欢迎大家来看我写的A+B呀

##A+B

感谢大李老板友情赞助防ak，赛前打包票两个人尝试没有人能过嘻嘻嘻

$A={\sum }_{x_1=1}^n\cdot \cdot \cdot {\sum }_{x_m=1}^n{x}_1^3{\sum }_{d|gcd(x_1,x_2,\cdot \cdot \cdot ,x_m)}\mu (d)$

$={\sum }_{d=1}^n\mu (d){\sum }_{x_1=1,d|x_1}^n\cdot \cdot \cdot {\sum }_{x_m=1,d|x_m}^n\ast x_1^3$

$={\sum }_{d=1}^n\mu (d)\ast \lfloor \frac{n}{d}{\rfloor }^{m-1}\ast {\sum }_{x_1}^{\frac{n}{d}}{x}_1^3\ast d^3$

$={\sum }_{d=1}^n{d}^3\mu (d)\lfloor \frac{n}{d}{\rfloor }^{m-1}\ast f(\lfloor \frac{n}{d}\rfloor )$

$f(n)=\frac{n^2(n+1{)}^2}{4}$其中 $\sum d^3\mu (d)$

$用线性筛+前缀和预处理o(n)$当然直接暴力好像也不是不可以

$\lfloor \frac{n}{d}{\rfloor }^{m-1}f(\lfloor \frac{n}{d}\rfloor )$

$可以用数论分块+快速幂 $

$\mu (d)是莫比乌斯函数$

$复杂度：o(n+q\ast \sqrt{n}\ast \log m)$

$B={\sum }_{i=1}^m{\sum }_{x_1=1}^n\cdot \cdot \cdot {\sum }_{x_m=1}^n(x_i^2+x_i\ast {\sum }_{i\ne j,j=1}^n{x}_j)\ast {\sum }_{d|gcd(x_1,x_2,\cdot \cdot \cdot ,x_m)}\mu (d)$

$={\sum }_{i=1}^m{\sum }_{x_1=1}^n\cdot \cdot \cdot {\sum }_{x_m=1}^n(x_i^2)\ast {\sum }_{d|gcd(x_1,x_2,\cdot \cdot \cdot ,x_m)}\mu (d)+{\sum }_{i=1}^m{\sum }_{x_1=1}^n\cdot \cdot \cdot {\sum }_{x_m=1}^n{x}_i\ast {\sum }_{i\ne j,j=1}^n{x}_j\ast {\sum }_{d|gcd(x_1,x_2,\cdot \cdot \cdot ,x_m)}\mu (d)$

$=P+Q$

由于对称性，知$\forall i\in [1,m],有$

${\sum }_{x_1=1}^n\cdot \cdot \cdot {\sum }_{x_m=1}^n(x_i^2){\sum }_{d|gcd(x_1,x_2,\cdot \cdot \cdot ,x_m)}\mu (d)$

$={\sum }_{x_1=1}^n\cdot \cdot \cdot {\sum }_{x_m=1}^n(x_1^2){\sum }_{d|gcd(x_1,x_2,\cdot \cdot \cdot ,x_m)}\mu (d)$

$P=m\ast {\sum }_{x_1=1}^n\cdot \cdot \cdot {\sum }_{x_m=1}^n(x_1^2){\sum }_{d|gcd(x_1,x_2,\cdot \cdot \cdot ,x_m)}\mu (d)$

$=m\ast {\sum }_{d=1}^n\mu (d)\ast {\sum }_{x_1=1,d|x_1}^n\cdot \cdot \cdot {\sum }_{x_m=1,d|x_m}^n(x_1^2)$

$=m\ast {\sum }_{d=1}^n\mu (d)\lfloor \frac{n}{d}{\rfloor }^{m-1}{\sum }_{x_1=1}^{\frac{n}{d}}{x}_1^2\ast d^2$

$=m\ast {\sum }_{d=1}^n\mu (d)d^2\lfloor \frac{n}{d}{\rfloor }^{m-1}\ast g(\lfloor \frac{n}{d}\rfloor )$

$其中，g(n)=\frac{n(n+1)(2n+1)}{6}$

$预处理\sum d^2\mu (d)$

$数论分块+快速幂$

$o(n+q\sqrt{n}logm)$

$ 同理Q$

$Q=m(m-1){\sum }_{x_1=1}^n\cdot \cdot \cdot {\sum }_{x_m=1}^n{x}_1{x}_2{\sum }_{d|gcd(x_1,x_2,\cdot \cdot \cdot ,x_m)}\mu (d)$

$不理解的可以手推(x+y+z{)}^2$

${\sum }_{x=1}^n{\sum }_{y=1}^n{\sum }_{z=1}^n(x+y+z{)}^2$

$=3{\sum }_{x=1}^n{\sum }_{y=1}^n{\sum }_{z=1}^n{x}^2+6\ast {\sum }_{x=1}^n{\sum }_{y=1}^n{\sum }_{z=1}^{n}xy$

或者借助完全图来理解变量之间的对称

$Q=m(m-1){\sum }_{x_1=1}^n\cdot \cdot \cdot {\sum }_{x_m=1}^n{x}_1{x}_2{\sum }_{d|gcd(x_1,x_2,\cdot \cdot \cdot ,x_m)}\mu (d)$

$=m(m-1){\sum }_{d=1}^n\mu (d)\lfloor \frac{n}{d}{\rfloor }^{m-2}\ast {\sum }_{x_1=1}^{\frac{n}{d}}{\sum }_{x_2=1}^{\frac{n}{d}}{d}^2{x}_1{x}_2$

$=m(m-1){\sum }_{d=1}^n\mu (d)d^2\lfloor \frac{n}{d}{\rfloor }^{m-2}\ast ({\sum }_{x_1=1}^{\frac{n}{d}}{x}_1{)}^2$

$=m(m-1){\sum }_{d=1}^n\mu (d)d^2\lfloor \frac{n}{d}{\rfloor }^{m-2}\ast (\frac{\lfloor \frac{n}{d}\rfloor (1+\lfloor \frac{n}{d}\rfloor )}{2}{)}^2$

$=m(m-1){\sum }_{d=1}^n\mu (d)d^2\lfloor \frac{n}{d}{\rfloor }^m(\frac{(1+\lfloor \frac{n}{d}\rfloor )}{2}{)}^2$

$数论分块+快速幂$