Python中的Pascal三角(11的幂)

这要简单得多:

ROWS = 5

def row( a, b ) :
   return ' ' * (b-a-1) + ' '.join([i for i in str(11**a)])

for i in range(ROWS) :
    print row( i, ROWS )

输出:

    1
   1 1
  1 2 1
 1 3 3 1
1 4 6 4 1

如果你想让它在python3中工作,在 print( .. ) 打电话来。

每行可以缩进 len(row) - i 然后对每个字符进行空格,使每行的长度是原来的两倍,以平衡其他正确的对齐方式。

inc = int(input('Input number of rows: '))
n = 0
row = []
while n <= inc:
    m = 11 ** n
    row.append(m)
    n += 1
for i in range(0, len(row)):
    indent = " " * (len(row) - i)
    spaced_row = " ".join(list(str(row[i])))
    row[i] = indent + spaced_row

result = '\n'.join(row)
print(result)

输出

Input number of rows: 3
    1
   1 1
  1 2 1
 1 3 3 1

你很接近,你可以 str.center 每行:

inc = int(input('Input number of rows: '))
n = 0
row = []
while n < inc:
    m = 11 ** n
    row.append(m)
    n += 1
for i in range(0, len(row)):
    row[i] = ' '.join(list(str(row[i])))
for i in range(0, len(row)):
    row[i] = row[i].center(len(row[-1]),' ')
result = '\n'.join(row)
print(result)

输出:

Input number of rows: 4
   1   
  1 1  
 1 2 1 
1 3 3 1

较短的版本是:

inc = int(input('Input number of rows: '))
max_len = 2 * len(str(11**inc)) - 1
row = (' '.join([*str(11**p)]).center(max_len,' ') for p in range(inc))
print(*row, sep='\n')

输出:

输入行数:4
1个
11个
1 2 1号
1 3 3 1