2019年9月2日
The Preliminary Contest for ICPC Asia Nanjing 2019 C. Tsy's number 5
摘要： https://nanti.jisuanke.com/t/41300 题意:求$\sum_{i=1}^n\phi(i)\phi(j)2^{\phi(i)\phi(j)}$ $f_i=\sum_{k=1}^n[\phi(k)==i]$ $\sum_{i=1}^n\phi(i)\phi(j)2^{\ph 阅读全文
posted @ 2019-09-02 14:36 walfy 阅读(234) 评论(0) 推荐(0)