تحديات برمجيةالتحدي الخامس - حل التمرين الثالث بلغة بايثون

n = 0

while n <= 0 or n % 2 == 0:
    n = int(input("Enter and odd number as the number of lines: "))

for i in range(1, n + 1):

    if i <= n / 2:
        print(' ' * int(n / 2), end='')
        print('*', end='')

        for j in range(1, int(n / 2)):
            if i == 1 or i == int(n / 2) + 1:
                print('*', end='')
            else:
                print(' ', end='')

        print('*', end='')

    elif i == int(n / 2) + 1:
        print('*' * n, end='')

    else:
        print('*', end='')

        for j in range(1, int(n / 2)):
            if i == n:
                print('*', end='')
            else:
                print(' ', end='')

        print('*', end='')

    print()

n = 0

while n <= 0 or n % 2 == 0:
    n = int(input("Enter and odd number as the number of lines: "))

for i in range(1, n + 1):

    if i <= n / 2:

        for k in range(1, int(n / 2) + 1):
            print(' ', end='')

        print('*', end='')

        for j in range(1, int(n / 2)):
            if i == 1 or i == int(n / 2) + 1:
                print('*', end='')
            else:
                print(' ', end='')

        print('*', end='')

    elif i == int(n / 2) + 1:
        for j in range(1, n + 1):
            print('*', end='')

    else:
        print('*', end='')

        for j in range(1, int(n / 2)):
            if i == n:
                print('*', end='')
            else:
                print(' ', end='')

        print('*', end='')

    print()

Enter an odd number as the number of lines: 9
    *****
    *   *
    *   *
    *   *
*********
*   *
*   *
*   *
*****