Stargazer的博客

传送门莫反得到∑i=1a∑j=1b∑k=1calcm(i,j)blcm(j,k)clcm(i,k)\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^c\frac{a}{lcm(i,j)}\frac b{lcm(j,k)}\frac c{lcm(i,k)}∑i=1a​∑j=1b​∑k=1c​lcm(i,j)a​lcm(j,k)b​lcm(i,k)c​然后按照旧试题那道题的做法...