BZOJ DZY Loves Math系列

⑤（BZOJ 3560）

$\Sigma_{i_1|a_1}\Sigma_{i_2|a_2}\Sigma_{i_3|a_3}\Sigma_{i_4|a_4}...\Sigma_{i_n|a_n}\phi(i_1i_2i_3i_4...i_n)$
$\phi()$是积性函数
$\phi(p^k)=p^{k-1}*(p-1)$
设当前质数为p，对于第i个数，假设它分解质因数后p的次数为ai，那么p的答案就是
$[(1+p^1+...+p^{a1})(1+p^1+...+p^{a2})...(1+p^1+...+p^{an})-1]\frac{p-1}{p}+1$

乘起来就好了.....

 

$\Sigma_{i=1}^n\Sigma_{j=1}^mlcm(i,j)^{gcd(i,j)}$
$=\Sigma_{i=1}^n\Sigma_{j=1}^m (\frac{i*j}{gcd(i,j)})^{gcd(i,j)}$
枚举gcd(i,j)=d
$=\Sigma_{d=1}^n\Sigma_{i=1}^{\lfloor \frac{n}{d}\rfloor}\Sigma_{j=1}^{\lfloor \frac{m}{d}\rfloor}(d*i*j)^d*(gcd(i,j)==1)$
$=\Sigma_{d=1}^n\Sigma_{i=1}^{\lfloor \frac{n}{d}\rfloor}\Sigma_{j=1}^{\lfloor \frac{m}{d}\rfloor}\Sigma_{k|i且k|j}(d*i*j)^d$
$=\Sigma_{d=1}^nd^d\Sigma_{t=1}^{\lfloor\frac{n}{d}\rfloor}\mu(t)[\Sigma_{i=1}^{\lfloor\frac{n}{dt}\rfloor}(it)^d\Sigma_{j=1}^{\lfloor\frac{m}{dt}\rfloor}(jt)^d]$
$=\Sigma_{d=1}^nd^d\Sigma_{t=1}^{\lfloor\frac{n}{d}\rfloor}\mu(t)*t^{2d}[\Sigma_{i=1}^{\lfloor\frac{n}{dt}\rfloor}i^d\Sigma_{j=1}^{\lfloor\frac{m}{dt}\rfloor}j^d]$

 

 

$\Sigma _{i=1}^n\Sigma _{j=1}^i\mu(lcm(i,j)^{gcd(i,j)})$
$=\Sigma_{k=1}^n\Sigma_{i=1}^{\lfloor\frac{n}{k}\rfloor}\Sigma_{j=1}^i\mu((ijk)^{k}*gcd(i,j)==1)$
$∵$k>1时 $\mu(x^k)=0$
$∴ =\Sigma_{i=1}^n\Sigma_{j=1}^i\mu(ij)*e(gcd(i,j))$
$∵gcd(i,j)==1$
$∴\mu(ij)=\mu(i)*\mu(j)$
$=\Sigma_{i=1}^n\mu(i)*\Sigma_{j=1}^i\mu(j)*\Sigma_{k|i且k|j}\mu(k)$
$=\Sigma_{i=1}^n\mu(i)*\Sigma_{k|i}\mu(k)\Sigma_{j=1}^{\lfloor\frac{i}{k}\rfloor}\mu(jk)$
$\mu(i)≠0$时 再枚举k是i的约数 发现数量只有$5*10^7$
复杂度变成了
什么复杂度
O(能过)就好了...