查看原题'" class='mdui-btn mdui-btn-raised'>点击加载点击跳转设$n<m$,$\sigma(d)$为$d$的约数和先不考虑$\sigma(\gcd(i,j))\le a$$$\sum{i=1}^n\sum{j=1}^m \sigma(\gcd(i,j))\=\sum{d=1}^n \sigma(d) \sum{i=1}^n\sum{j=1}^m [\gcd(i,j)=d]\=\sum{d=1}^n \sigma(d) \sum{i=1}^{\left \lfloor \frac nd \right \rfloor}\sum{j=1}^{\left \lf……