例题
C. DZY Loves Math IV
求$\sum\limits_{i=1}^n\sum\limits_{j=1}^m\phi(ij)$
开始吧！
设gr=n
$g=\prod\limits_{p\in n}p$
$gcd=gcd(n,i)$
$phi(gi)=gcd*phi(\frac{gi}{gcd})=gcd*phi(\frac{g}{gcd})*phi(i)$
$\sum\limits_{i=1}^n\sum\limits_{j=1}^m\phi(ij)$
$=\sum\limits_{i=1}^nS(i,m)$
$S(n,m)=\sum\limits_{i=1}^m\phi(ni)$
$=r\sum\limits_{i=1}^mgcd*\phi(\frac{g}{gcd})\phi(i)$
$=r\sum\limits_{i=1}^m\sum\limits_{d|gcd}\phi(d)\phi(\frac{g}{gcd})\phi(i)$
$=r\sum\limits_{i=1}^m\sum\limits_{d|gcd}\phi(\frac{g}{\frac{gcd}{d}})\phi(i)$
$=r\sum\limits_{i=1}^m\sum\limits_{d|gcd}\phi(\frac{g}{d})\phi(i)$
$=r\sum\limits_{d|g}\phi(\frac{g}{d})\sum\limits_{i=1}^{\frac{m}{d}} \phi(di)$
$=r\sum\limits_{d|g}\phi(\frac{g}{d})S(d,\frac{m}{d})$
递归形式了，复杂度与质因子个数有关。