摘要: 懒惰即美德斐波那契数列>>> fibs=[0,1]>>> for i in range(8): fibs.append(fibs[-2]+fibs[-1])>>> fibs[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]orfibs=[0,1]num=input('How many... 阅读全文
posted @ 2015-08-04 10:08 whats 阅读(348) 评论(0) 推荐(0) 编辑