积性函数+容斥
2019ICPC南昌邀请赛网络赛 G.tsy's number
题意
求$\sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n\frac{\phi(i)\phi(j^2)\phi(k^3)}{\phi(i)\phi(j)\phi(k)}\phi(gcd(i,j,k))$ 共T组数据，$T\leq 10000$，$1\leq n \leq 10^7$
题解
枚举gcd(i,j,k) = d，然后容斥一下 $$ \begin{align*} ans &= \sum_{d=1}^n\phi(d)\sum_{i=1}^{[\frac{n}{d}]}\sum_{j=1}^{[\frac{n}{d}]}\sum_{k=1}^{[\frac{n}{d}]} [gcd(i,j,k)==1]\frac{\phi(id)\phi((jd)^2)\phi((kd)^3)}{\phi(id)\phi(jd)\phi(kd)}\ &= \sum_{d=1}^n\phi(d)\sum_{s=1}^{[\frac{n}{d}]}\mu(s)\sum_{i=1}^{[\frac{n}{ds}]}\sum_{j=1}^{[\frac{n}{ds}]}\sum_{k=1}^{[\frac{n}{ds}]} \frac{\phi(ids)\phi((jds)^2)\phi((kds)^3)}{\phi(ids)\phi(jds)\phi(kds)}\ &= \sum_{d=1}^n\phi(d)\sum_{s=1}^{[\frac{n}{d}]}\mu(s)\sum_{i=1}^{[\frac{n}{ds}]}\frac{\phi(ids)}{\phi(ids)}\sum_{j=1}^{[\frac{n}{ds}]}\frac{\phi((jds)^2))}{\phi(jds)} \sum_{k=1}^{[\frac{n}{ds}]}\frac{\phi((kds)^3)}{\phi(kds)}\ \end{align*} $$ 可以归纳证明， $$ \sum_{i=1}^{[\frac{n}{d}]}\frac{\phi((id)^k)}{\phi(id)} = d^{(k-1)}* \sum_{i=1}^{[\frac{n}{d}]}i^{(k-1)} $$ 所以 $$ \begin{align*} ans &= \sum_{d=1}^n\phi(d)\sum_{s=1}^{[\frac{n}{d}]}\mu(s)* (ds)^3[\frac{n}{ds}] \sum_{i=1}^{[\frac{n}{ds}]}i \sum_{i=1}^{[\frac{n}{ds}]}i^2 \end{align} $$ 设$T=sd$,则 $$ \begin{align} ans &= \sum_{T=1}^n (T^3* [\frac{n}{T}]* \sum_{i=1}^{[\frac{n}{T}]}i* \sum_{i=1}^{[\frac{n}{T}]}i^2)* \sum_{d|T}\phi(d)* \mu(T/d) \end{align*} $$ 前面一部分和它的前缀和显然可以预处理，后面是两个积性函数的狄利克雷卷积，设为f(T)，p为与T互质的质数，则有 $$ \begin{align*} f(p^kT) &= \sum_{i=0}^k\sum_{d|T}\phi(p^id)* \mu(\frac{p^{k-i}* T}{d})\ &= (\phi(p^k)-\phi(p^{k-1}))* f(T) \end{align*} $$ 可以用线性筛预处理，然后就是$\sqrt n$的套路了，总复杂度为$O(n+T*\sqrt n)$.
来源：oschina
链接：https://my.oschina.net/u/4305544/blog/3563874