摘要: '''求斐波那契数列截至第n项时,所有的偶数和'''def solution(n): if n == 0 or n == 1: return 0 total = 0 i, j = 0, 1 # a[0] = 0, a[1] = 1 m = 2 # 从第二项开始 while m <= n: s = i 阅读全文
posted @ 2021-11-22 14:21 keep2021 阅读(51) 评论(0) 推荐(0) 编辑