对于本题的问题，有
$$ \sum_{p\in P}\sum_{x=1}^n\sum_{y=1}^m[gcd(x,y)=p] \\ =\sum_{p\in P}\sum_{x=1}^{\lfloor\frac{n}{p}\rfloor}\sum_{y=1}^{\lfloor\frac{m}{p}\rfloor}[gcd(x,y)=1] \\ =\sum_{p\in P}\sum_{x=1}^{\lfloor\frac{n}{p}\rfloor}\sum_{y=1}^{\lfloor\frac{m}{p}\rfloor}\sum_{d\mid x,d\mid y}\mu(d) \\=\sum_{p\in P} \sum_{d=1}^{\lfloor \frac{min(n,m)}{p} \rfloor}\mu(d) \lfloor\frac{n}{pd}\rfloor\lfloor\frac{m}{pd}\rfloor \\\text{令t=pd} \\=\sum_{t=1}{min(n,m)} \lfloor\frac{n}{t}\rfloor\lfloor\frac{m}{t}\rfloor\sum_{p\in P,p\mid t}\mu(\frac{t}{p}) \\令f(n)=[n\in P],且 \\g(n)=\sum_{p\in P,p\mid n}\mu(\frac{n}{p}) \\则g=f*u $$FIELD NOTES
2257推导
$$
\sum_{x=1}^n\sum_{m=1}^n [gcd(x,y)=1]
\\
= \sum_{x=1}^n\sum_{m=1}^n\sum_{d\mid gcd(x,y)} \mu(d)
\\
= \sum_{x=1}^n\sum_{m=1}^n\sum_{d\mid x,d\mid y} \mu(d)
\\
\text{此时x,y便独立开来，调换求和顺序}
\\=\sum_{d=1}^{min(n,m)} \mu(d) \lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor
$$