求三角形

n = int(input())
count1 = (1 + n) * n // 2
count2 = n *  n
lis1 = [str(i).zfill(2) for i in range(1,count1 + 1)]
lis2 = [str(i).zfill(2) for i in range(1,count2 + 1)]for i in range(0,len(lis2),n):print(''.join(lis2[i:i + n]))print()for i in range(1,n + 1):temp = lis1[:i]lis1 = lis1[i:]print(' ' * (2 * (n - i)) + ''.join(temp))

后面一部分和数字直角三角形异曲同工