python 给定n，返回n以内的斐波那契数列

方式一：函数

 def fabs(n):
     a, b = 0, 1
     while b < n:
         print(b, end=' ')
         a, b = b, a+b
 
 fabs(1000)

方式二：列表

 result = [0, 1]
 
 def fabs(n):
     while n-result[-1] > result[-2]:
         result.append(result[-2] + result[-1])
 
 fabs(100)
 print(result)

方式三：类

 class Fabs:
 
     def __init__(self, max):
         self.max = max
         self.a, self.b = 0, 1
 
     def __iter__(self):
         return self
 
     def next(self):
         if self.b<self.max:
             r = self.b
             self.a, self.b = self.b, self.a+self.b
             return r
 
 f1 = Fabs(1000)
 m = f1.next()
 while m:
     print(m, end=' ')
     m = f1.next()

方式四：生成器

 def fabs(n):
     a, b = 0, 1
     while b<n:
         yield b
         a, b = b, a+b
 
 if __name__ == '__main__':
     for i in fabs(1000):
         print(i, end=' ')