Question

想象一下，我们有w个h网格，其中的图块从左上角的1开始编号。 w（宽度）和h（高度）已存储在函数中。您可以访问这些存储的值，只要您将其称为w和h即可。编写要返回的程序：用户给出的区号的列号。开始计算第1列的列。

import subprocess
    def template(w, h, t):

        #w and h are memory locations that already contain values
        tile = t

        sum = ((t - 1) // w) + 1 
#END OF YOUR CODE

failed = 0
width = 1
while width <= 3:
    height = 1
    while height <= 3:
        tile = 1
        while tile  <=  width * height:
            result  =  template(width, height, tile)
            col = (tile - 1) % width + 1
            if (result == col):
                print "On a " + str(width) + " by " + str(height) +" board, with tile " + str(tile) +", col is " + str(col) + ", you got it RIGHT!"
            else:
                print "On a " + str(width) + " by " +:str(height)%+" board, with tile " + str(tile)%+", col is " + str(col) + ", you got: " + str(result)
                failed = 1
            tile += 1
        height += 1
    width += 1



if (failed == 0):
    print "Your code is CORRECT!"

    print "Please check your code, at least one test case did not pass."

我认为我几乎就在那里，但这个公式并不完全正确，而且我没有想法。

Answer 1

要创建网格，请使用list comprehension。

grid=[list(range(x,x+w))for x in range(1,w*h,w)]

要查找t的列号，请查找t的剩余部分除以w：

t%w

所以函数将是：

def template(w,h,t):
    grid=[list(range(x,x+w))for x in range(1,w*h,w)]
    return t%w

示例：

template(6,5,22)

输出：

Answer 2

见下面的伪代码，这种方法适合你，

# divide t by h, the tile should reside in the next row
tileRow = (t/h)+1
# get the reminder of the division, that's the column number
tileColumn = t%h

请参阅我在下面尝试的示例代码

>>> w = 5
>>> h = 10
>>> t =36
>>> tileRow = (t/h)+1
>>> tileRow
4     # the tile is in the 4th row
>>> tileColumn = t%h
>>> tileColumn
6     # the tile is in the 6th column
>>>

您可能还需要检查图块编号是否在范围内，在上面的例子中是w x h（50）

如果您需要进一步澄清，请发表评论。如果这可以解决您的问题，您可以接受并投票给答案

W by H grid tiles？

2 个答案: