求
$$
-\sum_{i=1}^n\sum_{j=1}^ni\cdot j\cdot \gcd(i,j)\bmod p\\
-n\leq10^{10},5\times10^8\leq p\leq1.1\times10^9,\text{p 是质数}
+\sum_{i=1}^n\sum_{j=1}^ni\cdot j\cdot \gcd(i,j)\bmod p
+$$
+
+$$
+n\leq10^{10},5\times10^8\leq p\leq1.1\times10^9,p \text{是质数}
$$
看似是一道和 $\gcd$ 有关的题，不过由于带有系数，并不容易化简
$$
\begin{aligned}
-&& \sum_{i=1}^n\sum_{j=1}^ni\cdot j\cdot \gcd(i,j)\\
-&=&\sum_{i=1}^n\sum_{j=1}^ni\cdot j
+& \sum_{i=1}^n\sum_{j=1}^ni\cdot j\cdot \gcd(i,j)\\
+&=\sum_{i=1}^n\sum_{j=1}^ni\cdot j
\sum_{d \mid i,d \mid j}\varphi(d)\\
-&=&\sum_{d=1}^n\sum_{i=1}^n
+&=\sum_{d=1}^n\sum_{i=1}^n
\sum_{j=1}^n[d \mid i,d \mid j]\cdot i\cdot j
\cdot\varphi(d)\\
-&=&\sum_{d=1}^n
+&=\sum_{d=1}^n
\sum_{i=1}^{\left\lfloor\frac{n}{d}\right\rfloor}
\sum_{j=1}^{\left\lfloor\frac{n}{d}\right\rfloor}
d^2\cdot i\cdot j\cdot\varphi(d)\\
-&=&\sum_{d=1}^nd^2\cdot\varphi(d)
+&=\sum_{d=1}^nd^2\cdot\varphi(d)
\sum_{i=1}^{\left\lfloor\frac{n}{d}\right\rfloor}i
\sum_{j=1}^{\left\lfloor\frac{n}{d}\right\rfloor}j\\
-&=&\sum_{d=1}^nF^2\left(\left\lfloor\frac{n}{d}\right\rfloor\right)\cdot d^2\varphi(d)
+&=\sum_{d=1}^nF^2\left(\left\lfloor\frac{n}{d}\right\rfloor\right)\cdot d^2\varphi(d)
\left(F(n)=\frac{1}{2}n\left(n+1\right)\right)\\
\end{aligned}
$$
$$
\begin{aligned}
-&&\sum_{d=1}^{n}d^3\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}ij\cdot[\gcd(i,j)=1]\\
-&=&\sum_{d=1}^{n}d^3\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}ij\sum_{t\mid \gcd(i,j)}\mu(t)\\
-&=&\sum_{d=1}^{n}d^3\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}ij\sum_{t=1}^{\lfloor\frac{n}{d}\rfloor}\mu(t)[t\mid \gcd(i,j)]\\
-&=&\sum_{d=1}^{n}d^3\sum_{t=1}^{\lfloor\frac{n}{d}\rfloor}t^2 \mu(t)\sum_{i=1}^{\lfloor\frac{n}{td}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{td}\rfloor}ij[1\mid \gcd(i,j)]\\
-&=&\sum_{d=1}^{n}d^3\sum_{t=1}^{\lfloor\frac{n}{d}\rfloor}t^2 \mu(t)\sum_{i=1}^{\lfloor\frac{n}{td}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{td}\rfloor}ij
+&\sum_{d=1}^{n}d^3\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}ij\cdot[\gcd(i,j)=1]\\
+&=\sum_{d=1}^{n}d^3\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}ij\sum_{t\mid \gcd(i,j)}\mu(t)\\
+&=\sum_{d=1}^{n}d^3\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}ij\sum_{t=1}^{\lfloor\frac{n}{d}\rfloor}\mu(t)[t\mid \gcd(i,j)]\\
+&=\sum_{d=1}^{n}d^3\sum_{t=1}^{\lfloor\frac{n}{d}\rfloor}t^2 \mu(t)\sum_{i=1}^{\lfloor\frac{n}{td}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{td}\rfloor}ij[1\mid \gcd(i,j)]\\
+&=\sum_{d=1}^{n}d^3\sum_{t=1}^{\lfloor\frac{n}{d}\rfloor}t^2 \mu(t)\sum_{i=1}^{\lfloor\frac{n}{td}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{td}\rfloor}ij
\end{aligned}
$$
$$
\begin{aligned}
-&&\sum_{d=1}^{n}d^3\sum_{t=1}^{\lfloor\frac{n}{d}\rfloor}t^2 \mu(t)\sum_{i=1}^{\lfloor\frac{n}{td}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{td}\rfloor}ij\\
-&=&\sum_{d=1}^{n}d^3\sum_{t=1}^{\lfloor\frac{n}{d}\rfloor}t^2 \mu(t)\cdot F^2\left(\left\lfloor\frac{n}{td}\right\rfloor\right)\\
