欧拉计划 11 Python0n
Project Euler 11 Pyth0n
我正在尝试解决这个难题，但我无法计算出适用于示例但无法解决问题的逻辑！
我认为这与元组的方向列表的负数有关。
如有任何建议，我们将不胜感激！
这是我遇到问题的代码:)
如何求同方向相邻四个数的最大乘积。如下所述？
# Largest product in a grid
# Problem 11
# In the 20×20 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 20×20 grid?
grid = [
[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, 1, 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]
]
straight = [(1,0), (0,1), (-1,0), (0,-1)]
diagonal = [(1,1), (-1,1), (1,-1), (-1,-1)]
def prod(lst):
""" returns the product of a list """
p = 1
for i in lst:
if i != 0: # ignore zeros..
p *= i
return p
def traverse(start, steps, grid, direction):
""" Input a starting point and number of steps to Traverse """
x, y = start[0]-1, start[1]-1
print(x, y)
j, k = direction[0], direction[1]
print(j, k)
try:
for i in range(steps):
print(grid[x+i+j][y+i+k])
yield grid[x+i+j][y+i+k]
except IndexError:
pass
print()
# Example
d = traverse((6, 8), 4, grid, (1, 1))
p = prod([x for x in d])
print('26 x 63 x 78 x 14 = ', 1788696, p == 1788696, '\nActual:', p) # [26, 63, 78, 14] 1788696
print()
#Answer
d = traverse((15, 3), 4, grid, (1, -1))
p = prod([x for x in d])
print('87 x 97 x 94 x 89 = ', 70600674, p == 70600674, '\nActual:', p) # [87, 97, 94, 89] 70600674
你们非常接近。这是经过编辑的有效代码：
def traverse(start, steps, grid, direction):
""" Input a starting point and number of steps to Traverse """
x, y = start[0], start[1] # EDIT: remove "- 1"
print(x, y)
j, k = direction[0], direction[1]
print(j, k)
try:
for i in range(steps):
print(grid[x+i*j][y+i*k]) # EDIT: i*j instead of i+j
yield grid[x+i*j][y+i*k] # EDIT: i*j instead of i+j
except IndexError:
pass
print()
# Example
d = traverse((6, 8), 4, grid, (1, 1))
p = prod([x for x in d])
print('26 x 63 x 78 x 14 = ', 1788696, p == 1788696, '\nActual:', p) # [26, 63, 78, 14] 1788696
print()
#Answer
d = traverse((15, 3), 4, grid, (-1, 1)) # EDIT: fix direction
p = prod([x for x in d])
print('87 x 97 x 94 x 89 = ', 70600674, p == 70600674, '\nActual:', p) # [87, 97, 94, 89] 70600674
我评论了我对您的代码所做的所有编辑以使其正常工作。如果您有任何问题，请在下方评论。
我知道这是一个迟到的条目，但我是使用 numpy 数组完成的。这是代码：
import numpy as np
s = '''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'''
ss = np.array([[int(x) for x in s.split()][i:i+20] for i in range(0,400,20)]).reshape(20, 20)
#process 4 elements to right, down, left-down-diagonal, right-down-diagonal
#check right: row: 1 thru 20 (index 0:19); columns: 1 thru 16 (index 0:15)
mr = max(np.prod(ss[i,j:j+4]) for i in range(20) for j in range(16))
#check down: row: 1 thru 16 (index 0:15); columns: 1 thru 20 (index 0:19)
mc = max(np.prod(ss[i:i+4,j]) for i in range(16) for j in range(20))
#check right-down-diagonal: row: 1 thru 16 (index 0:15); columns: 1 thru 16 (index 0:15). row,col increments by 1 to go right-down-diagonal
mx = max(np.prod([ss[i+k,j+k] for k in range(4)]) for i in range(16) for j in range(16))
#check left-down-diagonal: row: 1 thru 20 (index 0:19); columns: 4 thru 20 (index 3:19). row increments by 1 and col decrements by 1 to go right-down-diagonal
