最近好多分享这个问题的代码，题目说的是用面向对象或者函数式编程，下面是PYTHON的实现示例
代码如下:
#!/usr/bin/python
#encoding:utf8
'''
The Game of "FizzBuzzWhizz"
author : wang.jiankui89@gmail.com
mobile : 130-2199-5152
'''
import sys
class teacher:
def __init__(self, student_num):
self.student_num = student_num
def gameStart(self, numList):
for i in range (1, self.student_num + 1):
stu = student(i)
print stu.answer(numList)
class student:
def __init__(self, my_num):
self.my_num = my_num
def judgeMod(self, numList):
modRes = ""
for num in numList:
if self.my_num % num == 0:
modRes += rule[num]
return modRes if modRes else self.my_num
def judgeContain(self, first_num):
conRes = ""
if str(first_num) in str(self.my_num) :
conRes = rule[first_num]
return conRes
def answer(self, numList):
conRes = self.judgeContain(numList[0])
return conRes if conRes else self.judgeMod(numList)
def getOps():
'''parse options'''
if len(sys.argv) != 4 :
print "use as fizzBuzzWhizz.py [0-9] [0-9] [0-9]"
sys.exit()
else:
first_num = int(sys.argv[1])
second_num = int(sys.argv[2])
third_num = int(sys.argv[3])
return first_num, second_num, third_num
def main():
first_num, second_num, third_num = getOps()
global rule
rule = {first_num:"Fizz", second_num:"Buzz", third_num:"Whizz"}
student_num = 100
tea = teacher(100)
tea.gameStart( (first_num, second_num, third_num) )
if __name__ == "__main__":
main()
