P5518 [MtOI2019]幽灵乐团 / 莫比乌斯反演基础练习题

瞎扯

建议在阅读题解之前欣赏这首由普莉兹姆利巴姐妹带来的的合奏。

Q：你参加省选吗？不是说好了考完 NOIP 就退吗。
A：对啊。
Q：那你学这玩意干啥？
A：对啊，我学这玩意干啥？

写这题的动机？
一是一直很喜欢的曲子，感觉快退役了，圆个梦。
二是写了很多题解了，之前认为最优秀的是 NOI嘉年华的题解，但被叉掉之后不知道该怎么改了，于是删了。其他的都太不精致，都不满意。想在退役之前留下一篇最优秀的题解，于是瞅准了这题。
再有，就是想争口气吧。

最后扯一句，题面里将露娜萨（Lunasa）误写成了 （Lusana）。

简述

原题面：Luogu

给定参数 \(p\)，有 \(T\) 组数据，每次给定参数 \(A,B,C\)，求：

\[\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}\left(\dfrac{\operatorname{lcm}(i,j)}{\gcd(i,k)}\right)^{f(type)} \]

其中 \(f(type)\) 的取值如下：

\[f(type) = \begin{cases} 1 &type = 0\\ i\times j\times k &type = 1\\ \gcd(i,j,k) &type = 2 \end{cases}\]

\(1\le A,B,C\le 10^5\)，\(10^7\le p\le 1.05\times 10^9\)，\(p\in \mathbb{P}\)，\(T=70\)。
2.5S，128MB。

分析

原式

先化下原式，原式等于：

\[\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}\left(\dfrac{i\times j }{\gcd(i,j)\times \gcd(i,k)}\right)^{f(type)} \]

发现每一项仅与两个变量有关，设：

\[\begin{aligned} f_1(a,b,c) &= \prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} i^{f(type)}\\ f_2(a,b,c) &= \prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} \gcd(i,j)^{f(type)} \end{aligned}\]

发现 \(\prod\) 可以随意交换，则原式等价于：

\[\dfrac{f_1(a,b,c)\times f_1(b,a,c)}{f_2(a,b,c)\times f_2(a,c,b)} \]

考虑在 \(type\) 取值不同时，如何快速求得 \(f_1\) 与 \(f_2\)。
一共有 \(6\) 个需要推导的式子，不妨就叫它们 \(1\sim 6\) 面了（

---

type = 0

\[\begin{aligned} f_1(a,b,c) &= \prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} i\\ f_2(a,b,c) &= \prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} \gcd(i,j) \end{aligned}\]

对于 1 面，显然有：

\[\prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} i = \left(\prod_{i=1}^{a}i\right)^{b\times c} \]

预处理阶乘 + 快速幂即可，单次计算时间复杂度 \(O(\log n)\)。

---

再考虑 2 面，套路地枚举 \(\gcd\)，显然有：

\[\large \begin{aligned} &\prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} \gcd(i,j)\\ =&\left(\prod_{i=1}^{a}\prod_{j=1}^{b}\gcd(i,j)\right)^c\\ =& \left(\prod_{d=1} d^{\left(\sum\limits_{i=1}^{a}\sum\limits_{j=1}^{b}[\gcd(i,j) = d]\right)}\right)^c \end{aligned}\]

指数是个套路，可以看这里：P3455 [POI2007]ZAP-Queries。于是有：

\[\begin{aligned} &\sum\limits_{i=1}^{a}\sum\limits_{j=1}^{b}[\gcd(i,j) = d]\\ =& \sum\limits_{i=1}^{\left\lfloor\frac{a}{d}\right\rfloor}\sum\limits_{j=1}^{\left\lfloor\frac{b}{d}\right\rfloor}[\gcd(i,j) = 1]\\ =& \sum\limits_{i=1}^{\left\lfloor\frac{a}{d}\right\rfloor}\sum\limits_{j=1}^{\left\lfloor\frac{b}{d}\right\rfloor}\sum_{k\mid \gcd(i,j)}\mu (k)\\ =& \sum_{k=1}\mu(k)\left\lfloor\dfrac{a}{kd}\right\rfloor\left\lfloor\dfrac{b}{kd}\right\rfloor \end{aligned}\]

代回原式，略做处理，则原式等于：

\[\large \begin{aligned} &\left(\prod_{d=1} d^{\left(\sum\limits_{k=1}\mu(k)\left\lfloor\frac{a}{kd}\right\rfloor\left\lfloor\frac{b}{kd}\right\rfloor\right)}\right)^c\\ =& \left(\prod_{d=1} \left(d^{\sum\limits_{k=1}\mu(k)}\right)^{\left\lfloor\frac{a}{kd}\right\rfloor\left\lfloor\frac{b}{kd}\right\rfloor}\right)^c\\ =& \prod_{d=1} \left(\prod_{k=1}^{\left\lfloor\frac{n}{d}\right\rfloor}d^{\left(\mu(k)\left\lfloor\frac{a}{kd}\right\rfloor\left\lfloor\frac{b}{kd}\right\rfloor\right)}\right)^c \end{aligned}\]

