随笔分类 -  牛客刷题笔记

C/C++
摘要:下列程序输出:67 解析: 解法1: f(n)=0, cnt自加一次,f(n)=1, cnt自加一次,即: n=0 cnt = 1; n=1 cnt = 1; n=2 cnt = f(1) + f(0) = 1+1+1 = 3; n=3 cnt = f(2) + f(1) = 3+1+1 = 5; 阅读全文

posted @ 2018-05-22 12:23 MrRS 阅读(376) 评论(0) 推荐(0)

导航