算是反演的板子题了吧……
然而我刚学反演所以还是写一写题解吧。
我们要求$sum limits _{x=1}^{N} sum limits _{y=1}^{M} left [ gcd(x,y)in prime ight ]$
枚举质数：$sum limits _{gin prime} sum limits _{x=1}^{left lfloor frac{N}{g} ight floor} sum limits _{y=1}^{left lfloor frac{M}{g} ight floor} left [ gcd(x,y)=1 ight ]$
反演得原式=$sum limits _{gin prime} sum limits _{x=1}^{left lfloor frac{N}{g} ight floor} sum limits _{y=1}^{left lfloor frac{M}{g} ight floor} sum limits _{t mid gcd(x,y)} u(t)$
将最后一个求和提前得原式=$sum limits _{gin prime} sum limits _{t=1}^{minleft ( left lfloor frac{N}{g} ight floor,left lfloor frac{M}{g} ight floor ight )} u(t) sum limits _{x=1}^{left lfloor frac{N}{g*t} ight floor} sum limits _{y=1}^{left lfloor frac{M}{g*t} ight floor} 1$
我自己推到这里就颓题解了
令T=g*t,得$sum limits _{T=1}^{min(n,m)} left ( left lfloor frac{N}{T} ight floor left lfloor frac{M}{T} ight floor * sum limits _{tmid T,tin prime} u(frac{T}{t}) ight )$
后面的部分可以用一个类似埃筛的东西预处理，前面的对于每次询问整除分块就行了。