好难啊..
首先，本题的突破口在于直线的性质。不论是几维的空间，两点一定能确定一条直线
选取两个点作为最左下和最右上的点！
假设现在是二维空间，选取了$(x1,y1)$和$(x2,y2)$两个点，那么它们连线上经过的点数就是$gcd(x2-x1,y2-y1)-1$
选取的方案数为$\left ( _{gcd-1}^{c-2} \right )$
发现这个方案数只和坐标的差值有关，我们直接枚举差值就行
接下来就是反演了，为了方便叙述，以$n=2$为例
$\sum\limits_{x=1}^{m_{1}}\sum\limits_{y=1}^{m_{2}} \left ( _{c-2}^{gcd(x,y)-1} \right ) (m_{1}-x)(m_{2}-y)$
$\sum\limits_{k=1}^{m} \left ( _{c-2}^{k-1} \right ) \sum\limits_{x=1}^{m_{1}} \sum\limits_{y=1}^{m_{2}} [gcd(x,y)==k] (m_{1}-x)(m_{2}-y)$
$\sum\limits_{k=1}^{m} \left ( _{c-2}^{k-1} \right ) \sum\limits_{x=1}^{\left \lfloor \frac{m_{1}}{k} \right \rfloor} \sum\limits_{y=1}^{\left \lfloor \frac{m_{2}}{k} \right \rfloor} [gcd(x,y)==1] (m_{1}-xk)(m_{2}-yk)$
$\sum\limits_{k=1}^{m} \left ( _{c-2}^{k-1} \right ) \sum\limits_{x=1}^{\left \lfloor \frac{m_{1}}{k} \right \rfloor} \sum\limits_{y=1}^{\left \lfloor \frac{m_{2}}{k} \right \rfloor} \sum\limits_{d|k}\mu(d)(m_{1}-xk)(m_{2}-yk)$
$\sum\limits_{k=1}^{m} \left ( _{c-2}^{k-1} \right ) \sum\limits_{d=1}^{\left \lfloor \frac{m_{1}}{k} \right \rfloor} \mu(d) \sum\limits_{x=1}^{\left \lfloor \frac{m_{1}}{kd} \right \rfloor} (m_{1}-xkd) \sum\limits_{y=1}^{ \left \lfloor \frac{m_{2}}{kd} \right \rfloor} (m_{2}-ykd)$
令$Q=kd$
$\sum\limits_{Q=1}^{m} \left ( \sum\limits_{k|Q} \left ( _{c-2}^{k-1} \right ) \mu(\frac{Q}{k}) \right ) \left ( \sum\limits_{x=1}^{\left \lfloor \frac{m_{1}}{Q} \right \rfloor} (m_{1}-xQ) \right ) \left ( \sum\limits_{y=1}^{ \left \lfloor \frac{m_{2}}{Q} \right \rfloor} (m_{2}-yQ) \right ) $
利用等差数列公式 $\left ( \sum\limits_{x=1}^{\left \lfloor \frac{m}{Q} \right \rfloor} (m-xQ) \right ) = \left ( \left \lfloor \frac{m}{Q} \right \rfloor m- \frac{ (1+\left \lfloor \frac{m}{Q} \right \rfloor )\left \lfloor \frac{m}{Q} \right \rfloor Q}{2} \right ) $
$\sum\limits_{Q=1}^{m} \left ( \sum\limits_{k|Q} \left ( _{c-2}^{k-1} \right ) \mu(\frac{Q}{k}) \right ) \left ( \left \lfloor \frac{m_{1}}{Q} \right \rfloor m_{1}- \frac{ (1+\left \lfloor \frac{m_{1}}{Q} \right \rfloor )\left \lfloor \frac{m_{1}}{Q} \right \rfloor Q}{2} \right ) \left ( \left \lfloor \frac{m_{2}}{Q} \right \rfloor m_{2}- \frac{ (1+\left \lfloor \frac{m_{2}}{Q} \right \rfloor )\left \lfloor \frac{m_{2}}{Q} \right \rfloor Q}{2} \right ) $
推广到更高维上
$\sum\limits_{Q=1}^{m} \left ( \sum\limits_{k|Q} \left ( _{c-2}^{k-1} \right ) \mu(\frac{Q}{k}) \right ) \prod_{t=1}^{n} \left ( \left \lfloor \frac{m_{t}}{Q} \right \rfloor m_{t}- \frac{ (1+\left \lfloor \frac{m_{t}}{Q} \right \rfloor )\left \lfloor \frac{m_{t}}{Q} \right \rfloor Q}{2} \right ) $
预处理出$f(Q,c)=\sum\limits_{k|Q} \left ( _{c-2}^{k-1} \right ) $
用整除分块即可
时间$O(Tn^{3}\sqrt {m})$