题解 P6156 简单题

题意

给定 $n,k$，求

$$\sum _{i=1}^n\sum _{j=1}^n(i+j{)}^k{\mu }^2(gcd(i,j))gcd(i,j)$$

数据范围：$1\le n\le 5\times {10}^6,1\le k\le {10}^{18}$

题解

$$\begin{array}{rl}\sum _{i=1}^n\sum _{j=1}^n(i+j{)}^k{\mu }^2(gcd(i,j))gcd(i,j)& =\sum _{d=1}^n{\mu }^2(d)d^{k+1}\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }(i+j{)}^k[gcd(i,j)=1]\\ & =\sum _{d=1}^n{\mu }^2(d)d^{k+1}\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }(i+j{)}^k\sum _{s|i}\sum _{s|j}\mu (s)\\ & =\sum _{d=1}^n{\mu }^2(d)d^{k+1}\sum _{s=1}^{\lfloor \frac{n}{d}\rfloor }\mu (s)s^k\sum _{i=1}^{\lfloor \frac{n}{sd}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{sd}\rfloor }(i+j{)}^k\\ & =\sum _{q=1}^n{q}^k\sum _{d|q}{\mu }^2(d)d\mu (\frac{q}{d})\sum _{i=1}^{\lfloor \frac{n}{q}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{q}\rfloor }(i+j{)}^k\\ & =\sum _{q=1}^n\sum _{i=1}^{\lfloor \frac{n}{q}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{q}\rfloor }(i+j{)}^k{q}^k\sum _{d|q}{\mu }^2(d)d\mu (\frac{q}{d})\end{array}$$ 令 $S(n)=\sum _{i=1}^n\sum _{j=1}^n(i+j{)}^k,g(n)=n^k\sum _{d|n}{\mu }^2(d)d\mu (\frac{n}{d})$

$$\begin{array}{rl}\sum _{i=1}^n\sum _{j=1}^n(i+j{)}^k{\mu }^2(gcd(i,j))gcd(i,j)& =\sum _{q=1}^{n}S(\lfloor \frac{n}{q}\rfloor )g(q)\end{array}$$

对于 $g(n)$，线性筛出 $i^k$，$O(n\log n)$ 预处理即可。

考虑 $S(n)$ 怎么求，令 $f(n)=\sum _{i=1}^n{i}^k$，所以 $S(n)=S(n-1)+(2n{)}^k+2[f(2n-1)-f(n+1)]$，直接推即可。

左边整除分块，右边直接 $O(n\log n)$ 暴力搞出来，时间复杂度 $O(n\log n)$。