随笔分类 -  类欧几里得

摘要:A \[ \sum\limits_{i=0}^{n}\sum\limits_{1\le cj\le ai+b}i^pj^q=\sum\limits_{i=0}^{n}\sum\limits_{j=1}^{\lfloor \frac{ai+b}{c}\rfloor}i^pj^q \] 令$F(n)=\ 阅读全文
posted @ 2021-08-16 21:25 jack_yyc 阅读(81) 评论(0) 推荐(0)