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和大家拥抱----乐正绫《万圣街》
T1. P5572 [CmdOI2019] 简单的数论题
和毒瘤之神的考验类似的毒瘤题。
首先先化简原式：
\[\begin{array}{l}
\sum^n_{i = 1} \sum^m_{j = 1} \varphi(\dfrac{{\rm lcm}(i,j)}{\gcd(i,j)}) = \sum^n_{i = 1}\sum^m_{j = 1} \varphi(\dfrac{i \times j}{\gcd(i,j)^2})\\=
\sum^n_{d = 1}\sum^{n}_{i = 1}\sum^m_{j = 1} \varphi(\dfrac{i \times j}{d^2})[\gcd(i,j) == d]\\=
\sum^n_{d = 1}\sum^{\lfloor \frac{n}{d} \rfloor}_{i = 1}\sum^{\lfloor \frac{m}{d} \rfloor}_{j = 1} \varphi(i \times j)[\gcd(i,j) == 1]\\=
\sum^n_{d = 1}\sum^{\lfloor \frac{n}{d} \rfloor}_{i = 1}\sum^{\lfloor \frac{m}{d} \rfloor}_{j = 1} \varphi(i) \times \varphi(j)[\gcd(i,j) == 1]\\=
\sum^n_{d = 1}\sum^{\lfloor \frac{n}{d} \rfloor}_{i = 1}\sum^{\lfloor \frac{m}{d} \rfloor}_{j = 1} \varphi(i) \times \varphi(j) \sum_{p \mid \gcd(i,j)} \mu(p)\\=
\sum^n_{d = 1}\sum^{\lfloor \frac{n}{d} \rfloor}_{p = 1} \mu(p) \sum^{\lfloor \frac{n}{d} \rfloor}_{i = 1} \sum^{\lfloor \frac{m}{d} \rfloor}_{j=1} \mu(i) \mu(j) [p \mid i] [p \mid j] \\=
\sum^n_{d = 1}\sum^{\lfloor \frac{d}{n} \rfloor}_{p = 1} \mu(p) \sum^{\lfloor \frac{n}{pd} \rfloor}_{i = 1} \varphi(ip) \sum^{\lfloor \frac{m}{pd} \rfloor}_{j = 1} \varphi(jp)\\=
\sum^n_{T = 1} \sum_{p | T} \mu(p) \sum^{\lfloor \frac{n}{T} \rfloor}_{i = 1} \mu(ip) \sum^{\lfloor \frac{m}{T} \rfloor}_{j = 1} \varphi(jp)
\end{array}
\]
然后我们设 \(G(x,y) = \sum^x_{i = 1} \varphi(iy)\)。那么原式子就变为 \(\sum_{T = 1}^n \sum_{p | T} \mu(p) G(\lfloor \frac{n}{T} \rfloor,p) G(\lfloor \frac{m}{T} \rfloor,p)\)。
然后我们定义 \(H(x,y,z) = \sum_{T = 1}^z \sum_{p | T} \mu(p) G(x,p) G(y,p)\)。
接着我们就可以用毒瘤之神的考验的套路了。用根号分支来求 \(H\)，具体的我们设置一个阈值 \(B\)，将所有 \(\le B\) 的 \(H\) 都提前预处理出来，然后对于 \(\ge B\) 的部分，我们可以在查询时利用整除分块计算。
T2. P3911 最小公倍数之和
又是推式子题，好欸好欸好欸！虽然说大多数步骤都是类似的，但是还是有许多细节。
首先问们定义 \(c_i\) 表示 \(i\) 的出现次数，\(n\) 为最大值。那么就有：
\[\begin{array}{l}
\sum^n_{i = 1} \sum^m_{j = 1}c_i \times c_j \times \rm{lcm}(i,j)\\=
\sum^n_{i = 1} \sum^m_{j = 1}c_i \times c_j \times \dfrac{i \times j}{\gcd(i,j)}\\=
\sum^k_{i = 1} \sum^n_{i = 1} \sum^m_{j = 1} c_i \times c_j \times \dfrac{i \times j}{k}[\gcd(i,j) == k]\\=
\sum^k_{i = 1} \sum^{\lfloor \frac{n}{k} \rfloor}_{i = 1} \sum^{\lfloor \frac{m}{k} \rfloor}_{j = 1} c_{ik} \times c_{jk} \times i \times j \times k[\gcd(i,j) == 1]\\=
\sum^k_{i = 1} \sum^{\lfloor \frac{n}{k} \rfloor}_{i = 1} \sum^{\lfloor \frac{m}{k} \rfloor}_{j = 1} \sum_{d \mid \gcd(i,j)} \mu(d) c_{ik} \times c_{jk} \times i \times j \times k\\=
\sum^k_{i = 1}\sum^n_{kd = 1}\mu(d) \times d^2 \sum^{\lfloor \frac{n}{kd} \rfloor}_{i = 1}\sum^{\lfloor \frac{m}{kd} \rfloor}_{j = 1}c_{ikd} \times c_{jkd} \times i \times j \times k\\=
\sum^n_{T = 1}T(\sum_{d|T} \mu(d) \times d) \sum^{\lfloor \frac{n}{T} \rfloor}_{i = 1}\sum^{\lfloor \frac{m}{T} \rfloor}_{j = 1}c_{iT} \times c_{jT} \times i \times j\\=
\sum^n_{T = 1}T(\sum_{d|T} \mu(d) \times d) (\sum^{\lfloor \frac{n}{T} \rfloor}_{i = 1} c_{iT} \times i)^2
\end{array}
\]
然后第一个预处理，后面的暴力算即可。
T3. CF911G Mass Change Queries
我们发现 \(a_i \le 100\)，所以我们可以建出 \(100\) 棵线段树，然后我们只需要维护这些线段树即可。我们把所有 \(a_i = x\) 的 \(i\) 放到第 \(x\) 个线段树上。
然后我们考虑怎么维护更改操作。其实就是把 \(x\) 线段树上在 \([l,r]\) 之间的点合并到 \(y\) 上，这样子对于所有在 \([l,r]\) 的点并且 \(a_i = x\) 的 \(i\) 就全部移到了 \(y\) 线段树上。
