莫比乌斯反演一些式子的推导(附题目)

一、一些符号和等式

$1.a|b$意为 $a$整除 $b$
$2.[a=b]=\begin{cases} 1,a=b\\0,a ≠b \end{cases}$
$3.\lfloor a \rfloor$意为对 $a$向下取整
$4.\lfloor \frac{\lfloor n/d \rfloor}{k} \rfloor = \lfloor \frac{n}{dk} \rfloor$

二、狄利克雷卷积

关于狄利克雷卷积，请戳这里
然后莫比乌斯反演中需要知道的是
$\varphi$为欧拉函数， $\mu$为莫比乌斯函数， $id(n)=n$， $\varepsilon(n)=[n=1]$， $I(n)=1$
① $\varphi*I=id$
② $\mu*I=\varepsilon$
在①等式两边同时 $*\mu$，我们可以得到 $\varphi*I*\mu=\mu*id$
再根据②，我们有 $\varphi*\varepsilon=\mu*id$
对于任意数论函数都有 $f=f*\varepsilon$，故最终我们有：
③ $\mu*id=\varphi$
这里面③是非常重要的，它揭示了欧拉函数和莫比乌斯函数的关系，在有些式子的化简当中起到至关重要的作用

三、莫比乌斯反演

对于莫比乌斯反演的题目，我们只要记住上面的①式就行，写开之后就是这样
$\displaystyle\sum_{d|n}\mu(d)=[n=1]$
那么 $[gcd(i,j)=1]$就有
$\displaystyle\sum_{d|gcd(i,j)}\mu(d)=[gcd(i,j)=1]$
亦可以写作
$\displaystyle\sum_{d|i,d|j}\mu(d)=[gcd(i,j)=1]$
再做一下拓展求 $[gcd(i,j)=k]$
因 $[gcd(i,j)=k]\Leftrightarrow[gcd(\frac{i}{k},\frac{j}{k})=1]$，故有
$\displaystyle\sum_{d|\frac{i}{k}，d|\frac{j}{k}}\mu(d)=[gcd(i,j)=k]$

四、经典式子的推导

1. $\displaystyle\sum_{i=1}^{n}\displaystyle\sum_{j=1}^{n}[gcd(i,j)=1]$

推导过程

$\begin{aligned} \displaystyle\sum_{i=1}^{n}\displaystyle\sum_{j=1}^{n}[gcd(i,j)=1] & = \displaystyle\sum_{i=1}^{n} \displaystyle\sum_{j=1}^{n} \displaystyle\sum_{d|i, d|j} \mu(d) \\ & = \displaystyle\sum_{d=1}^{n} \mu(d) \displaystyle\sum_{d|i} \displaystyle\sum_{d|j} 1\\ & = \displaystyle\sum_{d=1}^{n}\mu(d)\displaystyle\sum_{i=1}^{\lfloor n/d \rfloor}\displaystyle\sum_{j=1}^{\lfloor n/d \rfloor}1\\ & = \displaystyle\sum_{d=1}^{n}\mu(d) \lfloor \frac{n}{d} \rfloor ^ 2 \end{aligned}$

其中的第二步，先枚举 $d$
再看下一步，因为 $d|i,d|j$，因此 $i$和 $j$一定是 $d$的整数倍
于是我们枚举 $i=1, 2, ..., \lfloor n/d \rfloor$，那么 $i*d$就是原来的 $i$值，这样就能保证 $d|i$
代码实现比较简单，求一个 $\mu(i)$的前缀和，加一个数论分块就行