像「SDOI2017」数字表格 一样，考虑枚举 \(t=kd\) 和 \(d\)，则原式等于：

\[\large \prod_{t=1}^{n}\left(\left(\prod_{d|t} d^{\mu{\left(\frac{t}{d}\right)}}\right)^{\left\lfloor\frac{a}{t}\right\rfloor\left\lfloor\frac{b}{t}\right\rfloor}\right)^c \]

设：

\[\large g_0(t) = \prod_{d|t}d^{\mu\left(\frac{t}{d}\right)} \]

线性筛预处理 \(\mu\) 后，\(g_0(t)\) 可以用埃氏筛预处理，时间复杂度 \(O(n\log n)\)。再代回原式，原式等于：

\[\large \prod_{t=1}^{a}\left(g_0(t)^{\left\lfloor\frac{a}{t}\right\rfloor\left\lfloor\frac{b}{t}\right\rfloor}\right)^c \]

预处理 \(g_0(t)\) 的前缀积和前缀积的逆元，复杂度 \(O(n\log n)\)。
数论分块 + 快速幂计算即可，单次时间复杂度 \(O(\sqrt n\log n)\)。

---

type = 1

\[\begin{aligned} f_1(a,b,c) &= \prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} i^{i\times j\times k}\\ f_2(a,b,c) &= \prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} \gcd(i,j)^{i\times j\times k} \end{aligned}\]

考虑 3 面，把 \(\prod k\) 扔到指数位置，有：

\[\large \prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} i^{i\times j\times k} = \prod_{i=1}^{a}\prod_{j=1}^{b}i^{\left(i\times j\times \sum\limits_{k = 1}^{c} k\right)} \]

再把 \(\prod j\) 也扔到指数位置，引入 \(\operatorname{sum}(n) = \sum_{i=1}^{n} i = \frac{n(n+1)}{2}\)，原式等于：

\[\left(\prod_{i=1}^{a}i^i\right)^{\operatorname{sum}(b)\times \operatorname{sum}(c)} \]

预处理 \(i^i\) 的前缀积，复杂度 \(O(n\log n)\)。
指数可以 \(O(1)\) 算出，然后快速幂，单次时间复杂度 \(O(\log n)\)。

根据费马小定理，指数需要对 \(p - 1\) 取模。注意 \(p-1\) 不是质数，计算 \(\operatorname{sum}\) 时不能用逆元，但乘不爆 LL，直接算就行。

---

再考虑 4 面，发现 \(k\) 与 \(\gcd\) 无关，则同样把 \(\prod k\) 扔到指数位置，则有：

\[\large \begin{aligned} &\prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} \gcd(i,j)^{i\times j\times k}\\ =& \left(\prod_{i=1}^a\prod_{j=1}^b\gcd(i,j)^{i\times j}\right)^{\operatorname{sum}(c)} \end{aligned}\]

套路地枚举 \(\gcd\)，原式等于：

\[\large \left(\prod_{d=1}d^{\left(\sum\limits_{i=1}^a \sum\limits_{j=1}^b i\times j[\gcd(i,j)=d]\right)}\right)^{\operatorname{sum}(c)} \]

大力化简指数，有：

\[\large \begin{aligned} &\sum\limits_{i=1}^a \sum\limits_{j=1}^b i\times j[\gcd(i,j)=d]\\ =& d^2 \sum\limits_{i=1}^{\left\lfloor\frac{a}{d}\right\rfloor} \sum\limits_{j=1}^{\left\lfloor\frac{b}{d}\right\rfloor} i\times j[\gcd(i,j)=1\\ =& d^2 \sum\limits_{i=1}^{\left\lfloor\frac{a}{d}\right\rfloor} i \sum\limits_{j=1}^{\left\lfloor\frac{b}{d}\right\rfloor} j\sum\limits_{t|\gcd(i,j)}\mu(t)\\ =& d^2 \sum\limits_{i=1}^{\left\lfloor\frac{a}{d}\right\rfloor} i \sum\limits_{j=1}^{\left\lfloor\frac{b}{d}\right\rfloor} j\sum\limits_{k|\gcd(i,j)}\mu(k)\\ =& d^2 \sum\limits_{k=1}\mu(k)\sum\limits_{i=1}^{\left\lfloor\frac{a}{d}\right\rfloor} i[k|i] \sum\limits_{j=1}^{\left\lfloor\frac{b}{d}\right\rfloor} j[k|j]\\ =& d^2 \sum\limits_{k=1}k^2\mu(k)\sum\limits_{i=1}^{\left\lfloor\frac{a}{kd}\right\rfloor} i\sum\limits_{j=1}^{\left\lfloor\frac{b}{kd}\right\rfloor} j\\ =& d^2\sum\limits_{k=1}k^2\mu(k)\operatorname{sum}{\left(\left\lfloor\frac{a}{kd}\right\rfloor\right)} \operatorname{sum}{\left(\left\lfloor\frac{b}{kd}\right\rfloor\right)}\\ \end{aligned}\]

