(Python3练手100例第011题兔子问题)

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本文通过一个经典的兔子繁殖问题介绍了斐波那契数列的概念，并提供了Python实现的方法来计算任意月份的兔子总数。通过递归函数演示了数列的增长规律。

#coding:utf-8
# ex011.py
#11 古典问题：有一对兔子，从出生后第3个月起每个月都生一对兔子，小兔子长到第三个月后每个月又生一对兔子，
# 假如兔子都不死，问每个月的兔子总数为多少对？
# 0 1 1 2 3 5 即为斐波那契数列 fib
'''
>>> [fib(n) for n in range(13)]
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144]
>>> fib(5)
5
>>> fib(20)
6765
>>> fib(12)
144
'''
def fib(n):
    if n<=0:
        return 0
    elif n<=2:
        return 1
    else:
        return fib(n-2)+fib(n-1)


if __name__ == '__main__':
    try:
        n=int(input("input month:"))
        print(fib(n))
    except:
        print("Error: an int is allow!")
'''
test(使用doctest模块测试):
python -m doctest -v ex011.py
Trying:
    [fib(n) for n in range(13)]
Expecting:
    [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144]
ok
Trying:
    fib(5)
Expecting:
    5
ok
Trying:
    fib(20)
Expecting:
    6765
ok
Trying:
    fib(12)
Expecting:
    144
ok
1 items had no tests:
    ex011.fib
1 items passed all tests:
   4 tests in ex011
4 tests in 2 items.
4 passed and 0 failed.
Test passed.
'''
#:eof