2. $\displaystyle\sum_{i=1}^{n}\displaystyle\sum_{j=1}^{n}[gcd(i,j)=k]$

推导过程

$\begin{aligned} \displaystyle\sum_{i=1}^{n}\displaystyle\sum_{j=1}^{n}[gcd(i,j)=k] & = \displaystyle\sum_{i=1}^{n}\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{d|\frac{i}{k},d|\frac{j}{k}}\mu(d) \\ & = \displaystyle\sum_{i=1}^{\lfloor n/k \rfloor}\displaystyle\sum_{j=1}^{\lfloor n/k \rfloor}\displaystyle\sum_{d|i,d|j}\mu(d) \\ & = \displaystyle\sum_{d=1}^{n}\mu(d)\displaystyle\sum_{i=1}^{\lfloor n/dk \rfloor}\displaystyle\sum_{j=1}^{\lfloor n/dk \rfloor}1\\ & = \displaystyle\sum_{d=1}^{n}\mu(d) \lfloor \frac{n}{dk} \rfloor ^ 2 \end{aligned}$
其中第二步我们将 $i$和 $j$同时缩小了 $k$倍，于是有 $\displaystyle\sum_{i=1}^{\lfloor n/k \rfloor}\displaystyle\sum_{j=1}^{\lfloor n/k \rfloor}\displaystyle\sum_{d|i,d|j}\mu(d)$
其余部分和上一条一样的

练习题

3. $\displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n} gcd(i, j)$

推导过程

$\begin{aligned} \displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n} gcd(i, j) & = \displaystyle\sum_{d = 1}^{n} \displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n}d[gcd(i, j)= d] \\ & = \displaystyle\sum_{d = 1}^{n} d \displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n}[gcd(i, j) = d] \\ & = \displaystyle\sum_{d = 1}^{n} d \displaystyle\sum_{i = 1}^{ \lfloor n / d \rfloor } \displaystyle\sum_{j = 1}^{ \lfloor n / d \rfloor } [gcd(i, j) = 1] \\ & = \displaystyle\sum_{d = 1} d \displaystyle\sum_{i = 1}^{ \lfloor n / d \rfloor } \displaystyle\sum_{j = 1}^{ \lfloor n / d \rfloor } \displaystyle\sum_{t | i, t|j} \mu(t) \\ & = \displaystyle\sum_{d = 1} d \displaystyle\sum_{t = 1}^{ \lfloor n / d \rfloor } \mu(t) \lfloor \frac{n}{dt} \rfloor ^ 2 \end{aligned}$
到这里，我们已经能够在 $O(n)$的时间内计算，但是还不够，我们仍可以继续优化下去
令 $T=td$，要使 $\lfloor \frac{n}{dt} \rfloor \gt 0$，于是我们枚举 $T = td = 1, 2, ..., n$
$\displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n} gcd(i, j) = \displaystyle\sum_{T = 1}^{n} \lfloor \frac{n}{T} \rfloor ^ 2 \displaystyle\sum_{t | T} \mu(d)(\frac{T}{d})$
注意， $\displaystyle\sum_{t | T} \mu(d)(\frac{T}{d}) = (\mu*id)(T)=\varphi(T)$
所以最终我们有
$\displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n} gcd(i, j) = \displaystyle\sum_{T = 1}^{n} \lfloor \frac{n}{T} \rfloor ^ 2 \varphi(T)$
这样就能够 $O(\sqrt n)$计算出我们的答案
记住这里的优化方式，以后会经常碰到，当你发现优化不下去的时候，考虑代换

练习题

4. $\displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n} lcm(i, j)$

推导过程

首先 $lcm(i, j) = \frac{ij}{gcd(i,j)}$
$\begin{aligned} \displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n} lcm(i, j) & = \displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n} \frac{ij}{gcd(i, j)}\\ & = \displaystyle\sum_{d = 1}^{n} \displaystyle\sum_{i = 1}^{\lfloor n/d \rfloor} \displaystyle\sum_{j = 1}^{\lfloor n/d \rfloor} \frac{idjd}{d}[gcd(i, j) = 1] \\ & = \displaystyle\sum_{d = 1}^{n} d \displaystyle\sum_{i = 1}^{\lfloor n/d \rfloor} \displaystyle\sum_{j = 1}^{\lfloor n/d \rfloor} ij \displaystyle\sum_{t | i,t | j} \mu(t) \\ & = \displaystyle\sum_{d = 1}^{n} d \displaystyle\sum_{t = 1}^{\lfloor n/d \rfloor} \mu(t) \displaystyle\sum_{i = 1} ^ {\lfloor n / dt \rfloor} it \displaystyle\sum_{j = 1} ^ {\lfloor n / dt \rfloor} jt \\ & = \displaystyle\sum_{d = 1} ^ {n} d \displaystyle\sum_{t = 1} ^ {\lfloor n / d \rfloor} \mu(t)t ^ 2 \displaystyle\sum_{i = 1} ^ {\lfloor n / dt \rfloor} i \displaystyle\sum_{j = 1} ^ {\lfloor n / dt \rfloor} j \\ \end{aligned}$