指数化不动了，代回原式，原式等于：

\[\large \left(\prod_{d=1}d^{\left(d^2\sum\limits_{k=1}k^2\mu(k)\operatorname{sum}{\left(\left\lfloor\frac{a}{kd}\right\rfloor\right)} \operatorname{sum}{\left(\left\lfloor\frac{b}{kd}\right\rfloor\right)}\right)}\right)^{\operatorname{sum}(c)} \]

同 2 面的情况，先展开一下，再枚举 \(t=kd\) 和 \(d\)，原式等于：

\[\large \begin{aligned} &\left(\prod_{d=1}\left(\prod_{k=1}^{\left\lfloor\frac{n}{d}\right\rfloor}d^{\left(d^2 k^2\mu(k)\right)}\right)^{\left(\operatorname{sum}{\left(\left\lfloor\frac{a}{kd}\right\rfloor\right)} \operatorname{sum}{\left(\left\lfloor\frac{b}{kd}\right\rfloor\right)}\right)}\right)^{\operatorname{sum}(c)}\\ =& \prod_{t=1}\left(\left(\prod_{d|t}d^{\left(d^2\left(\frac{t}{d}\right)^2\mu\left(\frac{t}{d}\right)\right)}\right)^{\operatorname{sum}{\left(\left\lfloor\frac{a}{t}\right\rfloor\right)} \operatorname{sum}{\left(\left\lfloor\frac{b}{t}\right\rfloor\right)}}\right)^{\operatorname{sum}(c)}\\ =& \prod_{t=1}\left(\left(\prod_{d|t}d^{\left(t^2\mu\left(\frac{t}{d}\right)\right)}\right)^{\operatorname{sum}{\left(\left\lfloor\frac{a}{t}\right\rfloor\right)} \operatorname{sum}{\left(\left\lfloor\frac{b}{t}\right\rfloor\right)}}\right)^{\operatorname{sum}(c)} \end{aligned}\]

与二面相同，设：

\[\large g_1(t) = \prod_{d|t}d^{\left(t^2\mu\left(\frac{t}{d}\right)\right)} \]

\(g_1(t)\) 可以用埃氏筛套快速幂筛出，时间复杂度 \(O(n\log^2 n)\)。再代回原式，原式等于：

\[\prod_{t=1}\left(g_1(t)^{\operatorname{sum}{\left(\left\lfloor\frac{a}{t}\right\rfloor\right)} \operatorname{sum}{\left(\left\lfloor\frac{b}{t}\right\rfloor\right)}}\right)^{\operatorname{sum}(c)} \]

同样预处理 \(g_1(t)\) 的前缀积及其逆元，时间复杂度 \(O(n\log n)\)。
整除分块 + 快速幂即可，单次时间复杂度 \(O(\sqrt n\log n)\)。

注意指数的取模。

---

type = 2

\[\begin{aligned} f_1(a,b,c) &= \prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} i^{\gcd(i,j,k)}\\ f_2(a,b,c) &= \prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} \gcd(i,j)^{\gcd(i,j,k)} \end{aligned}\]

