題目地址

題意簡述

給定兩個整數 $n,p$，求下面式子的值：

$\underset{i}{\overset{}{\sum }}$

在模 $p$意義下的值。

$n\leq 10^{10},p\leq 1.1\times 10^9$， $p$為質數。

• 分析

我們根據套路，來列舉 $gcd$，原式變成：

$\sum_{d=1}^n\sum_{i=1}^n\sum_{j=1}^nijd[gcd(i,j)=d]\\ =\sum_{d=1}^nd\sum_{i=1}^n\sum_{j=1}^nij[gcd(i,j)=d]\\ =\sum_{d=1}^nd\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}id\times jd[gcd(i,j)=1]\\ =\sum_{d=1}^nd^3\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}ij[gcd(i,j)=1]$

後面我們用莫比烏斯反演的套路，原式變成：

$= \sum_{d=1}^nd^3\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}ij\sum_{w|i,w|j}\mu(w)\\ = \sum_{d=1}^nd^3\sum_{w=1}^{\lfloor\frac{n}{d}\rfloor}\mu(w)\sum_{i=1}^{\lfloor\frac{n}{dw}\rfloor}iw\sum_{j=1}^{\lfloor\frac{n}{dw}\rfloor}jw\\ = \sum_{d=1}^nd^3\sum_{w=1}^{^{\lfloor\frac{n}{d}\rfloor}}\mu(w)w^2\sum_{i=1}^{\lfloor\frac{n}{dw}\rfloor}i\sum_{j=1}^{\lfloor\frac{n}{dw}\rfloor}j$