【最大公约数】欧拉反演+积性函数拆分

题意

求\(\sum\limits_{i=1}^n \gcd(i,n)\) ,其中 \(n<=2^{32}\) 。

\(\begin{aligned}\sum\gcd(i,n)&=\sum\limits_{j=1}^{n}j\sum\limits_{i=1}^{n}[gcd(i,n)==1]\\&=\sum\limits_{j|n}j\ast\varphi(n/j) \\&= \sum_{d \mid n} \varphi(d)*\frac{n}{d}\end{aligned}\)

\(f(n) = \sum_{d \mid n} \varphi(d)*\frac{n}{d}\) 是狄利克雷卷积 \(\varphi\ast Id\) ，所以 \(f(n)\) 是积性函数

\(f(n) = f({p_1}^{c_1}) f({p_2}^{c_2}) ... f({p_n}^{c_n})\)

则只用求每一个 \(f(p^c)\)

所以

\(\begin{aligned} f(p^c) &= \sum_{i=0}^{c} \varphi(p^i) \frac{p^c}{p^i} \\ &= \sum_{i=0}^{c} \varphi(p^i) p^{c-i} \end{aligned}\)

而

\(\varphi(p^c)=p^c(1-\frac{1}{p}),(c\not=0)\)

\(\varphi(p^c)=1,(c=0)\)

所以

\(\begin{aligned} f(p^c)&=\sum_{i=0}^c p^i(1-\frac{1}{p})p^{c-i}\\&= p^c+\sum_{i=1}^c (p^c-p^{c-1})\\&=(c+1)p^c-c*p^{c-1}\\&(c\not=0)\end{aligned}\\\)

对n进行唯一因式分解后得到答案