摘要: 1 前置cheese 排序不等式 设$a_1\leq a_2\leq ...\leq a_n,b_1\leq b_2\leq ...\leq b_n$,则 \(\sum_{i=1}^na_ib_{n-i+1}\leq \sum_{i=1}^na_ib_{d_i}\leq\sum_{i=1}^na_i 阅读全文
posted @ 2020-10-04 13:25 刘子闻 阅读(131) 评论(1) 推荐(1)