5 , 66 , 73 , 99 , 26 , 97 , 17 , 78 , 78 , 96 , 83 , 14 , 88 , 34 , 89 , 63 , 72 ,
21 , 36 , 23 , 9 , 75 , 0 , 76 , 44 , 20 , 45 , 35 , 14 , 0 , 61 , 33 , 97 , 34 , 31 , 33 , 95 ,
78 , 17 , 53 , 28 , 22 , 75 , 31 , 67 , 15 , 94 , 3 , 80 , 4 , 62 , 16 , 14 , 9 , 53 , 56 , 92 ,
16 , 39 , 5 , 42 , 96 , 35 , 31 , 47 , 55 , 58 , 88 , 24 , 0 , 17 , 54 , 24 , 36 , 29 , 85 , 57 ,
86 , 56 , 0 , 48 , 35 , 71 , 89 , 7 , 5 , 44 , 44 , 37 , 44 , 60 , 21 , 58 , 51 , 54 , 17 , 58 ,
19 , 80 , 81 , 68 , 5 , 94 , 47 , 69 , 28 , 73 , 92 , 13 , 86 , 52 , 17 , 77 , 4 , 89 , 55 , 40 ,
4 , 52 , 8 , 83 , 97 , 35 , 99 , 16 , 7 , 97 , 57 , 32 , 16 , 26 , 26 , 79 , 33 , 27 , 98 , 66 ,
88 , 36 , 68 , 87 , 57 , 62 , 20 , 72 , 3 , 46 , 33 , 67 , 46 , 55 , 12 , 32 , 63 , 93 , 53 , 69 ,
4 , 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 , 4 , 36 , 16 ,
20 , 73 , 35 , 29 , 78 , 31 , 90 , 01 , 74 , 31 , 49 , 71 , 48 , 86 , 81 , 16 , 23 , 57 , 5 , 54 ,
1 , 70 , 54 , 71 , 83 , 51 , 54 , 69 , 16 , 92 , 33 , 48 , 61 , 43 , 52 , 1 , 89 , 19 , 67 , 48
};

int main(){
int product = 0 ;
int tmp1,tmp2;

for ( int i = 0 ; i < 16 ; i ++ ){
for ( int j = 0 ; j < 16 ; j ++ ){
tmp1 = grid[i][j] * grid[i + 1 ][j + 1 ] * grid[i + 2 ][j + 2 ] * grid[i + 3 ][j + 3 ];
tmp2 = grid[i][j + 3 ] * grid[i + 1 ][j + 2 ] * grid[i + 2 ][j + 1 ] * grid[i + 3 ][j];
if (tmp1 > product)
product = tmp1;
if (tmp2 > product)
product = tmp2;
}
}

for ( int i = 0 ; i < 20 ; i ++ ){
for ( int j = 0 ; j < 16 ; j ++ ){
tmp1 = grid[i][j] * grid[i][j + 1 ] * grid[i][j + 2 ] * grid[i][j + 3 ];
tmp2 = grid[j][i] * grid[j + 1 ][i] * grid[j + 2 ][i] * grid[j + 3 ][i];
if (tmp1 > product)
product = tmp1;
if (tmp2 > product)
product = tmp2;
}
}

cout << product << endl;
return 0 ;
}

我们发现，向下和向右可以一起遍历，左斜和右斜可以一起遍历

当然，题目还有一点，要把原题中给的字符转换为整型

我直接在编辑器里手动转换了，要编写函数转的话也很方便。