洛谷 P3768 简单的数学题 莫比乌斯反演+杜教筛

\[ \begin{split} \sum_{i=1}^n\sum_{j=1}^n ij(i,j)&=\sum_{d=1}^nd\sum_{i=1}^n\sum_{j=1}^n ij[(i,j)=d]\\ &=\sum_{d=1}^nd\sum_{i=1}^{\lfloor \dfrac{n}{d}\rfloor}\sum_{j=1}^{\lfloor \dfrac{n}{d}\rfloor} id\times jd[(i,j)=1]\\ &=\sum_{d=1}^nd^3\sum_{i=1}^{\lfloor \dfrac{n}{d}\rfloor}\sum_{j=1}^{\lfloor \dfrac{n}{d}\rfloor} ij[(i,j)=1]\\ &=\sum_{d=1}^nd^3\sum_{i=1}^{\lfloor \dfrac{n}{d}\rfloor}\sum_{j=1}^{\lfloor \dfrac{n}{d}\rfloor} ij \sum_{k\mid i,k\mid j} \mu(k)\\ &=\sum_{d=1}^nd^3\sum_{k=1}^{\lfloor \dfrac{n}{d}\rfloor} \mu(k) \sum_{i=1}^{\lfloor \dfrac{n}{kd}\rfloor}\sum_{j=1}^{\lfloor \dfrac{n}{kd}\rfloor} ik\times jk \\ &=\sum_{d=1}^nd^3\sum_{k=1}^{\lfloor \dfrac{n}{d}\rfloor} \mu(k) k^2\sum_{i=1}^{\lfloor \dfrac{n}{kd}\rfloor}\sum_{j=1}^{\lfloor \dfrac{n}{kd}\rfloor} ij \\ &=\sum_{T=1}^n \sum_{d\mid T}d^3 \mu\left({T \over d}\right) \left({T \over d}\right)^2\sum_{i=1}^{\lfloor \dfrac{n}{T}\rfloor}\sum_{j=1}^{\lfloor \dfrac{n}{T}\rfloor} ij \\ \end{split} \]