考虑 5 面，手段同上，大力反演化简一波，再调换枚举对象，则有：

\[\large \begin{aligned} &\prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} i^{\gcd(i,j,k)}\\ =&\prod_{d=1}\prod\limits_{i=1}^{a}i^{\left(d\sum\limits_{j=1}^{b}\sum\limits_{k=1}^{c}[\gcd(i,j,k)=d]\right)}\\ =& \prod_{d=1}\prod\limits_{i=1}^{a}i^{\left(d\sum\limits_{j=1}^{\left\lfloor\frac{b}{d}\right\rfloor}\sum\limits_{k=1}^{\left\lfloor\frac{c}{d}\right\rfloor}[\gcd(\frac{i}{d},j,k)=1]\right)}\\ =& \prod_{d=1}\prod\limits_{i=1}^{\left\lfloor\frac{a}{d}\right\rfloor}(id)^{\left(d\sum\limits_{j=1}^{\left\lfloor\frac{b}{d}\right\rfloor}\sum\limits_{k=1}^{\left\lfloor\frac{c}{d}\right\rfloor}[\gcd(i,j,k)=1]\right)}\\ =& \prod_{d=1}\prod\limits_{i=1}^{\left\lfloor\frac{a}{d}\right\rfloor}(id)^{\left(d\sum\limits_{j=1}^{\left\lfloor\frac{b}{d}\right\rfloor}\sum\limits_{k=1}^{\left\lfloor\frac{c}{d}\right\rfloor}\sum\limits_{x|\gcd(i,j,k)}{\mu(x)}\right)}\\ =& \prod_{d=1}\prod\limits_{i=1}^{\left\lfloor\frac{a}{d}\right\rfloor}(id)^{\left(d\sum\limits_{x=1}\mu(x)[x|i]\sum\limits_{j=1}^{\left\lfloor\frac{b}{d}\right\rfloor}[x|j]\sum\limits_{k=1}^{\left\lfloor\frac{c}{d}\right\rfloor}[x|k]\right)}\\ =& \prod_{d=1}\prod_{x=1}\prod\limits_{i=1}^{\left\lfloor\frac{a}{d}\right\rfloor}(id)^{\left(d\times \mu(x)[x|i]\sum\limits_{j=1}^{\left\lfloor\frac{b}{d}\right\rfloor}[x|j]\sum\limits_{k=1}^{\left\lfloor\frac{c}{d}\right\rfloor}[x|k]\right)}\\ =& \prod_{d=1}\prod_{x=1}\prod\limits_{i=1}^{\left\lfloor\frac{a}{xd}\right\rfloor}(ixd)^{\left(d\times \mu(x){\left\lfloor\frac{b}{xd}\right\rfloor}{\left\lfloor\frac{c}{xd}\right\rfloor}\right)}\\ =& \prod_{t = 1}\prod_{d|T}\prod_{i=1}^{\left\lfloor\frac{a}{t}\right\rfloor}(it)^{\left(d\times \mu\left(\frac{t}{d}\right){\left\lfloor\frac{b}{t}\right\rfloor}{\left\lfloor\frac{c}{t}\right\rfloor}\right)}\\ =& \prod_{t = 1}\prod_{d|T}\left(t^{\left\lfloor\frac{a}{t}\right\rfloor}\prod_{i=1}^{\left\lfloor\frac{a}{t}\right\rfloor}i\right)^{d\times \mu\left(\frac{t}{d}\right){\left\lfloor\frac{b}{t}\right\rfloor}{\left\lfloor\frac{c}{t}\right\rfloor}}\\ \end{aligned}\]

引入 \(\operatorname{fac}(n) = \prod_{i=1}^{n} i\)，再根据枚举对象调整一下指数，原式等于：

\[\large \begin{aligned} &\prod_{t = 1}\prod_{d|t}\left(t^{\left\lfloor\frac{a}{t}\right\rfloor}\times \operatorname{fac}\left(\left\lfloor\frac{a}{t}\right\rfloor\right)\right)^{\left(d\times \mu\left(\frac{t}{d}\right){\left\lfloor\frac{b}{t}\right\rfloor}{\left\lfloor\frac{c}{t}\right\rfloor}\right)}\\ =& \prod_{t = 1}\left(\prod_{d|t}\left(t^{\left\lfloor\frac{a}{t}\right\rfloor}\times \operatorname{fac}\left(\left\lfloor\frac{a}{t}\right\rfloor\right)\right)^{d\times \mu\left(\frac{t}{d}\right)}\right)^{{\left\lfloor\frac{b}{t}\right\rfloor}{\left\lfloor\frac{c}{t}\right\rfloor}}\\ =& \prod_{t = 1}\left(\left(t^{\left\lfloor\frac{a}{t}\right\rfloor}\times \operatorname{fac}\left(\left\lfloor\frac{a}{t}\right\rfloor\right)\right)^{\sum\limits_{d|t}d\times \mu\left(\frac{t}{d}\right)}\right)^{{\left\lfloor\frac{b}{t}\right\rfloor}{\left\lfloor\frac{c}{t}\right\rfloor}} \end{aligned}\]

指数中出现了一个经典的狄利克雷卷积的形式，对其进行反演。
将 \((\operatorname{Id}\ast \mu) (n)= \varphi (n)\) 代入原式，原式等于：

\[\large \begin{aligned} &\prod_{t = 1}\left(t^{\left\lfloor\frac{a}{t}\right\rfloor}\times \operatorname{fac}\left(\left\lfloor\frac{a}{t}\right\rfloor\right)\right)^{\varphi(t){\left\lfloor\frac{b}{t}\right\rfloor}{\left\lfloor\frac{c}{t}\right\rfloor}}\\ =& \prod_{t = 1}\left(t^{\varphi(t)\left\lfloor\frac{a}{t}\right\rfloor{\left\lfloor\frac{b}{t}\right\rfloor}{\left\lfloor\frac{c}{t}\right\rfloor}}\times \operatorname{fac}\left(\left\lfloor\frac{a}{t}\right\rfloor\right)^{\varphi(t){\left\lfloor\frac{b}{t}\right\rfloor}{\left\lfloor\frac{c}{t}\right\rfloor}}\right) \end{aligned}\]

