python3实现如下所示的打印:

#size 3

----c----
--c-b-c--
c-b-a-b-c
--c-b-c--
----c----

#size 5

--------e--------
------e-d-e------
----e-d-c-d-e----
--e-d-c-b-c-d-e--
e-d-c-b-a-b-c-d-e
--e-d-c-b-c-d-e--
----e-d-c-d-e----
------e-d-e------
--------e--------

#size 10

------------------j------------------
----------------j-i-j----------------
--------------j-i-h-i-j--------------
------------j-i-h-g-h-i-j------------
----------j-i-h-g-f-g-h-i-j----------
--------j-i-h-g-f-e-f-g-h-i-j--------
------j-i-h-g-f-e-d-e-f-g-h-i-j------
----j-i-h-g-f-e-d-c-d-e-f-g-h-i-j----
--j-i-h-g-f-e-d-c-b-c-d-e-f-g-h-i-j--
j-i-h-g-f-e-d-c-b-a-b-c-d-e-f-g-h-i-j
--j-i-h-g-f-e-d-c-b-c-d-e-f-g-h-i-j--
----j-i-h-g-f-e-d-c-d-e-f-g-h-i-j----
------j-i-h-g-f-e-d-e-f-g-h-i-j------
--------j-i-h-g-f-e-f-g-h-i-j--------
----------j-i-h-g-f-g-h-i-j----------
------------j-i-h-g-h-i-j------------
--------------j-i-h-i-j--------------
----------------j-i-j----------------
------------------j------------------

代码如下:

def print_rangoli(size):
    endchar = chr(96+size)
    for i in range(2*size-1):
        midlechar =  chr(abs(i-size+1)+97)
        str = midlechar
        while midlechar!=endchar:
            tmp = chr(ord(midlechar)+1)
            str= tmp+'-'+str+'-'+tmp
            midlechar = tmp
        print(str.center(4*size-3,'-'))

if __name__ == '__main__':
    n = int(input())
    print_rangoli(n)

用到的函数包括:

chr   ord,这两个函数分别实现字符与asc2的互转

abs,取绝对值

posted on 2019-06-30 16:28  二豆  阅读(204)  评论(0编辑  收藏  举报