From df2bad0f0d531c5ac7e97b92e0d1ddc671774e71 Mon Sep 17 00:00:00 2001
From: FFjet <Leoleepaytser@hotmail.com>
Date: Fri, 26 Apr 2019 15:56:06 +0800
Subject: [PATCH] Update mobius.md

---
 docs/math/mobius.md | 55 +++++++++++++++++++++++++++--------------------------
 1 file changed, 28 insertions(+), 27 deletions(-)

diff --git a/docs/math/mobius.md b/docs/math/mobius.md
index bce4edb8..f5427521 100644
--- a/docs/math/mobius.md
+++ b/docs/math/mobius.md
@@ -683,7 +683,7 @@ signed main(){
 求
 
 $$
-\sum_{i=1}^n\sum_{j=1}^ni\cdot j\cdot gcd(i,j)\bmod p\\
+\sum_{i=1}^n\sum_{j=1}^ni\cdot j\cdot \gcd(i,j)\bmod p\\
 n\leq10^{10},5\times10^8\leq p\leq1.1\times10^9,\text{p 是质数}
 $$
 
@@ -692,22 +692,23 @@ $$
 我们利用 $\varphi\ast1=ID$ 反演
 
 $$
-\begin{split}
-&\sum_{i=1}^n\sum_{j=1}^ni\cdot j
+\begin{eqnarray}
+&& \sum_{i=1}^n\sum_{j=1}^ni\cdot j\cdot \gcd(i,j)\\
+&=&\sum_{i=1}^n\sum_{j=1}^ni\cdot j
 \sum_{d|i,d|j}\varphi(d)\\
-&\sum_{d=1}^n\sum_{i=1}^n
+&=&\sum_{d=1}^n\sum_{i=1}^n
 \sum_{j=1}^n[d|i,d|j]\cdot i\cdot j
 \cdot\varphi(d)\\
-&\sum_{d=1}^n
+&=&\sum_{d=1}^n
 \sum_{i=1}^{\left\lfloor\frac{n}{d}\right\rfloor}
 \sum_{j=1}^{\left\lfloor\frac{n}{d}\right\rfloor}
 d^2\cdot i\cdot j\cdot\varphi(d)\\
-&\sum_{d=1}^nd^2\cdot\varphi(d)
+&=&\sum_{d=1}^nd^2\cdot\varphi(d)
 \sum_{i=1}^{\left\lfloor\frac{n}{d}\right\rfloor}i
 \sum_{j=1}^{\left\lfloor\frac{n}{d}\right\rfloor}j\\
-&\sum_{d=1}^nF^2\left(\left\lfloor\frac{n}{d}\right\rfloor\right)\cdot d^2\varphi(d)
+&=&\sum_{d=1}^nF^2\left(\left\lfloor\frac{n}{d}\right\rfloor\right)\cdot d^2\varphi(d)
 \left(F(n)=\frac{1}{2}n\left(n+1\right)\right)\\
-\end{split}
+\end{eqnarray}
 $$
 
 对 $\sum_{d=1}^nF\left(\left\lfloor\frac{n}{d}\right\rfloor\right)^2$ 做数论分块，$d^2\varphi(d)$ 的前缀和用杜教筛处理：
@@ -798,38 +799,38 @@ $$
 我们证明一下
 
 $$
-\begin{split}
-&g(n)=\sum_{i=1}^n\mu(i)t(i)f\left(\left\lfloor\frac{n}{i}\right\rfloor\right)\\
-=&\sum_{i=1}^n\mu(i)t(i)
+\begin{eqnarray}
+&&g(n)=\sum_{i=1}^n\mu(i)t(i)f\left(\left\lfloor\frac{n}{i}\right\rfloor\right)\\
+&=&\sum_{i=1}^n\mu(i)t(i)
 \sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}t(j)
 g\left(\left\lfloor\frac{\left\lfloor\frac{n}{i}\right\rfloor}{j}\right\rfloor\right)\\
-=&\sum_{i=1}^n\mu(i)t(i)
+&=&\sum_{i=1}^n\mu(i)t(i)
 \sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}t(j)
 g\left(\left\lfloor\frac{n}{ij}\right\rfloor\right)\\
-=&\sum_{T=1}^n
+&=&\sum_{T=1}^n
 \sum_{i=1}^n\mu(i)t(i)
 \sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}[ij=T]
 t(j)g\left(\left\lfloor\frac{n}{T}\right\rfloor\right)
-&\text{【先枚举 ij 乘积】}\\
-=&\sum_{T=1}^n
+&&\text{【先枚举 ij 乘积】}\\
+&=&\sum_{T=1}^n
 \sum_{i|T}\mu(i)t(i)
 t\left(\frac{T}{i}\right)g\left(\left\lfloor\frac{n}{T}\right\rfloor\right)
-&\text{【}\sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}[ij=T] \text{对答案的贡献为 1，于是省略】}\\
-=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)
+&&\text{【}\sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}[ij=T] \text{对答案的贡献为 1，于是省略】}\\
+&=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)
 \sum_{i|T}\mu(i)t(i)t\left(\frac{T}{i}\right)\\
-=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)
+&=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)
 \sum_{i|T}\mu(i)t(T)
-&\text{【t 是完全积性函数】}\\
-=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T)
+&&\text{【t 是完全积性函数】}\\
+&=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T)
 \sum_{i|T}\mu(i)\\
-=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T)
+&=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T)
 \varepsilon(T)
-&\text{【}\mu\ast 1= \varepsilon\text{】}\\
-=&g(n)t(1)
-&\text{【当且仅当 T=1,}\varepsilon(T)=1\text{时】}\\
-=&g(n)
-&&& \square
-\end{split}
+&&\text{【}\mu\ast 1= \varepsilon\text{】}\\
+&=&g(n)t(1)
+&&\text{【当且仅当 T=1,}\varepsilon(T)=1\text{时】}\\
+&=&g(n)
+&& \square
+\end{eqnarray}
 $$
 
 =======
-- 
2.11.0