预处理 \(t^{\varphi(t)}\) 的前缀积及逆元，阶乘的前缀积及阶乘逆元，\(\pmod {p-1}\) 下的 \(\varphi(t)\) 的前缀和（指数
），时间复杂度 \(O(n\log n)\)。
同样整除分块 + 快速幂即可，单次时间复杂度 \(O(\sqrt n\log n)\)。

---

然后是最掉 sans 的 6 面。有 \(\gcd(i,j,k) = \gcd(\gcd(i,j), k)\)，考虑先枚举 \(\gcd(i,j)\)，然后套路化式子，则有：

\[\large \begin{aligned} &\prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} \gcd(i,j)^{\gcd(i,j,k)}\\ =& \prod_{d=1}\prod_{i=1}^{a}\prod_{j=1}^{b}\prod_{k=1}^{c} [\gcd(i,j)=d] d^{\gcd(d,k)}\\ =& \prod_{d=1} \left(d^{\left(\sum\limits_{i=1}^{a}\sum\limits_{j=1}^{b} [\gcd(i,j)=d]\right)}\right)^{\sum\limits_{k=1}^{c}\gcd(d,k)} \end{aligned}\]

先考虑最外面的指数，这也是个套路，可以参考 一个例子。用 \(\operatorname{Id} = \varphi \ast 1\) 反演，显然有：

\[\large \begin{aligned} &\sum\limits_{k=1}^{c}\gcd(d,k)\\ =& \sum\limits_{k=1}^{c}\sum_{x|\gcd(d,k)}\varphi(x)\\ =& \sum_{x=1}\varphi(x)[x|d]\sum_{k=1}^{c}[x|k]\\ =& \sum_{x|d}\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor \end{aligned}\]

再考虑里面的指数，发现这式子在 2 面已经推了一遍了，于是直接拿过来用，有：

\[\sum\limits_{i=1}^{a}\sum\limits_{j=1}^{b}[\gcd(i,j) = d]=\sum_{y=1}\mu(y)\left\lfloor\dfrac{a}{yd}\right\rfloor\left\lfloor\dfrac{b}{yd}\right\rfloor \]

将化简后的两个指数代入原式，原式等于：

\[\large \begin{aligned} &\prod_{d=1} \left(d^{\left(\sum\limits_{y=1}\mu(y)\left\lfloor\frac{a}{yd}\right\rfloor\left\lfloor\frac{b}{yd}\right\rfloor\right)}\right)^{\sum\limits_{x|d}\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor}\\ =& \prod_{d=1} \left(\prod\limits_{y=1}d^{\left(\mu(y)\left\lfloor\frac{a}{yd}\right\rfloor\left\lfloor\frac{b}{yd}\right\rfloor\right)}\right)^{\sum\limits_{x|d}\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor} \end{aligned}\]

与 2、4 面同样套路地，考虑枚举 \(t=yd\) 和 \(d\)，再略作调整，原式等于：

\[\large \begin{aligned} &\prod_{d=1} \left(\prod\limits_{y=1}d^{\left(\mu(y)\left\lfloor\frac{a}{yd}\right\rfloor\left\lfloor\frac{b}{yd}\right\rfloor\right)}\right)^{\sum\limits_{x|d}\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor}\\ =& \prod_{t=1}\prod_{d|t} d^{\left(\mu\left(\frac{t}{d}\right)\left\lfloor\frac{a}{t}\right\rfloor\left\lfloor\frac{b}{t}\right\rfloor\sum\limits_{x|d}\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor\right)}\\ =& \prod_{t=1}\left(\prod_{d|t} d^{\left(\mu\left(\frac{t}{d}\right)\sum\limits_{x|d}\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor\right)}\right)^{\left\lfloor\frac{a}{t}\right\rfloor\left\lfloor\frac{b}{t}\right\rfloor}\\ =& \prod_{t=1}\left(\prod_{d|t} \prod_{x|d}d^{\left(\mu\left(\frac{t}{d}\right)\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor\right)}\right)^{\left\lfloor\frac{a}{t}\right\rfloor\left\lfloor\frac{b}{t}\right\rfloor} \end{aligned}\]

发现要同时枚举 \(d\) 和 \(x\)，化不动了。
从题解里学到一个比较神的技巧，考虑把 \(d\) 拆成 \(x\) 和 \(\frac{d}{x}\) 分别计算贡献再相乘，即分别计算下两式：

\[\large \begin{aligned} &\prod_{t=1}\left(\prod_{d|t} \prod_{x|d}x^{\left(\mu\left(\frac{t}{d}\right)\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor\right)}\right)^{\left\lfloor\frac{a}{t}\right\rfloor\left\lfloor\frac{b}{t}\right\rfloor}\\ &\prod_{t=1}\left(\prod_{d|t} \prod_{x|d}{\left(\frac{d}{x}\right)}^{\left(\mu\left(\frac{t}{d}\right)\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor\right)}\right)^{\left\lfloor\frac{a}{t}\right\rfloor\left\lfloor\frac{b}{t}\right\rfloor} \end{aligned}\]

