摘要: 思路: $\Sigma_{i=1}^n\Sigma_{j=1}^mgcd(i,j)==p(p是素数)$ $\Sigma_{p是素数}^{p<=n}\Sigma_{i=1}^{\lfloor \frac{n}{p} \rfloor}\Sigma_{j=1}^{\lfloor \frac{m}{p} \ 阅读全文
posted @ 2017-04-07 23:32 SiriusRen 阅读(163) 评论(0) 推荐(0) 编辑