我们记 $sum_1(n) = 1 + 2 + ... + n = n(n+1)/2$
其中 $\displaystyle\sum_{i = 1} ^ {\lfloor n / dt \rfloor} i = 1 + 2 + ... + \lfloor n / dt \rfloor = sum_1(\lfloor n / dt \rfloor)$
$\displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n} lcm(i, j) = \displaystyle\sum_{d = 1} ^ {n} d \displaystyle\sum_{t = 1} ^ {\lfloor n / d \rfloor} \mu(t)t ^ 2 sum_1(\lfloor n / dt \rfloor) ^ 2 \\$
记 $f(n)=\displaystyle\sum_{t = 1} ^ {n} \mu(t)t ^ 2 sum_1(\lfloor n / t \rfloor) ^ 2$，那么 $f(n)$可以 $O(\sqrt n)$求，于是
$\displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n} lcm(i, j) = \displaystyle\sum_{d = 1} ^ n df(\lfloor n / d \rfloor)$
最后总体就是 $O(\sqrt n)O(\sqrt n)=O(n)$

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练习题

5. $\displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n} ijgcd(i, j)$

推导过程

$\begin{aligned} \displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n} ijgcd(i, j) & = \displaystyle\sum_{d = 1} ^ {n} d \displaystyle\sum_{i = 1} ^ {n} \displaystyle\sum_{j = 1} ^ {n} ij[gcd(i, j)=d] \\ & = \displaystyle\sum_{d = 1} ^ {n} d \displaystyle\sum_{i = 1} ^ {\lfloor n / d \rfloor} \displaystyle\sum_{j = 1} ^ {\lfloor n / d \rfloor} idjd \displaystyle\sum_{t|i,t|j} \mu(t) \\ & = \displaystyle\sum_{d = 1} ^ {n} d ^ 3 \displaystyle\sum_{t = 1} ^ {\lfloor n / d \rfloor} \mu(t) \displaystyle\sum_{i = 1} ^ {\lfloor n / dt \rfloor} it \displaystyle\sum_{j = 1} ^ {\lfloor n / dt \rfloor} jt \\ & = \displaystyle\sum_{d = 1} ^ {n} d ^ 3 \displaystyle\sum_{t = 1} ^ {\lfloor n / d \rfloor} \mu(t)t ^ 2 \displaystyle\sum_{i = 1} ^ {\lfloor n / dt \rfloor} i \displaystyle\sum_{j = 1} ^ {\lfloor n / dt \rfloor} j \\ & = \displaystyle\sum_{d = 1} ^ {n} d ^ 3 \displaystyle\sum_{t = 1} ^ {\lfloor n / d \rfloor} \mu(t)t ^ 2 sum_1(\lfloor n / dt \rfloor) ^ 2 \\ \end{aligned}$
到这里，我们已经能够 $O(n)$求得答案，但是别慌，我们仍然可以继续优化
又是熟悉的代换，令 $T = dt$，于是
$\begin{aligned} \displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n} ijgcd(i, j) & = \displaystyle\sum_{T = 1} ^ {n} sum_1(\lfloor n / T \rfloor) ^ 2 \displaystyle\sum_{t | T} \mu(t) t ^ 2 (\frac{T}{t}) ^ 3 \\ & = \displaystyle\sum_{T = 1} ^ {n} sum_1(\lfloor n / T \rfloor) ^ 2 T ^ 2 \displaystyle\sum_{t | T} \mu(t)(\frac{T}{t}) \end{aligned}$
又是熟悉的 $\mu*id=\varphi$
于是最终
$\displaystyle\sum_{i = 1}^{n} \displaystyle\sum_{j = 1}^{n} ijgcd(i, j) = \displaystyle\sum_{T = 1} ^ {n} sum_1(\lfloor n / T \rfloor) ^ 2 T ^ 2 \varphi(T)$
复杂度 $O(\sqrt n)$

