无法在 else 中为模式打印定义逻辑

Question

卡住了预期的图案打印

下面是我的代码，我已经打印了第一部分，但在打印其他部分时卡住了 需要一些输入如何为其他部分构建逻辑

n=29
breaking= int((n+1)/2)
lst  = [ len(str(num).zfill(2)) * ' ' for num in reversed(range(1,n+1)) ]
cnt=-1
for row in range(1,n+1):

  if row <= breaking:
    lst[cnt]=str(row).zfill(2) + '   '
    print(' '.join(lst))
    cnt = cnt - 1 
    
  else:  

预期输出：

                                          01   
                                       02    01   
                                    03    02    01   
                                 04    03    02    01   
                              05    04    03    02    01   
                           06    05    04    03    02    01   
                        07    06    05    04    03    02    01   
                     08    07    06    05    04    03    02    01   
                  09    08    07    06    05    04    03    02    01   
               10    09    08    07    06    05    04    03    02    01   
            11    10    09    08    07    06    05    04    03    02    01   
         12    11    10    09    08    07    06    05    04    03    02    01   
      13    12    11    10    09    08    07    06    05    04    03    02    01   
   14    13    12    11    10    09    08    07    06    05    04    03    02    01   
15    14    13    12    11    10    09    08    07    06    05    04    03    02    01
   14    13    12    11    10    09    08    07    06    05    04    03    02    01
      13    12    11    10    09    08    07    06    05    04    03    02    01
         12    11    10    09    08    07    06    05    04    03    02    01
            11    10    09    08    07    06    05    04    03    02    01
               10    09    08    07    06    05    04    03    02    01
                  09    08    07    06    05    04    03    02    01
                     08    07    06    05    04    03    02    01
                        07    06    05    04    03    02    01
                           06    05    04    03    02    01
                              05    04    03    02    01
                                 04    03    02    01
                                    03    02    01
                                       02    01
                                          01

Answer 1

虽然很乱，但是很管用。请在评论中告诉我，我会解释。

for i in range(1, n//2 + 1):
    print(" "*(n+1-2*i) + "  ".join(["{:02d}".format(i - j) for j in range(i)]))
for i in range(n//2 + 1, 0, -1):
    print(" "*(n+1-2*i) + "  ".join(["{:02d}".format(i - j) for j in range(i)]))

Answer 2

Here's what you can do:

代码

n = int(input('Enter max value : '))
num_width = len(str(n))
seperation = '    '
tot_width = num_width * n + len(seperation) * (n-1)

for i in list(range(1, n + 1)) + list(range(n - 1, 0, -1)):
    line = seperation.join([str.rjust(str(j), num_width, '0') for j in range(i, 0, -1)])
    print(str.center(line, tot_width))

输出

说明

First for loop iterates over max-value allowed in the current line to be printed.

Second for loop iterates over each number to be printed (i.e. from 1 to max-value in the current line in reverse order.

代码逐行解释

n = int(input('Enter max value : '))        # Get the max value for the pattern

# Know the length of highest value in the pattern so that we can make all the digits
# in the pattern of that width by adding zeroes
# for eg. if max number is 15, width is 2, max number is 178, width is 3..
num_width = len(str(n))

# The seperation to be added between 2 digits
seperation = '    '

# calculating max width of the pattern (which occurs at the middlemost line)
tot_width = n  *  num_width                   +      len(seperation) * (n-1)
#           (n digits of 'num-width' length)  +    (n-1) number of seperations between them

# i will go initially from 1 to n, which is : list(range(1, n + 1))
# then will decrease from n to 1,  which is : list(range(n - 1, 0, -1))
for i in list(range(1, n + 1)) + list(range(n - 1, 0, -1)):
    digits = []
    for j in range(i, 0, -1):  # now j will iterate from current value of 'i' to '1' in decreasing order

        # each digit in current line is created with a width of 'num-width' by adding preceding zeroes
        digits.append(str.rjust(str(j), num_width, '0'))

    # Finally, join all the digits with 'seperation' in between to create the line
    line = seperation.join(digits)
    print(str.center(line, tot_width))         # now center the line by adding spaces on both the sides, and print it

此代码普遍适用于 n

的任何值