Potyczki Algorytmiczne 2009

Trial Round: Rectangles \[ans=\sum_{1\leq i\leq j\leq n}[ij\leq n]=\frac{\lfloor\sqrt{n}\rfloor+\sum_{1\leq i,j\leq n}[ij\leq n]}{2}=\frac{\lfloor\sqr
posted @ 2022-01-25 21:38  Claris  阅读(259)  评论(0编辑  收藏  举报