「杜教筛」

例题

C. DZY Loves Math IV

求$sumlimits_{i=1}^nsumlimits_{j=1}^mphi(ij)$

开始吧！

设gr=n

$g=prodlimits_{pin n}p$

$gcd=gcd(n,i)$

$phi(gi)=gcd*phi(frac{gi}{gcd})=gcd*phi(frac{g}{gcd})*phi(i)$

$sumlimits_{i=1}^nsumlimits_{j=1}^mphi(ij)$

$=sumlimits_{i=1}^nS(i,m)$

$S(n,m)=sumlimits_{i=1}^mphi(ni)$

$=rsumlimits_{i=1}^mgcd*phi(frac{g}{gcd})phi(i)$

$=rsumlimits_{i=1}^msumlimits_{d|gcd}phi(d)phi(frac{g}{gcd})phi(i)$

$=rsumlimits_{i=1}^msumlimits_{d|gcd}phi(frac{g}{frac{gcd}{d}})phi(i)$

$=rsumlimits_{i=1}^msumlimits_{d|gcd}phi(frac{g}{d})phi(i)$

$=rsumlimits_{d|g}phi(frac{g}{d})sumlimits_{i=1}^{frac{m}{d}} phi(di)$

$=rsumlimits_{d|g}phi(frac{g}{d})S(d,frac{m}{d})$

递归形式了，复杂度与质因子个数有关。