---

先考虑 \(x\) 的情况，首先把枚举 \(x\) 调整到最外层。设 \(\operatorname{lim}=\max(a,b,c)\)，则原式等于：

\[\large \begin{aligned} &\prod_{x=1} \prod_{t=1}^{\operatorname{lim}}[x|t]\left(\prod_{d|t} [x|d]{x}^{\left(\mu\left(\frac{t}{d}\right)\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor\right)}\right)^{\left\lfloor\frac{a}{t}\right\rfloor\left\lfloor\frac{b}{t}\right\rfloor}\\ =& \prod_{x=1} \prod_{t=1}^{\left\lfloor\frac{\operatorname{lim}}{x}\right\rfloor}\left(\prod_{d|t} {x}^{\left(\mu\left(\frac{tx}{dx}\right)\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor\right)}\right)^{\left\lfloor\frac{a}{tx}\right\rfloor\left\lfloor\frac{b}{tx}\right\rfloor}\\ =& \prod_{x=1} \prod_{t=1}^{\left\lfloor\frac{\operatorname{lim}}{x}\right\rfloor}\prod_{d|t} {x}^{\left(\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor\left\lfloor\frac{a}{tx}\right\rfloor\left\lfloor\frac{b}{tx}\right\rfloor\mu\left(\frac{t}{d}\right)\right)} \end{aligned}\]

把 \(\prod {t}\) 挪到指数位置，原式等于：

\[\large \begin{aligned} &\prod_{x=1} {x}^{\left(\sum\limits_{t=1}^{\left\lfloor\frac{\operatorname{lim}}{x}\right\rfloor}\sum\limits_{d|t}\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor\left\lfloor\frac{a}{tx}\right\rfloor\left\lfloor\frac{b}{tx}\right\rfloor\mu\left(\frac{t}{d}\right)\right)}\\ =& \prod_{x=1} {x}^{\left(\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor\sum\limits_{t=1}^{\left\lfloor\frac{\operatorname{lim}}{x}\right\rfloor}\left\lfloor\frac{a}{tx}\right\rfloor\left\lfloor\frac{b}{tx}\right\rfloor\sum\limits_{d|t}\mu\left(\frac{t}{d}\right)\right)} \end{aligned}\]

指数中又出现了一个经典的狄利克雷卷积的形式，对其进行反演。
将 \((\mu \ast 1) (n)= \epsilon (n)=[n=1]\) 代入原式，原式等于：

\[\large \begin{aligned} &\prod_{x=1} {x}^{\left(\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor\sum\limits_{t=1}^{\left\lfloor\frac{\operatorname{lim}}{x}\right\rfloor}\left\lfloor\frac{a}{tx}\right\rfloor\left\lfloor\frac{b}{tx}\right\rfloor[t=1]\right)}\\ =& \prod_{x=1} {x}^{\left(\varphi(x)\left\lfloor\frac{a}{x}\right\rfloor\left\lfloor\frac{b}{x}\right\rfloor\left\lfloor\frac{c}{x}\right\rfloor\right)} \end{aligned}\]

得到了一个非常优美的式子，而且发现这个式子是 5 面最终答案的一部分。同 5 面的做法，直接整除分块即可。

---

再考虑 \(\frac{d}{x}\) 的情况，同上先把枚举 \(x\) 放到最外层，并调整一下指数，则原式等于：

\[\large \begin{aligned} &\prod_{x=1} \prod_{t=1}^{\operatorname{lim}}[x|t]\left(\prod_{d|t} [x|d]{\left(\frac{d}{x}\right)}^{\left(\mu\left(\frac{t}{d}\right)\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor\right)}\right)^{\left\lfloor\frac{a}{t}\right\rfloor\left\lfloor\frac{b}{t}\right\rfloor}\\ =& \prod_{x=1} \prod_{t=1}^{\left\lfloor\frac{\operatorname{lim}}{x}\right\rfloor}\left(\prod_{d|tx} [x|d]{\left(\frac{d}{x}\right)}^{\left(\mu\left(\frac{tx}{d}\right)\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor\right)}\right)^{\left\lfloor\frac{a}{tx}\right\rfloor\left\lfloor\frac{b}{tx}\right\rfloor}\\ =& \prod_{x=1} \left(\prod_{t=1}^{\left\lfloor\frac{\operatorname{lim}}{x}\right\rfloor}\left(\prod_{d|tx} [x|d]{\left(\frac{d}{x}\right)}^{\mu\left(\frac{tx}{d}\right)}\right)^{\left\lfloor\frac{a}{tx}\right\rfloor\left\lfloor\frac{b}{tx}\right\rfloor}\right)^{\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor} \end{aligned}\]

