[笔记]杜教筛核心原理

∑ni=1∑mj=1gcd(i,j)$\sum _{i=1}^n\sum _{j=1}^{m}gcd(i,j)$
=∑nd=1d∗f(d)$=\sum _{d=1}^{n}d\ast f(d)$
=∑nd=1d∗∑d|jμ(j/d)∗F(d)$=\sum _{d=1}^{n}d\ast \sum _{d|j}\mu (j/d)\ast F(d)$
=∑i×j<=ni∗μ(j)∗F(i∗j)$=\sum _{i\times j<=n}i\ast \mu (j)\ast F(i\ast j)$
=∑nT=1F(T)∗ϕ(T)$=\sum _{T=1}^{n}F(T)\ast \varphi (T)$
=∑nT=1[nT][mT]ϕ(T)$=\sum _{T=1}^n[\frac{n}{T}][\frac{m}{T}]\varphi (T)$

———-杜教筛：
已知问题：

S(n)=∑i=1nf(i)$$S(n)=\sum _{i=1}^{n}f(i)$$

取一个积性函数g使得f*g方便计算前缀和并做如下变换

(g∗f)(i)=∑d|if(d)∗g(i/d)$$(g\ast f)(i)=\sum _{d|i}f(d)\ast g(i/d)$$

∑i=1n(g∗f)(i)=∑i=1n∑d|if(d)∗g(i/d)$$\sum _{i=1}^n(g\ast f)(i)=\sum _{i=1}^n\sum _{d|i}f(d)\ast g(i/d)$$

=∑i∗j<=nf(i)∗g(j)$$=\sum _{i\ast j<=n}f(i)\ast g(j)$$

=∑i=1ng(i)∑j=1n/if(j)$$=\sum _{i=1}^{n}g(i)\sum _{j=1}^{n/i}f(j)$$

=∑i=1ng(i)∗S(n/i)$$=\sum _{i=1}^{n}g(i)\ast S(n/i)$$

就可以得到

S(n)=1∗S(n)=g(1)∗S(n)$$S(n)=1\ast S(n)=g(1)\ast S(n)$$

=∑i=1ng(i)∗S(n/i)−∑i=2ng(i)∗S(n/i)$$=\sum _{i=1}^{n}g(i)\ast S(n/i)-\sum _{i=2}^{n}g(i)\ast S(n/i)$$

=∑i=1n(g∗f)(i)−∑i=2ng(i)∗S(n/i)$$=\sum _{i=1}^n(g\ast f)(i)-\sum _{i=2}^{n}g(i)\ast S(n/i)$$

后面就是递归计算了，注意要记忆化S