练习题

这道练习题，因为数据比较大，需要用到杜教筛

五、拓展提高

学完了上面的内容之后，相信你应该已经基本掌握了比较经典的莫比乌斯反演
写了一些题，你会发现，莫比乌斯反演难，一是难在推导，二是难在一些预处理，比如要你预先处理好一些不常见的积性函数
下面的题目就按照难度逐个击破

1. $\displaystyle\sum_{i = 1} ^ N \displaystyle\sum_{j = 1} ^ N lcm(a_i, a_j)$

原题链接

给你 $N$个数 $a_1, a_2, ..., a_N$，其中 $1\leq a_i \leq 50000$，求 $\displaystyle\sum_{i = 1} ^ N \displaystyle\sum_{j = 1} ^ N lcm(a_i,a_j)$

推导过程

是不是看着很熟悉呢？经典推导中第四条
先对式子做基本的变形
$\displaystyle\sum_{i = 1} ^ N \displaystyle\sum_{j = 1} ^ N lcm(a_i, a_j) = \displaystyle\sum_{i = 1} ^ N \displaystyle\sum_{j = 1} ^ N \frac{a_i a_j}{gcd(a_i, a_j)}$
分析一下，光有 $a_i$是没用的，我们要将式子中 $a_i$转化为数字才行
所以我们不妨直接枚举 $i, j = 1, 2, 3, ..., n = 50000$，这样就有
$\displaystyle\sum_{i = 1} ^ N \displaystyle\sum_{j = 1} ^ N lcm(a_i, a_j) = \displaystyle\sum_{i = 1} ^ {n} \displaystyle\sum_{j = 1} ^ {n} cnt_i cnt_j \frac{ij}{gcd(i, j)}$
其中 $cnt_i,cnt_j$为数 $i,j$出现次数，于是我们继续
$\begin{aligned} \displaystyle\sum_{i = 1} ^ N \displaystyle\sum_{j = 1} ^ N lcm(a_i, a_j) & = \displaystyle\sum_{d = 1} ^ {n} \displaystyle\sum_{i = 1} ^ {n} \displaystyle\sum_{j = 1} ^ {n} cnt_i cnt_j \frac{ij}{d} [gcd(i, j)=d] \\ & = \displaystyle\sum_{d = 1} ^ {n} \displaystyle\sum_{i = 1} ^ {\lfloor n / d \rfloor} \displaystyle\sum_{j = 1} ^ {\lfloor n / d \rfloor} cnt_{id} cnt_{jd} \frac{i j d ^ 2}{d}[gcd(i, j) = 1] \\ & = \displaystyle\sum_{d = 1} ^ {n} d \displaystyle\sum_{i = 1} ^ {\lfloor n / d \rfloor} \displaystyle\sum_{j = 1} ^ {\lfloor n / d \rfloor} ijcnt_{id} cnt_{jd} \displaystyle\sum_{t | i, t | j} \mu(t) \\ & = \displaystyle\sum_{d = 1} ^ {n} d \displaystyle\sum_{t = 1} ^ {\lfloor n / d \rfloor} \mu(t) t ^ 2 \displaystyle\sum_{i = 1} ^ {\lfloor n / d t \rfloor} i cnt_{i d t} \displaystyle\sum_{j = 1} ^ {\lfloor n / d t \rfloor} j cnt_{j d t} \\ & = \displaystyle\sum_{d = 1} ^ {n} d \displaystyle\sum_{t = 1} ^ {\lfloor n / d \rfloor} \mu(t) t ^ 2 \left( \displaystyle\sum_{i = 1} ^ {\lfloor n / d t \rfloor} i cnt_{i d t} \right) ^ 2 \end{aligned}$
到这里我们令 $T=dt$
$\begin{aligned} \displaystyle\sum_{i = 1} ^ N \displaystyle\sum_{j = 1} ^ N lcm(a_i, a_j) & = \displaystyle\sum_{T = 1} ^ {n} \left( \displaystyle\sum_{i = 1} ^ {\lfloor n / T \rfloor} i cnt_{i T} \right) ^ 2 \displaystyle\sum_{t | T} \mu(t) t ^ 2 (\frac{T}{t}) \\ & = \displaystyle\sum_{T = 1} ^ {n} T \left( \displaystyle\sum_{i = 1} ^ {\lfloor n / T \rfloor} i cnt_{i T} \right) ^ 2 \displaystyle\sum_{t | T} \mu(t) t \\ \end{aligned}$
先看后面的 $\displaystyle\sum_{t | T} \mu(t)t$，我们记做 $f(T)$，我们可以 $O(nlogn)$预处理：