考虑枚举 \(dx\)，替换原来的 \(d\)，注意一下这里的倍数关系。原式等于：

\[\large \prod_{x=1} \left(\prod_{t=1}^{\left\lfloor\frac{\operatorname{lim}}{x}\right\rfloor}\left(\prod_{d|t}d^{\mu\left(\frac{t}{d}\right)}\right)^{\left\lfloor\frac{a}{tx}\right\rfloor\left\lfloor\frac{b}{tx}\right\rfloor}\right)^{\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor} \]

发现最内层的式子 \(\prod_{d|t}d^{\mu\left(\frac{t}{d}\right)}\)，即为二面处理过的 \(g_0(t)\)。直接代入，原式等于：

\[\large \prod_{x=1} \left(\prod_{t=1}^{\left\lfloor\frac{\operatorname{lim}}{x}\right\rfloor}g_0(t)^{\left\lfloor\frac{a}{tx}\right\rfloor\left\lfloor\frac{b}{tx}\right\rfloor}\right)^{\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor} \]

一个小结论，证明可以看 这里：

\[\forall a,b,c\in \mathbb{Z},\left\lfloor\dfrac{a}{bc}\right\rfloor = \left\lfloor{\dfrac{\left\lfloor\dfrac{a}{b}\right\rfloor}{c}}\right\rfloor \]

则原式等于：

\[\large \prod_{x=1} \left(\prod_{t=1}^{\left\lfloor\frac{\operatorname{lim}}{x}\right\rfloor}g_0(t)^{\left\lfloor\frac{\left\lfloor\frac{a}{x}\right\rfloor}{t}\right\rfloor\left\lfloor\frac{\left\lfloor\frac{b}{x}\right\rfloor}{t}\right\rfloor}\right)^{\varphi(x)\left\lfloor\frac{c}{x}\right\rfloor} \]

于是可以先对外层整除分块，再对内层整除分块。

然后就做完了，哈哈哈。

实现

一些实现上的小技巧：

• 逆元能预处理就预处理。
• 注意对指数取模，模数为 \(p-1\)。

//知识点：莫比乌斯反演 
/*
By:Luckyblock
用了比较清晰易懂的变量名，阅读体验应该会比较好。  
vsc 的自动补全真是太好啦！
*/
#include <algorithm>
#include <cctype>
#include <cstdio>
#include <cstring>
using std::min;
using std::max;
#define LL long long
const int Lim = 1e5;
const int kN = 1e5 + 10;
//=============================================================
LL A, B, C, mod, ans;
int T, p_num, p[kN];
bool vis[kN];
LL mu[kN], phi[kN], fac[kN], g[2][kN];
LL sumphi[kN], prodt_phi[kN], prodi_i[kN], prodg[2][kN];
LL inv[kN], inv_fac[kN], inv_prodt_phi[kN], inv_prodg[2][kN];
//=============================================================
inline int read() {
  int f = 1, w = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) {
    w = (w << 3) + (w << 1) + (ch ^ '0');
  }
  return f * w;
}
void Chkmax(int &fir_, int sec_) {
  if (sec_ > fir_) fir_ = sec_;
}
void Chkmin(int &fir_, int sec_) {
  if (sec_ < fir_) fir_ = sec_;
}
LL QPow(LL x_, LL y_) {
  x_ %= mod;
  y_ %= mod - 1;
  LL ret = 1;
  for (; y_; y_ >>= 1ll) {
    if (y_ & 1) ret = ret * x_ % mod;
    x_ = x_ * x_ % mod;
  }
  return ret;
}
LL Inv(LL x_) {
  return QPow(x_, mod - 2);
}
LL Sum(LL n_) {
  return ((n_ * (n_ + 1ll)) / 2ll) % (mod - 1);
}
void Euler() {
  vis[1] = true, mu[1] = phi[1] = 1; //初值
  for (int i = 2; i <= Lim; ++ i) {
    if (! vis[i]) {
      p[++ p_num] = i;
      mu[i] = -1;
      phi[i] = i - 1;
    }
    for (int j = 1; j <= p_num && i * p[j] <= Lim; ++ j) {
      vis[i * p[j]] = true;
      if (i % p[j] == 0) {
        mu[i * p[j]] = 0;
        phi[i * p[j]] = phi[i] * p[j];
        break;
      }
      mu[i * p[j]] = -mu[i];
      phi[i * p[j]] = phi[i] * (p[j] - 1);
    }
  }
}
void Prepare() {
  Euler();
  inv[1] = fac[0] = prodt_phi[0] = prodi_i[0] = 1;
  for (int i = 1; i <= Lim; ++ i) {
    g[0][i] = g[1][i] = 1;
    fac[i] = 1ll * fac[i - 1] * i % mod;
    sumphi[i] = (sumphi[i - 1] + phi[i]) % (mod - 1);
    prodi_i[i] = prodi_i[i - 1] * QPow(i, i) % mod;
    if (i > 1) inv[i] = (mod - mod / i) * inv[mod % i] % mod;

