Python--斐那锲波

 迭代实现

def fab(n):
    n1 = 1
    n2 = 1
    n3 = 1

    if n < 1:
        print('输入有误！')
        return -1
    while (n - 2) > 0:
        n3 = n2 + n1
        n1 = n2
        n2 = n3
        n -= 1

    return n3


daylop = int(input('请输入月数：'))
result = fab(daylop)
if result != -1:
    print('总共有%d对小兔子诞生！' % result)

递归

def fab(n):
    if n < 1:
        print('输入有误！')
        return -1
    if n == 1 or n ==2:
        return 1
    else:
        return fab(n-1) + fab(n-2)
dalop = int(input('请输入月数：'))

result = fab(dalop)

if result !=-1:
    print('总共或有%d对兔子'% result)