for (int t = 1; t <= n; t++)
        for (int T = t; T <= n; T += t)
            f[T] += mu[t] * t;

关于复杂度，这个是一个调和级数，即 $\frac{n}{1} + \frac{n}{2} + \frac{n}{3} + ... + 1$，是 $nlogn$级别的
再看 $\displaystyle\sum_{i = 1} ^ {\lfloor n / T \rfloor} i cnt_{i T}$，显然它也是是一个关于 $T$的函数
我们记 $s(T) = \displaystyle\sum_{i = 1} ^ {\lfloor n / T \rfloor} i cnt_{i T} = \displaystyle\sum_{T | t} \frac{t}{T} cnt_t = \frac{1}{T} \displaystyle\sum_{T | t} t cnt_t$
这样也能 $O(nlogn)$预处理

for (int T = 1; T <= n; T++) {
        for (int t = T; t <= n; t += T)
            s[T] += t * cnt[t];
        s[T] /= T;
    }

于是这题就能轻松搞定了

AC代码

2. $\displaystyle\sum_{i = 1} ^ {n} lcm(i, n)$

原题链接

推导过程

$\begin{aligned} \displaystyle\sum_{i = 1} ^ {n} lcm(i, n) & = \displaystyle\sum_{i = 1} ^ {n} \frac{in}{gcd(i, n)} \\ & = n \displaystyle\sum_{i = 1} ^ {n} \frac{i}{gcd(i, n)} \\ & = n \displaystyle\sum_{d | n} \displaystyle\sum_{i = 1} ^ {n / d} \frac{id}{d} [gcd(i, \frac{n}{d}) = 1] \\ & = n \displaystyle\sum_{d | n} \displaystyle\sum_{i = 1} ^ {n / d} i [gcd(i, \frac{n}{d}) = 1] \\ & = n \displaystyle\sum_{d | n} \displaystyle\sum_{i = 1} ^ {d} i [gcd(i, d) = 1] \\ & = n \displaystyle\sum_{d | n} \displaystyle\sum_{i = 1} ^ {d} i \displaystyle\sum_{t | i, t | d} \mu(t) \\ & = n \displaystyle\sum_{d | n} \displaystyle\sum_{t | d} \mu(t) \displaystyle\sum_{i = 1} ^ {d / t} i t \\ & = n \displaystyle\sum_{d | n} \displaystyle\sum_{t | d} \mu(t)t \displaystyle\sum_{i = 1} ^ {d / t} i \\ & = n \displaystyle\sum_{d | n} \displaystyle\sum_{t | d} \mu(t) t \frac{\frac{d}{t} (\frac{d}{t} + 1) }{2} \\ & = \frac{n}{2} \displaystyle\sum_{d | n} \left[ \displaystyle\sum_{t | d} \mu(t) \frac{d ^ 2}{t} + \displaystyle\sum_{t | d} \mu(t) d \right] \\ & = \frac{n}{2} \displaystyle\sum_{d | n} d \left[ \displaystyle\sum_{t | d} \mu(t) \frac{d}{t} + \displaystyle\sum_{t | d} \mu(t) \right] \\ & = \frac{n}{2} \displaystyle\sum_{d | n} d \left[ \left( \mu*id \right) \left( d \right) + \left( \mu*I \right) \left( d \right) \right] \\ & = \frac{n}{2} \displaystyle\sum_{d | n} d \left[ \varphi \left( d \right) + \varepsilon \left( d \right) \right] \\ \end{aligned}$
显然所有答案可以像上一题那样 $O(nlogn)$处理

