在以下这个20×20的网格中,四个处于同一对角线上的相邻数字用红色标了出来:
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
这四个数字的乘积是:26 × 63 × 78 × 14 = 1788696.
在这个20×20网格中,处于任何方向上(上,下,左,右或者对角线)的四个相邻数字的乘积的最大值是多少?
from functools import reduce
def prod_all(arr):
return reduce(lambda x,y: x * y, arr)
all_text = open('euler11.txt').read()
all_text = all_text.splitlines()
arr_list = []
for line in all_text:
arr_list.append([int(i) for i in line.split(' ')])
prod_max = 0
n = 4
# 横
for i in range(len(arr_list)):
j = 0
prod_n_1 = 0
while j < len(arr_list[i]) - n + 1:
if prod_n_1 == 0:
arr_sub = arr_list[i][j: j + n]
if 0 in arr_sub:
j = arr_sub.index(0) + j + 1
continue
else:
prod_ = prod_all(arr_sub)
else:
if arr_list[i][j + n - 1] == 0:
prod_n_1 = 0
j += n
continue
else:
prod_ = prod_n_1 * arr_list[i][j + n - 1]
if prod_ > prod_max:
prod_max = prod_
prod_n_1 = prod_ / arr_list[i][j]
j += 1
# 竖
for j in range(len(arr_list[0])):
i = 0
prod_n_1 = 0
while i < len(arr_list) - n + 1:
if prod_n_1 == 0:
arr_sub = [arr_list[i_][j] for i_ in range(i, i + n)]
if 0 in arr_sub:
i = arr_sub.index(0) + i + 1
continue
else:
prod_ = prod_all(arr_sub)
else:
if arr_list[i + n - 1][j] == 0:
prod_n_1 = 0
i += n
continue
else:
prod_ = prod_n_1 * arr_list[i + n - 1][j]
if prod_ > prod_max:
prod_max = prod_
prod_n_1 = prod_ / arr_list[i][j]
i += 1
# 斜上
for i in range(n, len(arr_list)):
j = 0
for j in range(0, len(arr_list[i]) - n + 1):
arr_sub = [arr_list[i - k][j + k] for k in range(n)]
prod_ = prod_all(arr_sub)
if prod_ > prod_max:
prod_max = prod_
# 斜下
for i in range(0, len(arr_list) - n + 1):
j = 0
for j in range(0, len(arr_list[i]) - n + 1):
arr_sub = [arr_list[i + k][j + k] for k in range(n)]
prod_ = prod_all(arr_sub)
if prod_ > prod_max:
prod_max = prod_
print(prod_max)
# 在20*20外围增加3层全0数据,变为26*26,一个for循环,4次判断
# 待续...