真是一道“简单”的数学题呢~
反演题, 化式子.
\[
ans=\sum_{i=1}^n\sum_{j=1}^nijgcd(i,j) \\ =\sum_{i=1}^n\sum_{j=1}^n\sum_{d=1}^nij[gcd(i,j)=d]\\ =\sum_{d=1}^nd\sum_{i=1}^n\sum_{i=1}^nij[gcd(i,j)=1]\\ =\sum_{d=1}^nd^3\sum_{i=1}^{\left \lfloor \frac nd \right \rfloor}\sum_{j=1}^{\left \lfloor \frac nd \right \rfloor}ij[gcd(i,j)=1] \\ =\sum_{d=1}^nd^3\sum_{i=1}^{\left \lfloor \frac nd \right \rfloor}i\sum_{j=1}^{\left \lfloor \frac nd \right \rfloor}j[gcd(i,j)=1] \\ =\sum_{d=1}^nd^3\sum_{i=1}^{\left \lfloor \frac nd \right \rfloor}i\sum_{j=1}^{\left \lfloor \frac nd \right \rfloor}j\sum_{t|i,j}\mu(t) \\ i=tp,j=tq,\\ =\sum_{d=1}^nd^3\sum_{t=1}^{\left \lfloor \frac nd \right \rfloor}t^2\cdot\mu(t)\sum_{p=1}^{\left \lfloor \frac n{td} \right \rfloor}p\sum_{q=1}^{\left \lfloor \frac n{td} \right \rfloor}q \\ \because \sum_{i=1}^n=\frac {n(n+1)}2\\ \therefore ans=\sum_{d=1}^nd^3\sum_{t=1}^{\left \lfloor \frac nd \right \rfloor}t^2\cdot\mu(t)[\frac{n(n+1)}{2}]
\]
然后我们令\(T=id\), 枚举\(T\),
\[
ans=\sum_{T=1}^n[\frac{\left \lfloor \frac nT \right \rfloor(\left \lfloor \frac nT \right \rfloor+1)}{2}]^2\sum_{d|T}d^3(\frac Td)^2\mu(\frac Td) \=\sum_{T=1}^n[\frac{\left \lfloor \frac nT \right \rfloor(\left \lfloor \frac nT \right \rfloor+1)}{2}]^2T^2\sum_{d|T}d\mu(\frac Td) \=\sum_{T=1}^n[\frac{\left \lfloor \frac nT \right \rfloor(\left \lfloor \frac nT \right \rfloor+1)}{2}]^2T^2(n*\mu)(T) \=\sum_{T=1}^n[\frac{\left \lfloor \frac nT \right \rfloor(\left \lfloor \frac nT \right \rfloor+1)}{2}]^2T^2\varphi(T)
\]
然后我们令\(X=[\frac{\left \lfloor \frac nT \right \rfloor(\left \lfloor \frac nT \right \rfloor+1)}{2}]^2, f(T)=T^2\varphi(T)\), 这样就变成了
\[
ans=\sum_{T=1}^nXf(T)
\]
\(X\)可以分块然后\(O(1)\)算, 那我们只要能求出\(f(T)\)的前缀和就行了.
那\(n<=10^{10}\)要用杜教筛. 我们想一下杜教筛的通式:
\[
s_f(x)=\frac{s_{f*g}(x)-\sum_{i=2}^ns_f(\left \lfloor \frac ni \right \rfloor)g(i)}{g(1)}
\]
其中\(g(x)\)和\((f*g)(x)\)是易求前缀和的函数.
看到这种乘积的前缀和我们又想到了之前做lcm之和的提出乘积中某一项的高端操作, 我们就试着让\(g(x)\)去卷\(n^2\)(就是\(f(x)=x^2\)啦).
然后
\[
g(x)=(n^2\cdot\varphi)(x) \(g*n^2)(x)=((n^2\cdot\varphi)*n^2)(x)=(n^2\cdot(\varphi*1))(x)=(n^2\cdot n)(x)=n^3(x)
\]
然后\(n^3(x)\)的前缀和也是有公式的可以\(O(1)\)算, 这样我们就令\(g(x)=n^2(x), (f*g)(x)=n^3(x)\), 然后扔到上面的杜教筛通式里做就好啦~
复杂度可能是\(O(\sqrt n*\n^{\frac 23}\)的, 不过这应该是最优的复杂度了..
代码应该不太好写, 懒得写了QAQ
