题解 | #兔子的数量#

1.递归：归适合于问题可以自然地划分为子问题的情况
n = int(input())
def fibs(n):
    if n == 1:
        return 2
    elif n == 2:
        return 3
    else:
        return fibs(n-2)+fibs(n-1)
result = fibs(n)
print(result)

2.迭代：迭代适合于问题的解决过程可以通过循环和迭代来实现的情况
n = int(input())
def fact(n)
if n == 1:
    return 2
elif n == 2:
    return 3
prev_prev == 2
prev == 3

for i in range(3,n+1):
	current = prev+prev_prev
	prev_prev = prev
	prev = current
return current
print(fact(n))