题目
In the 20x20 grid below, four numbers along a diagonal line have been marked in red.
The product of these numbers is 26×63×78×14=1788696.
What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20x20 grid?
解答
首先是把这个网格数给读成一个numpy的矩阵，这道题就变成了怎么样子来访问一个矩阵拿自己想要的元素。
import numpy as np
x = """
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
"""
y = x.split("\n")[1:-1]
a = []
for i in y:
a.append(i.split(" "))
d = np.array(a)
d = d.astype('uint32')
然后我们就按照要求，把行的，列的，往右下走的对角线，往左下走的对角线，分别遍历一次，按照窗口（这里是4个数）取个数算乘积。这里向左向右，向上向下是重复计算，所以只需要一个方向。所以这里只有4个方向需要计算，- | / \ ，正好是写一个木字。我们在这4个方向上计算出最大值出来，然后再取4个方向上的最大值，就是全局的最大值了。
def max_prod_grid(d, n):
nr = d.shape[0]
nc = d.shape[1]
mc = max(np.prod(d[i:i+n, j]) for i in range(nr-n) for j in range(nc))
mr = max(np.prod(d[i, j:j+n]) for i in range(nr) for j in range(nc-n))
mx = max(np.prod([d[i+k,j+k] for k in range(n)]) for i in range(nr-n) for j in range(nc-n))
my = max(np.prod([d[i+k,j-k] for k in range(n)]) for i in range(nr-n) for j in range(n-1,nc))
return max([mc, mr, mx, my])
