Gym105423D 做题记录
题目链接。来源。
一句话题意：如题意。
根据题意有：
\[\sum_{i=1}^{n}\sum_{j=1}^n\sum_{d\mid b_i,b_j} \left| a_i-a_j \right|\dfrac{b_ib_j}{d^2}[(b_i,b_j)=d]
\]
我们接下来引入一个莫反的推论：
\[[n=1]=\sum_{d\mid n}\mu(d)\\
\Rightarrow [(i,j)=1]=\sum_{d\mid(i,j)}\mu(d)\\
\Rightarrow [(i,j)=d]=[(\dfrac{i}{d},\dfrac{j}{d})=1]=\sum_{k\mid(\frac{i}{d},\frac{j}{d})}\mu(k)
\]
将其带入：
\[\sum_{i=1}^{n}\sum_{j=1}^n\sum_{d\mid b_i,b_j} \left| a_i-a_j \right|\dfrac{b_ib_j}{d^2}[(\dfrac{b_i}{d},\dfrac{b_j}{d})=1]
\\
\sum_{i=1}^{n}\sum_{j=1}^n\sum_{d\mid b_i,b_j} \left| a_i-a_j \right|\dfrac{b_i}{d}\dfrac{b_j}{d}[(\dfrac{b_i}{d},\dfrac{b_j}{d})=1]
\\
\sum_{i=1}^{n}\sum_{j=1}^n\sum_{d\mid b_i,b_j} \left| a_i-a_j \right|\sum_{k\mid(\frac{b_i}{d},\frac{b_j}{d})}\mu(k)\cdot \dfrac{b_i}{d}\cdot\dfrac{b_j}{d}
\]
考虑到 \(kd \mid b_i,b_j\)，于是令 \(T=kd\)。
\[\sum_{i=1}^{n}\sum_{j=1}^n\sum_{T\mid b_i,b_j} \left| a_i-a_j \right|\sum_{k\mid T}\mu(k)\cdot \dfrac{b_ib_j}{T^2}\cdot k^2
\\
\sum_{i=1}^{n}\sum_{j=1}^n\sum_{T\mid b_i,b_j} \dfrac{\left| a_i-a_j \right|b_ib_j}{T^2}\sum_{k\mid T}\mu(k) k^2
\]
定义 \(f(T)=\sum_{k\mid T}\mu(k)k^2\)。
\[\sum_{i=1}^{n}\sum_{j=1}^n\sum_{T\mid b_i,b_j} \dfrac{\left| a_i-a_j \right|b_ib_j}{T^2}f(T)
\\
2\sum_{i=1}^n\sum_{j=1}^{i-1}\sum_{T\mid b_i,b_j}(a_i-a_j)b_ib_j\cdot \dfrac{f(T)}{T^2}
\\
2\sum_{i=1}^n\sum_{T \mid b_i} \dfrac{b_if(T)}{T}(a_i\sum_{T\mid b_j,j<i}\dfrac{b_j}{T}-\sum_{T\mid b_j,j<i}\dfrac{a_jb_j}{T})
\]
用欧拉筛预处理 \(f(T)\) 和每个数的因数，把 \(\sum\limits_{T\mid b_j,j<i}\dfrac{b_j}{T}\) 和 \(\sum\limits_{T\mid b_j,j<i}\dfrac{a_jb_j}{T}\) 处理一下就可以了。
