前置芝士
整除分块
例题
二项式定理
\((x+y)^n=\sum_{i=0}^{n}C_{n}^{i}*x^{i}*y^{n-i}\)
关于\(gcd\)
\(\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)==x] \Leftrightarrow \sum_{i=1}^{\left \lfloor \frac{n}{x} \right \rfloor} \sum_{j=1}^{\left \lfloor \frac{m}{x} \right \rfloor}[gcd(i,j)==1]\)
\(\sum_{i=1}^{n}\sum_{j=1}^{m}i*j*[gcd(i,j)==x] \Leftrightarrow \sum_{i=1}^{\left \lfloor \frac{n}{x} \right \rfloor} \sum_{j=1}^{\left \lfloor \frac{m}{x} \right \rfloor}i*j*[gcd(i,j)==1]*x^{2}\)
\(\sum_{i=1}^{n}\sum_{j=1}^{m}[x|gcd(i,j)] \Leftrightarrow \left \lfloor \frac{n}{x} \right\rfloor \left \lfloor \frac{m}{x} \right\rfloor\)
\(\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)==1] \Leftrightarrow \sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d|gcd(i,j)}\mu(d)\)
---------------------------------------------\(\Leftrightarrow \sum_{d=1}^{n}\mu(d)*\sum_{i=1}^{n}\sum_{j=1}^{m}[d|gcd(i,j)]\)
狄利克雷卷积&莫比乌斯反演
先乱放了