斐波那契递归算法
作者:互联网
数列:1,2,3,5,8,13,21.........根据这样的规律,编程求出400万以内最大的斐波那契数,并求出他是第几个数:
# 递归
def fib(num):
if num == 1:
return 1
if num == 2:
return 2
return fib(num - 1) + fib(num - 2)
flag = 0
i = 1
while not flag:
if fib(i) > 4000000:
flag = 1
else:
i += 1
print(i,fib(i))
标签:fib,return,递归,斐波,flag,num,那契 来源: https://www.cnblogs.com/llzh/p/16373434.html