LG 3911 最小公倍数之和

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查看原题'" class='mdui-btn mdui-btn-raised'>点击加载点击跳转$\sum{i=1}^n\sum{j=1}^n lcm(A_i,A_j)$设$ai = \sum{j=1}^n [A_j=i]$那么$$ans = \sum_i^n \sum_j^n a_iaj \frac{ij}{\gcd(i,j)}\= \sum{d=1}^n \frac 1d \sum{i=1}^n \sum{j=1}^n a_iaj ij[\gcd(i,j)=d]\ = \sum{d=1}^n \sum{i=1}^{\frac nd} \sum{j=1}^{\frac nd}d \times……