03 2021 档案

摘要:1.1 cid \(\in\) [0, p) num = n / p mod = n % p my_first_i(cid) = cid * num + cid > n - m ? 1 : 0 my_last_i(cid) = (cid + 1) * num + cid >= n - m ? 1 : 阅读全文
posted @ 2021-03-13 22:42 llzhh 阅读(1428) 评论(0) 推荐(0)