    prodt_phi[i] = prodt_phi[i - 1] * QPow(i, phi[i]) % mod;
    inv_prodt_phi[i] = Inv(prodt_phi[i]);
  }

  for (int d = 1; d <= Lim; ++ d) {
    for (int j = 1; d * j <= Lim; ++ j) {
      int t = d * j;
      if (mu[j] == 1) {
        g[0][t] = g[0][t] * d % mod;
        g[1][t] = g[1][t] * QPow(1ll * d, 1ll * t * t) % mod;
      } else if (mu[j] == -1) {
        g[0][t] = g[0][t] * inv[d] % mod;
        g[1][t] = g[1][t] * Inv(QPow(1ll * d, 1ll * t * t)) % mod;
      }
    }
  }
  inv_prodg[0][0] = prodg[0][0] = 1;
  inv_prodg[1][0] = prodg[1][0] = 1;
  inv_prodt_phi[0] = 1;
  for (int i = 1; i <= Lim; ++ i) {
    for (int j = 0; j <= 1; ++ j) {
      prodg[j][i] = prodg[j][i - 1] * g[j][i] % mod;
      inv_prodg[j][i] = Inv(prodg[j][i]);
    }
  }
}
LL f1(LL a_, LL b_, LL c_, int type) {
  if (! type) return QPow(fac[a_], b_ * c_);
  if (type == 1) return QPow(prodi_i[a_], Sum(b_) * Sum(c_));
  LL ret = 1, lim = min(min(a_, b_), c_);
  for (LL l = 1, r = 1; l <= lim; l = r + 1) {
    r = min(min(a_ / (a_ / l), b_ / (b_ / l)), c_ / (c_ / l));
    ret = ret * QPow(prodt_phi[r] * inv_prodt_phi[l - 1], (a_ / l) * (b_ / l) % (mod - 1) * (c_ / l)) % mod;
    ret = ret * QPow(fac[a_ / l], (sumphi[r] - sumphi[l - 1] + mod - 1) % (mod - 1) * (b_ / l) % (mod - 1) * (c_ / l)) % mod;
  }
  return ret;
}
LL f2_2(LL a_, LL b_) { 
  LL ret = 1;
  for (LL l = 1, r = 1; l <= min(a_, b_); l = r + 1) {
    r = min(a_ / (a_ / l), b_ / (b_ / l));
    ret = ret * QPow(prodg[0][r] * inv_prodg[0][l - 1], 1ll * (a_ / l) * (b_ / l)) % mod;
  }
  return ret;
}
LL f2(LL a_, LL b_, LL c_, int type) {
  LL ret = 1;
  if (! type) {
    for (LL l = 1, r = 1; l <= min(a_, b_); l = r + 1) {
      r = min(a_ / (a_ / l), b_ / (b_ / l));
      LL val = QPow(prodg[0][r] * inv_prodg[0][l - 1], 1ll * (a_ / l) * (b_ / l));
      ret = (ret * QPow(val, c_)) % mod;
    }
  } else if (type == 1) {
    for (LL l = 1, r = 1; l <= min(a_, b_); l = r + 1) {
      r = min(a_ / (a_ / l), b_ / (b_ / l));
      LL val = QPow(prodg[1][r] * inv_prodg[1][l - 1], Sum(a_ / l) * Sum(b_ / l));
      ret = (ret * QPow(val, Sum(c_))) % mod;
    }
  } else {
    LL lim = min(min(a_, b_), c_);
    for (LL l = 1, r = 1; l <= lim; l = r + 1) {
      r = min(min(a_ / (a_ / l), b_ / (b_ / l)), c_ / (c_ / l));
      ret = ret * QPow(f2_2(a_ / l, b_ / l), (sumphi[r] - sumphi[l - 1] + mod - 1) % (mod - 1) * (c_ / l)) % mod;
      ret = ret * QPow(prodt_phi[r] * inv_prodt_phi[l - 1], (a_ / l) * (b_ / l) % (mod - 1) * (c_ / l)) % mod;
    }
  }
  return ret;
}
//=============================================================
int main() {
  T = read(), mod = read();
  Prepare();
  while (T -- ) {
    A = read(), B = read(), C = read();
    for (int i = 0; i <= 2; ++ i) {
      ans = f1(A, B, C, i) * f1(B, A, C, i) % mod;
      ans = ans * Inv(f2(A, B, C, i)) % mod * Inv(f2(A, C, B, i)) % mod;
      printf("%lld ", ans);  
    }
    printf("\n");
  }
  return 0;
}