「刷题」JZPKIL

　　这道反演题，真牛逼。

　　以下用$B$代表伯努利数，$l*g=f$代表狄利克雷卷积,先推式子。

　　对于给出的$n,x,y$求一百组数据的$ans$

$\begin{array}{rcl} ans & = & \sum\limits_{i=1}^ngcd(i,n)^xlcm(i,n)^y\end{array}$

 

$\begin{array}{rcl} & = & \sum\limits_{i=1}^ngcd(i,n)^x\frac{(in)^y}{gcd(i,n)^y}\end{array}$

 

$\begin{array}{rcl} & = & \sum\limits_{i=1}^ngcd(i,n)^{x-y}(in)^y\end{array}$

 

$\begin{array}{rcl} & = & n^y\sum\limits_{i=1}^ni^ygcd(i,n)^{x-y}\end{array}$

 

$\begin{array}{rcl} & = & n^y\sum\limits_{d|n}d^{x-y} \sum \limits_{i=1}^{\lfloor \frac{n}{d} \rfloor} (id)^y[gcd(i,\lfloor\frac{n}{d} \rfloor)=1]\end{array}$

 

$\begin{array}{rcl} & = & n^y\sum\limits_{d|n}d^x\sum\limits_{i=1}^{\lfloor \frac{n}{d} \rfloor}i^y\sum\limits_{t|gcd(i,\lfloor \frac{n}{d} \rfloor)}\mu(t)\end{array}$

 

$\begin{array}{rcl} & = & n^y\sum\limits_{d|n}d^x\sum\limits_{t|\lfloor\frac{n}{d}\rfloor}\mu(t)t^y\sum\limits_{i=1}^{\lfloor\frac{n}{td}\rfloor}i^y\end{array}$

 

 

 $\begin{array}{rcl}\sum\limits_{i=0}^{\lfloor\frac{n}{td}\rfloor}i^y & = & \frac{1}{y+1}\sum\limits_{i=0}^yC_{y+1}^iB_i(\lfloor\frac{n}{td}\rfloor)^{y-i+1}\end{array}$

 

$\begin{array}{rcl}R_i & = & \frac{C_{y+1}^iB_i}{y+1}\end{array}$

 

$\begin{array}{rcl}ans & = & n^y\sum\limits_{d|n}d^x\sum\limits_{t|\lfloor\frac{n}{d}\rfloor}\mu(t)t^y\sum\limits_{i=0}^yR_i(\lfloor\frac{n}{td}\rfloor)^{y-i+1}\end{array}$

 

$\begin{array}{rcl} & = & \sum\limits_{i=1}^yR_in^y\sum\limits_{d|n}d^x\sum\limits_{t|\lfloor\frac{n}{d}\rfloor}\mu(t)t^y(\lfloor\frac{n}{td}\rfloor)^{y-i+1}\end{array}$

 

$\begin{array}{rcl}f_{i,x,y}(n) & = & n^y\sum\limits_{d|n}d^x\sum\limits_{t|\lfloor\frac{n}{d}\rfloor}\mu(t)t^y(\lfloor\frac{n}{td}\rfloor)^{y-i+1}\end{array}$

 

分析$f_{i,x,y}(n)$。

 

$\begin{array}{rcl}l(x) & = & \mu(x)x^y \end{array}$

 

$\begin{array}{rcl}q_r(x) & = & x^r\end{array}$

 

$l,q$ 均为积性函数。

 

$\begin{array}{rcl} g(n) & = & \sum\limits_{d|n}\mu(d)d^yq(\lfloor\frac{n}{d}\rfloor)\end{array}$

 

 

$\begin{array}{rcl}g(n) & = & l(n)*q(n)\end{array}$

 

也为积性函数。

 

$\begin{array}{rcl}f(n) & = & \sum\limits_{d|n}q(d)g(\lfloor\frac{n}{d}\rfloor) \\ & = & q(n)*g(n) \end{array}$

 

所以$f_{i,x,y}(n)$是积性函数。

 

$\begin{array}{rcl}ans & = & \sum\limits_{i=0}^yR_if_{i,x,y}(n)\end{array}$

 

$n$为$1e18$考虑用$O(n^{1/4})$的$Pollard_Rho$算法对$n$进行质因分解。

 

$n=\_p^c$

 

$\begin{array}{rcl}f_{i,x,y}(p^c) & = & p^{cy}\sum\limits_{d|p^c}\sum\limits_{t|\lfloor\frac{p^c}{d}\rfloor}\mu(t)t^y(\lfloor\frac{p^c}{td}\rfloor)^{y-i+1}\end{array}$

 

$\begin{array}{rcl} & = & p^{cy}\sum\limits_{j=0}^cp^{jx}\sum\limits_{k=0}^{c-j}\mu(p^k)p^{ky}(p^{c-j-k})^{y-i+1}\end{array}$

 

当k=1或者0的时候，莫比乌斯函数不为0。

 

$\begin{array}{rcl} & = & p^{cy}\sum\limits_{j=0}^c p^{jx}[(p^{c-j})^{y-i+1}-p^y(p^{c-j-1})^{y-i+1}]\end{array}$

 

问题得到解决。

 

知识点：

莫比乌斯反演

狄利克雷卷积

积性函数

自然数幂和

伯努利数

$Miller\_Rabin$素数测试

$Pollard\_Rho$质因数分解

费马小定理

二次初探原理

生日悖论

 

有兴趣的可以尝试一下，是道好题。