斐波那契数列思路 递归 Python解答会超时 滚动数组 时间复杂度较低 Python也不会超时 代码1234567891011n = int(input())p,q,r = 0,1,1if n<2: print(n)else: for i in range(2,n): p = q q = r r = p + q print(r)