杜教筛之逆运算2
有了杜教筛的式子我们可以干很多事
\[\sum_{i=1}^nf*g(i)=\sum_{i=1}^nf(i)\sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}g(j)
\]
对于这种两种形式的式子可以用杜教筛式子秒杀
\[1.\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor f(i)\\
2.\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor^2 f(i)
\]
第一种式子
将杜教筛中\(g()\)特殊化为\(I():I(x)=1\)得
\[\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor f(i)=\sum_{i=1}^nI*f(i)
\]
第二种式子
给\(f()\)卷上\(id():id(x)=x\)得
\[\begin{align}
\sum_{i=1}^nid*f(i)&=\sum_{i=1}^nf(i)\sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}id(j)\\
&=\sum_{i=1}^nf(i)\frac{\lfloor\frac{n}{i}\rfloor\times(\lfloor\frac{n}{i}\rfloor+1)}{2}\\
&=\frac{\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor^2 f(i)+\sum_{i=1}^nI*f(i)}{2}
\end{align}
\]
\[\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor^2 f(i)=2\times\sum_{i=1}^nid*f(i)-\sum_{i=1}^nI*f(i)
\]
