查看原题'" class='mdui-btn mdui-btn-raised'>点击加载点击跳转前置定理：$$d(ij) = \sum{x|i} \sum{y|j}[gcd(x,y) = 1]$$所以：设$n<m$$$\sum{i=1}^n \sum{j=1}^m d(ij)\=\sum{i=1}^n \sum{j=1}^m \sum{x|i} \sum{y|j}[gcd(x,y) = 1]\=\sum\limits{x=1}^n\sum\limits{y=1}^m \left\lfloor\frac{n}{x}\right\rfloor \left\lfloor\frac{m}{y}\……