my = max(np.prod([ss[i+k,j-k] for k in range(4)]) for i in range(16) for j in range(3,20))
ans = max([mr,mc,mx,my])
print (ans)
data = '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'
number = (data.split())
获取从左到右方向相邻的四个数的乘积
def leftToRight(list, diff):
first = 0
last = len(list) - diff
product = []
for x in range(first, last):
multi = int(list[x]) * int(list[x + 1]) * int(list[x + 2]) * int(list[x + 3])
product.append(multi)
return product
从上到下方向求相邻四个数的乘积
def topToDown(list, diff):
first = 0
last = len(list) - (diff * 20)
product = []
for x in range(first, last):
multi = int(list[x]) * int(list[x + 20]) * int(list[x + 40]) * int(list[x + 60])
product.append(multi)
return product
求左对角线到右对角线方向相邻四个数的乘积
def rightDiagonal(list, diff):
first = 0
last = len(list) - (diff * 20)
product = []
for x in range(first, last):
if x % 20 < 17:
# print((list[x]), int(list[x + 21]), int(list[x + 42]), int(list[x + 63]))
multi = int(list[x]) * int(list[x + 21]) * int(list[x + 42]) * int(list[x + 63])
product.append(multi)
return product
取右对角线到左对角线方向相邻四个数的乘积
def leftDiagonal(list, diff):
first = 0
last = len(list) - (diff * 20)
product = []
for x in range(first, last):
if x % 20 > 2:
# print((list[x]), int(list[x + 19]), int(list[x + 38]), int(list[x + 57]))
multi = int(list[x]) * int(list[x + 19]) * int(list[x + 38]) * int(list[x + 57])
product.append(multi)
return product
获得最好的产品
new_list = leftToRight(number, 3) + topToDown(number, 3) + rightDiagonal(number, 3) + leftDiagonal(number, 3)
sort = list(reversed(sorted(new_list)))
highest = sort[0]
print(highest)
我正在尝试解决这个难题，但我无法计算出适用于示例但无法解决问题的逻辑！
我认为这与元组的方向列表的负数有关。
如有任何建议，我们将不胜感激！
这是我遇到问题的代码:)
如何求同方向相邻四个数的最大乘积。如下所述？
# Largest product in a grid
# Problem 11
# In the 20×20 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 20×20 grid?
grid = [
[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, 1, 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]
]
straight = [(1,0), (0,1), (-1,0), (0,-1)]
diagonal = [(1,1), (-1,1), (1,-1), (-1,-1)]
def prod(lst):
""" returns the product of a list """
p = 1
for i in lst:
if i != 0: # ignore zeros..
p *= i
return p
def traverse(start, steps, grid, direction):
""" Input a starting point and number of steps to Traverse """
x, y = start[0]-1, start[1]-1
print(x, y)
j, k = direction[0], direction[1]
print(j, k)
try:
for i in range(steps):
print(grid[x+i+j][y+i+k])
yield grid[x+i+j][y+i+k]
except IndexError:
pass
print()
# Example
d = traverse((6, 8), 4, grid, (1, 1))
p = prod([x for x in d])
print('26 x 63 x 78 x 14 = ', 1788696, p == 1788696, '\nActual:', p) # [26, 63, 78, 14] 1788696
print()
#Answer
d = traverse((15, 3), 4, grid, (1, -1))
p = prod([x for x in d])
print('87 x 97 x 94 x 89 = ', 70600674, p == 70600674, '\nActual:', p) # [87, 97, 94, 89] 70600674
你们非常接近。这是经过编辑的有效代码：
def traverse(start, steps, grid, direction):
""" Input a starting point and number of steps to Traverse """
x, y = start[0], start[1] # EDIT: remove "- 1"
print(x, y)
j, k = direction[0], direction[1]
print(j, k)
try:
for i in range(steps):
print(grid[x+i*j][y+i*k]) # EDIT: i*j instead of i+j
