[国家集训队] Crash的数字表格 / JZPTAB

题目所求即

$$\sum _{i=1}^n\sum _{j=1}^m\frac{ij}{gcd(i,j)}$$

这里没有出现$[gcd(x,y)=1]$，所以我们枚举$gcd$的值来硬凑，原式就等于

$$\sum _{d=1}^{min(n,m)}\sum _{i=1}^n\sum _{j=1}^m\frac{ij}{gcd(i,j)}[gcd(i,j)=d]$$

为了出现$[gcd(i,j)=1]$，直接将$i,j$变成$d$的倍数，原式就等于

$$\begin{gathered} \sum _{d=1}^{min(n,m)}\sum _{d|i}\sum _{d|j}\frac{ij}{d}[gcd(i,j)=d] \\ =\sum _{d=1}^{min(n,m)}\sum _{k=1}^{\frac{n}{d}}\sum _{t=1}^{\frac{m}{d}}ktd[gcd(k,t)=1] \\ =\sum _{d=1}^{min(n,m)}d\sum _{k=1}^{\frac{n}{d}}\sum _{t=1}^{\frac{m}{d}}kt\sum _{p|gcd(k,t)}u(p) \\ =\sum _{d=1}^{min(n,m)}d\sum _{p=1}^{min(\frac{n}{d},\frac{m}{d})}u(p)\sum _{p|k}\sum _{p|t}kt \\ =\sum _{d=1}^{min(n,m)}d\sum _{p=1}^{min(\frac{n}{d},\frac{m}{d})}u(p)\sum _{i=1}^{\frac{n}{dp}}ip\sum _{j=1}^{\frac{m}{dp}}jp \\ =\sum _{d=1}^{min(n,m)}d\sum _{p=1}^{min(\frac{n}{d},\frac{m}{d})}u(p)p^2(\sum _{i=1}^{\frac{n}{dp}}i)(\sum _{j=1}^{\frac{m}{dp}}j) \end{gathered}$$

最后两项用等差数列求和公式就好了，然后分块套分块即可

update 2024.8.6

也可以利用莫比乌斯反演的第二形式

$$\begin{gathered} \sum _{d=1}^{min(n,m)}d\sum _{p=1}^{min(\frac{n}{d},\frac{m}{d})}u(p)p^2(\sum _{i=1}^{\frac{n}{dp}}i)(\sum _{j=1}^{\frac{m}{dp}}j) \\ =\sum _{d=1}^{min(n,m)}d\sum _{p=1}^{min(\frac{n}{d},\frac{m}{d})}u(p)p^2\frac{(1+\frac{m}{dp})\frac{m}{dp}}{2}\frac{(1+\frac{n}{dp})\frac{n}{dp}}{2} \\ \stackrel{T=dp}{=}\sum _{d=1}^{min(n,m)}\sum _{p=1}^{min(\frac{n}{d},\frac{m}{d})}u(\frac{T}{d})\frac{T^2}{d}\frac{(1+\frac{m}{T})\frac{m}{T}}{2}\frac{(1+\frac{n}{T})\frac{n}{T}}{2} \\ =\sum _{T=1}^{min(n,m)}T\frac{(1+\frac{m}{T})\frac{m}{T}}{2}\frac{(1+\frac{n}{T})\frac{n}{T}}{2}\sum _{d|T}\frac{T}{d}u(\frac{T}{d}) \\ \stackrel{F(T)=\sum _{d|T}\frac{T}{d}u(\frac{T}{d})}{=}\sum _{T=1}^{min(n,m)}T\frac{(1+\frac{m}{T})\frac{m}{T}}{2}\frac{(1+\frac{n}{T})\frac{n}{T}}{2}F(T) \end{gathered}$$

现在考虑如何快速计算$F(T)$，显然由数据范围可以知道我们需要用线性筛，而线性筛弄的是积性函数，所以我们要先证明$F$为积性函数，即若$a,b$互质，则$F(ab)=F(a)F(b)$

证：

$$\begin{gathered} F(a)F(b)=\sum _{d|a}\frac{a}{d}u(\frac{a}{d})\sum _{k|b}\frac{b}{k}u(\frac{b}{k}) \\ =\sum _{d|a}\sum _{k|b}\frac{ab}{dk}u(\frac{a}{d})u(\frac{b}{k}) \\ \stackrel{\text{u是积性函数}}{=}\sum _{d|a}\sum _{k|b}\frac{ab}{dk}u(\frac{ab}{dk}) \\ \stackrel{a,b\mathrm{互}\mathrm{质}}{=}F(ab) \end{gathered}$$

然后用线性筛就好了

这道题目告诉我们只要出现了$gcd$就可以尝试枚举$gcd$的值，即使$gcd$出现在求式里面而不是作为条件