看题解才会的（指把题解瞅了个遍），没脸发题解
简单性质
\[\large \begin{aligned} &\varphi(ij)=ij\prod_{p\mid ij}\frac{p-1}{p}\\ &\varphi(\text{gcd}(i,j))=\text{gcd}(i,j)\prod_{p\mid i,p\mid j}\frac{p-1}{p}\\ &\varphi(ij)\times\varphi(\text{gcd}(i,j))\\ &=ij\prod_{p\mid ij}\frac{p-1}{p}\times\text{gcd}(i,j)\prod_{p\mid i,p\mid j}\frac{p-1}{p}\\ &=i\prod_{p\mid i}\frac{p-1}{p}\times j\prod_{p\mid j}\frac{p-1}{p}\\ &=\varphi(i)\varphi(j)\text{gcd}(i, j)\\ \end{aligned} \] \[\large{\varphi(ij)=\frac{\varphi(i)\varphi(j)\text{gcd}(i,j)}{\varphi(\text{gcd}(i,j))}} \]简单推导
\[\large \begin{aligned} &n\le m\\ &\sum_{i=1}^n\sum_{j=1}^m\varphi(ij)\\ &=\sum_{i=1}^n\sum_{j=1}^m\varphi(i)\varphi(j)\frac{\text{gcd}(i, j)}{\varphi(\text{gcd}(i,j))}\\ &=\sum_{d=1}^n\sum_{i=1}^n\sum_{j=1}^m\varphi(i)\varphi(j)\frac{d}{\varphi(d)}[\text{gcd}(i,j)=d]\\ &=\sum_{d=1}^n\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}\varphi(id)\varphi(jd)\frac{d}{\varphi(d)}[\text{gcd}(i,j)=1]\\ &=\sum_{d=1}^n\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}\varphi(id)\varphi(jd)\frac{d}{\varphi(d)}\sum_{k\mid i,k\mid j}\mu(k)\\ &=\sum_{d=1}^n\frac{d}{\varphi(d)}\sum_{k=1}^{\lfloor\frac{n}{d}\rfloor}\mu(k)\sum_{i=1}^{\lfloor\frac{n}{kd}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{kd}\rfloor}\varphi(ikd)\varphi(jkd)\rightarrow t=kd\\ &=\sum_{t=1}^n\sum_{k\mid t}\mu(k)\frac{\frac{t}{k}}{\varphi(\frac{t}{k})}\sum_{i=1}^{\lfloor\frac{n}{t}\rfloor}\varphi(it)\sum_{j=1}^{\lfloor\frac{m}{t}\rfloor}\varphi(jt)\\ &=\sum_{t=1}^n\sum_{k\mid t}\mu(\frac{t}{k})\frac{k}{\varphi(k)}\sum_{i=1}^{\lfloor\frac{n}{t}\rfloor}\varphi(it)\sum_{j=1}^{\lfloor\frac{m}{t}\rfloor}\varphi(jt) \end{aligned} \]毒瘤分析
\[\large \sum_{t=1}^n\sum_{k\mid t}\mu(\frac{t}{k})\frac{k}{\varphi(k)}\sum_{i=1}^{\lfloor\frac{n}{t}\rfloor}\varphi(it)\sum_{j=1}^{\lfloor\frac{m}{t}\rfloor}\varphi(jt) \]这个臃肿的式子无法满足我们的要求，只好看看能否更进一步
\[\large\text{osu}(x)=\sum_{k\mid t}\mu(\frac{t}{k})\frac{k}{\varphi(k)} \]这一段线筛预处理，\(\text{O(1)}\) 查询（注意里面的欧拉函数用逆元）
\[\large G(\lfloor\frac{n}{t}\rfloor,t)=\sum_{i=1}^{\lfloor\frac{n}{t}\rfloor}\varphi(it) \]明眼人都能看出参数来 =_=
\[\large { G(x,y)=\sum_{i=1}^{x}\varphi(iy)\\ G(x,y)=\sum_{i=1}^{x-1}\varphi(iy)+\varphi(xy)=G(x-1,y)+\varphi(xy)\\ } \]然后拿着最后的一溜串结果搞事情~
\[\large{ S(n,\lfloor\frac{n}{t}\rfloor,\lfloor\frac{m}{t}\rfloor)=\sum_{t=1}^n\sum_{k\mid t}\mu(\frac{t}{k})\frac{k}{\varphi(k)}\sum_{i=1}^{\lfloor\frac{n}{t}\rfloor}\varphi(it)\sum_{j=1}^{\lfloor\frac{m}{t}\rfloor}\varphi(jt) } \] \[\large \begin{aligned} &S(x,y,z)=\sum_{t=1}^x\sum_{k\mid t}\mu(\frac{t}{k})\frac{k}{\varphi(k)}\sum_{i=1}^{y}\varphi(it)\sum_{j=1}^{z}\varphi(jt)\\ &S(x,y,z)=S(x-1,y,z)+\text{osu}(x)\times G(y,x)\times G(z,x) \end{aligned} \]发现各个函数单独看也不是很难？
分个块勉强可以忍一下。（代码迫于枚举顺序参数位置稍有变动）