题意

求 $\large \sum_{i=1}^N\sum_{j=1}^Nmax(i,j)\cdot\sigma_1(ij)$
其中
$1\le N\le10^6$
$1\le T\le5\cdot10^4$
$\sigma_1(n)$表示 $n$的约数和

题目分析

令 $A=\sum_{i=1}^n\sum_{j=1}^ii\cdot\sigma_1(ij),B=\sum_{i=1}^ni\cdot\sigma_1(i^2)$,则 $Ans=2A-B$

且有
$\large \begin{aligned} A&=\sum_{i=1}^ni\cdot\sum_{j=1}^i\sum_{x|i}\sum_{y|j}x\cdot j/y[(x,y)=1]\\ &=\sum_{i=1}^ni\cdot\sum_{j=1}^i\sum_{x|i}\sum_{y|j}x\cdot j/y\sum_{d|(x,y)}\mu(d)\\ &=\sum_{d=1}^n\mu(d)\sum_{d|x}x\sum_{x|i}i\sum_{d|y}\sum_{y|j}^i\frac jy\\ &=\sum_{d=1}^n\mu(d)\sum_{x=1}^{\lfloor\frac nd\rfloor}dx\sum_{x|i}^{\lfloor\frac nd\rfloor}di\sum_{y=1}^{\lfloor\frac nd\rfloor}\sum_{y|j}^i\frac{dj}{dy}\\ &=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\sum_{x|i}x\sum_{y=1}^{\lfloor\frac nd\rfloor}\sum_{y|j}^iy\\ &=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\cdot\sigma_1(i)\sum_{j=1}^{i}\sigma_1(j)\\ &=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\cdot\sigma_1(i)\cdot S_{\sigma_1}(i)\\\end{aligned}$
$\large \begin{aligned} B&=\sum_{i=1}^ni\sum_{x|i}\sum_{y|i}x\cdot i/y[(x,y)=1]\\ &=\sum_{i=1}^ni\cdot\sum_{x|i}\sum_{y|i}x\cdot i/y\sum_{d|(x,y)}\mu(d)\\ &=\sum_{d=1}^n\mu(d)\sum_{d|i}i\sum_{d|x|i}\sum_{d|y|i}x\cdot \frac iy\\ &=\sum_{d=1}^n\mu(d)\sum_{i=1}^{\lfloor\frac nd\rfloor}di\sum_{x|i}\sum_{y|i}dx\cdot \frac {di}{dy}\\ &=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\sum_{x|i}x\sum_{y|i}\frac {i}{y}\\ &=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\sum_{x|i}x\sum_{y|i}y\\ &=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\sigma_1(i)^2\\ \end{aligned}$
$\large \begin{aligned} \therefore Ans=2A-B&= \end{aligned}$
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