-&=&\sum_{T=1}^{n}F^2\left(\left\lfloor\frac{n}{T}\right\rfloor\right) \sum_{d\mid T}d^3\left(\frac{T}{d}\right)^2\mu\left(\frac{T}{d}\right)\\
-&=&\sum_{T=1}^{n}F^2\left(\left\lfloor\frac{n}{T}\right\rfloor\right) T^2\sum_{d\mid T}d\cdot\mu\left(\dfrac{T}{d}\right)
+&\sum_{d=1}^{n}d^3\sum_{t=1}^{\lfloor\frac{n}{d}\rfloor}t^2 \mu(t)\sum_{i=1}^{\lfloor\frac{n}{td}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{td}\rfloor}ij\\
+&=\sum_{d=1}^{n}d^3\sum_{t=1}^{\lfloor\frac{n}{d}\rfloor}t^2 \mu(t)\cdot F^2\left(\left\lfloor\frac{n}{td}\right\rfloor\right)\\
+&=\sum_{T=1}^{n}F^2\left(\left\lfloor\frac{n}{T}\right\rfloor\right) \sum_{d\mid T}d^3\left(\frac{T}{d}\right)^2\mu\left(\frac{T}{d}\right)\\
+&=\sum_{T=1}^{n}F^2\left(\left\lfloor\frac{n}{T}\right\rfloor\right) T^2\sum_{d\mid T}d\cdot\mu\left(\dfrac{T}{d}\right)
\end{aligned}
$$
$$
\begin{aligned}
-&&g(n)=\sum_{i=1}^n\mu(i)t(i)f\left(\left\lfloor\frac{n}{i}\right\rfloor\right)\\
-&=&\sum_{i=1}^n\mu(i)t(i)
+&g(n)=\sum_{i=1}^n\mu(i)t(i)f\left(\left\lfloor\frac{n}{i}\right\rfloor\right)\\
+&=\sum_{i=1}^n\mu(i)t(i)
\sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}t(j)
g\left(\left\lfloor\frac{\left\lfloor\frac{n}{i}\right\rfloor}{j}\right\rfloor\right)\\
-&=&\sum_{i=1}^n\mu(i)t(i)
+&=\sum_{i=1}^n\mu(i)t(i)
\sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}t(j)
g\left(\left\lfloor\frac{n}{ij}\right\rfloor\right)\\
-&=&\sum_{T=1}^n
+&=\sum_{T=1}^n
\sum_{i=1}^n\mu(i)t(i)
\sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}[ij=T]
t(j)g\left(\left\lfloor\frac{n}{T}\right\rfloor\right)
-&&\text{【先枚举 ij 乘积】}\\
-&=&\sum_{T=1}^n
+&\text{【先枚举 ij 乘积】}\\
+&=\sum_{T=1}^n
\sum_{i \mid T}\mu(i)t(i)
t\left(\frac{T}{i}\right)g\left(\left\lfloor\frac{n}{T}\right\rfloor\right)
-&&\text{【}\sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}[ij=T] \text{对答案的贡献为 1，于是省略】}\\
-&=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)
+&\text{【}\sum_{j=1}^{\left\lfloor\frac{n}{i}\right\rfloor}[ij=T] \text{对答案的贡献为 1，于是省略】}\\
+&=\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)
\sum_{i \mid T}\mu(i)t(i)t\left(\frac{T}{i}\right)\\
-&=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)
+&=\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)
\sum_{i \mid T}\mu(i)t(T)
-&&\text{【t 是完全积性函数】}\\
-&=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T)
+&\text{【t 是完全积性函数】}\\
+&=\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T)
\sum_{i \mid T}\mu(i)\\
-&=&\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T)
+&=\sum_{T=1}^ng\left(\left\lfloor\frac{n}{T}\right\rfloor\right)t(T)
\varepsilon(T)
-&&\text{【}\mu\ast 1= \varepsilon\text{】}\\
-&=&g(n)t(1)
+&\text{【}\mu\ast 1= \varepsilon\text{】}\\
+&=g(n)t(1)
&&\text{【当且仅当 T=1,}\varepsilon(T)=1\text{时】}\\
-&=&g(n)
-&& \square
+&=g(n)
+& \square
\end{aligned}
$$
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