递归函数与斐波那契数列

#求阶乘的函数

def jiecheng(n):
if n ==1:
return 1

return n*jiecheng(n-1)

print(jiecheng(5))

#斐波那契数列

#0,1,1,2,3,5,8,13,21...........

def fei(n):

if n == 1:
return 0
if n == 2:
return 1
return int(fei(n-1))+int(fei(n-2))

# def fab(n):
# if n==1:
# return 1
# if n==0:
# return 0
# else:
# result=int(fab(n-1))+int(fab(n-2))
# return result
for i in range(1,11):
print(fei(i))
原文地址:https://www.cnblogs.com/lokerx/p/10760444.html