自变量互质的前缀和函数分析转载自 link###有这样一类问题，他们的形式常常是这个样子$$\begin{aligned

转载自 link

###有这样一类问题，他们的形式常常是这个样子

\begin{aligned} \sum_{i=1}^n{f(i)[gcd(i,j)=1]} \end{aligned}

###我们来对他进行变形

\begin{aligned} &\sum_{i=1}^n{f(i)[gcd(i,j)=1]}\\ =&\sum_{i=1}^n{f(i)e(gcd(i,j))}\\ =&\sum_{i=1}^n{f(i)(\mu*1)(gcd(i,j)}\\ =&\sum_{i=1}^n{f(i)\sum_{d|gcd(i,j)}\mu(d)}\\ =&\sum_{i=1}^n{f(i)\sum_{d|i,d|j}\mu(d)}\\ =&\sum_{d|j}{\mu(d)\sum_{d|i,1<=i<=n}f(i)}\\ =&\sum_{d|j}{\mu(d)\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}f(i*d)}\\ \end{aligned}

###如果 $f(i)=1$ 则

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

###！更加特殊的如果 $j=n$ 则

\begin{aligned} \sum_{i=1}^n{[gcd(i,n)=1]}=\sum_{d|j}{\mu(d)\frac{n}{d}}=(\mu*id)(n)=\phi(n)\\ \end{aligned}

###如果 $f(i)=i$ 则

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

###！更加特殊的如果 $j=n$ 则

\begin{aligned} &\sum_{i=1}^n{i[gcd(i,n)=1]}\\ =&\sum_{d|n}{\mu(d)d\frac{\frac{n}{d}(\frac{n}{d}+1)}{2}}\\ =&\frac{n}{2}\sum_{d|n}{\mu(d)(\frac{n}{d}+1)}\\ =&\frac{n}{2}(\sum_{d|n}{\mu(d)\frac{n}{d}}+\sum_{d|n}{\mu(d)})\\ =&\frac{n}{2}(\phi(n)+e(n))\\ \end{aligned}

###总结

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