20201201A组T3（过程推导）

20201201A组T3（过程推导）

\[\begin{equation}\begin{aligned}&\sum_{i=1}^{n}\sum_{j=1}^{m}gcd^n(i,j)\sum_{k=1}^{ij}[i \bot k][j \bot k]k\\ &=\sum_{i=1}^{n}\sum_{j=1}^{m}gcd^n(i,j)\varphi(ij)ij+[ij=1]\\&=\frac{1}{2}+\frac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{m}gcd^n(i,j)\frac{\varphi(i)\varphi(j)gcd(i,j)}{\varphi(gcd(i,j))}ij\\ &=\frac{1}{2}+\frac{1}{2}\sum_{d=1}^{\min(n,m)}\frac{d^{n+1}}{\varphi(d)}\sum_{d|i,i\le n}\sum_{d|j,j\le m}\varphi(i)\varphi(j)ij[gcd(i,j)=d]\\ &这里用到了莫比乌斯反演\\ &=\frac{1}{2}+\frac{1}{2}\sum_{d=1}^{\min(n,m)}\frac{d^{n+1}}{\varphi(d)}\sum_{d|i,i\le n}\sum_{d|j,j\le m}\varphi(i)\varphi(j)ij\sum_{u|\frac{\gcd(i,j)}{d}}\mu(u)\\ &=\frac{1}{2}+\frac{1}{2}\sum_{d=1}^{\min(n,m)}\frac{d^{n+1}}{\varphi(d)}\sum_{u=1}^{\min(n,m)}\mu(u)\sum_{du|i,i\le n}\varphi(i)i\sum_{du|j,j\le m}\varphi(j)j\\ &=\frac12+\frac12\sum_{T=1}^{n}\sum_{T|i}\varphi(i)i\sum_{T|j}\varphi(j)j\sum_{d|T}\frac{d^{n+1}}{\varphi(d)}\mu(\frac{T}{d})\end{aligned}\end{equation} \]

注意到\(\sum_{i|T}\varphi(i)i\)是可以预处理的

同时，\(\sum_{d|T}\frac{d^{n+1}}{\varphi(d)}\mu(\frac{T}{d})\)，可以用线筛搞定

（来自MHT）