yield grid[x+i*j][y+i*k] # EDIT: i*j instead of i+j
except IndexError:
pass
print()
# Example
d = traverse((6, 8), 4, grid, (1, 1))
p = prod([x for x in d])
print('26 x 63 x 78 x 14 = ', 1788696, p == 1788696, '\nActual:', p) # [26, 63, 78, 14] 1788696
print()
#Answer
d = traverse((15, 3), 4, grid, (-1, 1)) # EDIT: fix direction
p = prod([x for x in d])
print('87 x 97 x 94 x 89 = ', 70600674, p == 70600674, '\nActual:', p) # [87, 97, 94, 89] 70600674
我评论了我对您的代码所做的所有编辑以使其正常工作。如果您有任何问题，请在下方评论。
我知道这是一个迟到的条目，但我是使用 numpy 数组完成的。这是代码：
import numpy as np
s = '''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'''
ss = np.array([[int(x) for x in s.split()][i:i+20] for i in range(0,400,20)]).reshape(20, 20)
#process 4 elements to right, down, left-down-diagonal, right-down-diagonal
#check right: row: 1 thru 20 (index 0:19); columns: 1 thru 16 (index 0:15)
mr = max(np.prod(ss[i,j:j+4]) for i in range(20) for j in range(16))
#check down: row: 1 thru 16 (index 0:15); columns: 1 thru 20 (index 0:19)
mc = max(np.prod(ss[i:i+4,j]) for i in range(16) for j in range(20))
#check right-down-diagonal: row: 1 thru 16 (index 0:15); columns: 1 thru 16 (index 0:15). row,col increments by 1 to go right-down-diagonal
mx = max(np.prod([ss[i+k,j+k] for k in range(4)]) for i in range(16) for j in range(16))
#check left-down-diagonal: row: 1 thru 20 (index 0:19); columns: 4 thru 20 (index 3:19). row increments by 1 and col decrements by 1 to go right-down-diagonal
my = max(np.prod([ss[i+k,j-k] for k in range(4)]) for i in range(16) for j in range(3,20))
ans = max([mr,mc,mx,my])
print (ans)
data = '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'
number = (data.split())
获取从左到右方向相邻的四个数的乘积
def leftToRight(list, diff):
first = 0
last = len(list) - diff
product = []
for x in range(first, last):
multi = int(list[x]) * int(list[x + 1]) * int(list[x + 2]) * int(list[x + 3])
product.append(multi)
return product
从上到下方向求相邻四个数的乘积
def topToDown(list, diff):
first = 0
last = len(list) - (diff * 20)
product = []
for x in range(first, last):
multi = int(list[x]) * int(list[x + 20]) * int(list[x + 40]) * int(list[x + 60])
product.append(multi)
return product
求左对角线到右对角线方向相邻四个数的乘积
def rightDiagonal(list, diff):
first = 0
last = len(list) - (diff * 20)
product = []
for x in range(first, last):
if x % 20 < 17:
# print((list[x]), int(list[x + 21]), int(list[x + 42]), int(list[x + 63]))
multi = int(list[x]) * int(list[x + 21]) * int(list[x + 42]) * int(list[x + 63])
product.append(multi)
return product
取右对角线到左对角线方向相邻四个数的乘积
def leftDiagonal(list, diff):
first = 0
last = len(list) - (diff * 20)
product = []
for x in range(first, last):
if x % 20 > 2:
# print((list[x]), int(list[x + 19]), int(list[x + 38]), int(list[x + 57]))
multi = int(list[x]) * int(list[x + 19]) * int(list[x + 38]) * int(list[x + 57])
product.append(multi)
return product
获得最好的产品
new_list = leftToRight(number, 3) + topToDown(number, 3) + rightDiagonal(number, 3) + leftDiagonal(number, 3)
sort = list(reversed(sorted(new_list)))
highest = sort[0]
print(highest)