تحديات برمجيةالتحدي الثالث - حل التمرين الثاني بلغة بايثون

n = 0

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

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

    if i <= n / 2:
        print(' ' * (int(n / 2) - i + 1), end='')
        print('*', end='')
        print(' ' * ((i * 2) - 3), end='')

        if i != 1:
            print('*', end='')

    else:
        print(' ' * (i - int(n / 2) - 1), end='')
        print('*', end='')
        print(' ' * ((n * 2) - (i * 2) - 1), end='')

        if i != n:
            print('*', end='')

    print()

n = 0

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

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

    if i <= n / 2:

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

        print('*', end='')

        for k in range(1, (i * 2) - 2):
            print(' ', end='')

        if i != 1:
            print('*', end='')

    else:

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

        print('*', end='')

        for k in range(1, (n * 2) - (i * 2)):
            print(' ', end='')

        if i != n:
            print('*', end='')

    print()

n = 0

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

i = 1
while i <= n:

    if i <= n / 2:

        k = 1
        while k <= int(n / 2) - i + 1:
            print(' ', end='')
            k += 1

        print('*', end='')

        k = 1
        while k <= (i * 2) - 3:
            print(' ', end='')
            k += 1

        if i != 1:
            print('*', end='')

    else:

        k = 1
        while k < i - int(n / 2):
            print(' ', end='')
            k += 1

        print('*', end='')

        k = 1
        while k < (n * 2) - (i * 2):
            print(' ', end='')
            k += 1

        if i != n:
            print('*', end='')

    print()
    i += 1

Enter the number of lines: 9
    *
   * *
  *   *
 *     *
*       *
 *     *
  *   *
   * *
    *