发现根本不会。。复习一下

1.卷积

狄利克雷卷积

\[(f * g)(n) = \sum_{d|n}f(d)g(\frac {n} {d})\]

2.定义数论函数

\[\epsilon(n) = [n == 1]\]

\[id(n) = n\]

\[1(n) = 1\]

\[\varphi(n) = \sum_{d|n}1\]

性质

\[\sum_{i = 1}^{n} [(n, i) = 1]* i = \frac{[n = 1] + n * \varphi(n)}{2}\]

积性函数的点积和狄利克雷卷积也是积性函数

3.常见的数论函数卷积

\[\varphi * 1 = id\]

\[\mu * 1 = \epsilon\]

\[\mu * id = \varphi\]

\[1 * 1 = \sigma\]

\[id * 1 = \sigma_0\]

\[\epsilon * f = f\]

注:

\[d(n, m) = \sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i, j) = 1]\]

4.mobius反演

形式一

\[g = f * 1\]

\[f = g * \mu\]

形式二

\[g(n) = \sum_{n|d}f(d)\]

\[f(n) = \sum_{n|d}\mu(\frac{d}{n})g(d)\]

5.例子

\[\sum_{i=1}^{n}\sum_{j=1}^{m}gcd(i, j) ^ k\]

枚举\(d\)

\[\sum_{d=1}^{n} d^k \sum_{i=1}^{\frac{n}{d}} \sum_{j=1}^{\frac{m}{d}} [gcd(i, j)=1]\]

有

\[\sum_{d|n} \mu(d) = [n = 1]\]

代入可以交换求和顺序 并且计算倍数可以得到

\[\sum_{d=1}^{n} d^k \sum_{e=1}^{n} \mu(e) \frac{n}{de} \frac {m}{de}\]

设\(D = ed\)

\[\sum_{D=1}^{n} \sum_{d|D}d^k \mu(\frac{D}{d}) \frac{n}{de} \frac {m}{de}\]

记\(f(D) = \sum_{d|D}d^k \mu(\frac{D}{d})\) 为积性函数

\[f(D) = \prod_{p_i}f(p_i^{x_i})\]

\[= \prod_{p_i} p_i^{kx_i}\mu(1) + pi^{k(x_i-1)} \mu(pi)\]

\[= \prod_{p_i} p_i^{k(x_i - 1)}(pi^k-1)\]

线性筛就好了