【题解】加权约数和

$n\le 10^6$ 。多组数据。
$$\sum _{i=1}^n\sum _{j=1}^n\max (i,j)\times \sigma (ij)$$
考虑这样一件事情。我们暴力枚举 i 。（233
$$\begin{gathered} i\times \sum _{j=1}^i\sigma (ij) \\ =i\times \sum _{j=1}^i\sum _{x|i}\sum _{y|j}[(x,y)=1]\frac{iy}{x} \\ =i\times \sum _{j=1}^i\sum _{x|i}\sum _{y|j}[(\frac{i}{x},y)=1]xy \\ =i\times \sum _{j=1}^i\sum _{x|i}\sum _{y|j}xy\sum _{k|(\frac{i}{x},y)}\mu (k) \\ =i\times \sum _{k|i}\mu (k)\sum _{x|\frac{i}{k}}x\sum _{y=1}^{\frac{i}{k}}yk[\frac{i}{yk}] \\ =i\times \sum _{k|i}\mu (k)\times k\times \sigma (\frac{i}{k})\times \sum _{i=1}^{\frac{i}{k}}\sigma (i) \end{gathered}$$

$$\begin{gathered} \sum _{i=1}^{n}i\times \sum _{j=1}^i\sigma (ij) \\ =\sum _{i=1}^{n}i\times \sum _{k|i}\mu (k)\times k\times \sigma (\frac{i}{k})\times F(\frac{i}{k}) \\ =\sum _{i=1}^n\sigma (i)\times F(i)\times i\times \sum _{k=1}^{[\frac{n}{i}]}\mu (k)\times k^2 \end{gathered}$$

???
$$\begin{gathered} \sum _{i=1}^{n}i\times \sigma (i^2) \\ =\sum _{i=1}^{n}i\times \sum _{x|i}\sum _{y|i}xy\sum _{k|(\frac{i}{x},y)}\mu (k) \\ =\sum _{i=1}^{n}i\times \sum _{k|i}\mu (k)\times k\times \sigma (\frac{i}{k}{)}^2 \\ =\sum _{i=1}^n\sigma (i{)}^2\times i\times \sum _{k=1}^{[\frac{n}{i}]}\mu (k)\times k^2 \end{gathered}$$
however …

$O(T\sqrt{n})$ 过不去。

差分一下就好。