AC代码

3. $\displaystyle\sum_{i = 1} ^ {N} gcd\left( \lfloor \sqrt[3]{i} \rfloor, i \right)$

原题链接(2019年杭电多校第一场)

推导过程

记 $n = \sqrt[3] {N}$

$\begin{aligned} \displaystyle\sum_{i = 1} ^ {N} gcd\left( \lfloor \sqrt[3]{i} \rfloor, i \right) & = \displaystyle\sum_{i = 1} ^ {n} \displaystyle\sum_{j = i ^ 3} ^ { \min(\left( i + 1 \right) ^ 3 - 1, N) } gcd(i, j) \\ & = \displaystyle\sum_{i = 1} ^ {n - 1} \displaystyle\sum_{j = i ^ 3} ^ {\left( i + 1 \right) ^ 3 - 1} gcd(i, j) + \displaystyle\sum_{j = n ^ 3} ^ {N} gcd(n, j) \end{aligned}$
先求前半部分
不妨记 $f\left( n \right) = \displaystyle\sum_{i = 1} ^ n \displaystyle\sum_{j = i ^ 3} ^ { \left( i + 1 \right) ^ 3 - 1 } gcd\left( i, j \right)$，那么原式就是要求 $f\left( n - 1 \right)$
$\begin{aligned} f\left( n \right) & = \displaystyle\sum_{i = 1} ^ {n} \displaystyle\sum_{j = i ^ 3} ^ { \left( i + 1 \right) ^ 3 - 1 } gcd\left( i, j \right) \\ & = \displaystyle\sum_{d = 1} ^ {n} d \displaystyle\sum_{i = 1} ^ {n} \displaystyle\sum_{j = i ^ 3} ^ {\left( i + 1 \right) ^ 3 - 1} \left[ gcd\left( i, j \right) = d \right] \\ & = \displaystyle\sum_{d = 1} ^ {n} d \displaystyle\sum_{i = 1} ^ { \lfloor n / d \rfloor } \displaystyle\sum_{j = \lfloor (id) ^ 3 / d \rfloor} ^ { \lfloor \left[ \left( id + 1 \right) ^ 3 - 1 \right] / d \rfloor } \left[ gcd\left( i, j \right) = 1 \right] \\ & = \displaystyle\sum_{d = 1} ^ {n} d \displaystyle\sum_{i = 1} ^ { \lfloor n / d \rfloor } \displaystyle\sum_{j = \lfloor (id) ^ 3 / d \rfloor} ^ { \lfloor \left[ \left( id + 1 \right) ^ 3 - 1 \right] / d \rfloor } \displaystyle\sum_{t | i, t | j} \mu(t) \\ & = \displaystyle\sum_{d = 1} ^ {n} d \displaystyle\sum_{t = 1} ^ { \lfloor n / d \rfloor } \mu(t) \left( \displaystyle\sum_{i = 1} ^ { \lfloor n / dt \rfloor } \displaystyle\sum_{j = \lfloor (idt) ^ 3 / dt \rfloor} ^ { \lfloor \left[ \left( idt + 1 \right) ^ 3 - 1 \right] / dt \rfloor } \right) \\ \end{aligned}$
接下来老方法，令 $T = dt$
$\begin{aligned} f\left( n \right) & = \displaystyle\sum_{T = 1} ^ {n} \left( \displaystyle\sum_{i = 1} ^ { \lfloor n / T \rfloor } \displaystyle\sum_{j = \lfloor (iT) ^ 3 / T \rfloor} ^ { \lfloor \left[ \left( iT + 1 \right) ^ 3 - 1 \right] / T \rfloor } \right) \displaystyle\sum_{t | T} \mu(t) \left( \frac{T}{t} \right) \\ & = \displaystyle\sum_{T = 1} ^ {n} \left( \displaystyle\sum_{i = 1} ^ { \lfloor n / T \rfloor } \displaystyle\sum_{j = \lfloor (iT) ^ 3 / T \rfloor} ^ { \lfloor \left[ \left( iT + 1 \right) ^ 3 - 1 \right] / T \rfloor } \right) \varphi(T) \\ \end{aligned}$
再看括号里面的部分
$\begin{aligned} g (n, T) & = \displaystyle\sum_{i = 1} ^ { \lfloor n / T \rfloor } \displaystyle\sum_{j = \lfloor (iT) ^ 3 / T \rfloor} ^ { \lfloor \left[ \left( iT + 1 \right) ^ 3 - 1 \right] / T \rfloor } \\ & = \displaystyle\sum_{i = 1} ^ { \lfloor n / T \rfloor } \left( \lfloor \frac{ \left( iT + 1 \right) ^ 3 - 1 }{T} \rfloor - \lfloor \frac{ (iT) ^ 3 }{T} \rfloor + 1 \right) \\ & = \displaystyle\sum_{i = 1} ^ { \lfloor n / T \rfloor } \left( \lfloor \frac{ \left( iT + 1 \right) ^ 3 - (iT) ^ 3 - 1 }{T} \rfloor + 1 \right) \\ & = \displaystyle\sum_{i = 1} ^ { \lfloor n / T \rfloor } \left( 3 i ^ 2 T + 3 i + 1 \right) \\ & = 3 T \displaystyle\sum_{i = 1} ^ { \lfloor n / T \rfloor } i ^ 2 + 3\displaystyle\sum_{i = 1} ^ { \lfloor n / T \rfloor } i + \lfloor n / T \rfloor \end{aligned}$
可以预处理前缀和后 $O(1)$得到
那么 $f\left( n \right) = \displaystyle\sum_{T = 1} ^ {n} g \left(n, T \right) \varphi(T)$，就可以 $O(n)$得到

