python 斐波拉契数列

来源:互联网 发布:js 数组取最后一个 编辑:程序博客网 时间:2024/06/08 04:54

简单版

def fib(max):    n,befor,after=0,0,1    while n<max:        print(after)        # temp=befor        # befor=after        # after=temp+after       # befor=after        befor,after=after,befor+after        n=n+1fib(10) 

递归版

def fib2(max):    if max == 0:        return 0    elif max == 1:        return 1    else:        result=fib2(max-1)+fib2(max-2)        return result
生成器版

def fib(max):    n,befor,after=0,0,1    while n<max:        print(befor)        befor,after=after,befor+after        n=n+1g=fib(10)print(next(g))print(next(g))print(next(g))print(next(g))print(next(g))



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