最后来看原式的后半部分 $\displaystyle\sum_{j = n ^ 3} ^ {N} gcd(n, j)$
$\begin{aligned} \displaystyle\sum_{i = n ^ 3} ^ {N} gcd(n, i) & = \displaystyle\sum_{d | n} d \displaystyle\sum_{i = \lfloor n ^ 3 / d \rfloor} ^ { \lfloor N / d \rfloor } \left[ gcd\left( \frac{n}{d}, j \right) = 1 \right] \\ & = \displaystyle\sum_{d | n} d \displaystyle\sum_{i = \lfloor n ^ 3 / d \rfloor} ^ { \lfloor N / d \rfloor } \displaystyle\sum_{ t | \frac{n}{d}, t | i } \mu(t) \\ & = \displaystyle\sum_{d | n} d \displaystyle\sum_{td | n} \mu(t) \left( \displaystyle\sum_{i = \lfloor n ^ 3 / dt \rfloor} ^ { \lfloor N / dt \rfloor } \right) \Leftarrow T = dt \\ & = \displaystyle\sum_{T | n} \left( \displaystyle\sum_{i = \lfloor n ^ 3 / T \rfloor} ^ { \lfloor N / T \rfloor } \right) \displaystyle\sum_{ d | T } \mu(\frac{T}{d}) d \\ & = \displaystyle\sum_{T | n} \left( \displaystyle\sum_{i = \lfloor n ^ 3 / T \rfloor} ^ { \lfloor N / T \rfloor } \right) \varphi(T) \\ & = \displaystyle\sum_{T | n} \left( \lfloor N / T \rfloor - \lfloor n ^ 3 / T \rfloor + 1 \right) \varphi(T) \\ \end{aligned}$
于是我们后半部分可以 $O\left(\sqrt n \right)$求得
总体复杂度就是 $O\left( n \right)$

AC代码