定义两个数论函数运算\(*\),
若\(h=f*g\),则
\[h(n)=\sum_{d|n}f(d)g(\frac nd)\]
它满足一些性质:
\(f*g=g*f\)
\(f*(g*h)=(f*g)*h\)
因为\(\sum_{(ij)k=n}(f(i)g(j))h(k)=\sum_{i(jk)=n}f(i)(g(j)h(k))\)
\((f g)*h=f*h g*h\)
\((xf)*g=x(f*g)\)
设单位元\(\epsilon(n) = [n = 1]\),则\(\epsilon*f=f\)
对于每个\(f(1)\neq 0\)的函数\(f\),存在逆元\(g\)使得\(f*g=\epsilon\)
如何求一个函数的逆元呢?
令\(g(n)\)满足以下式子:
\[g(n)=\frac 1{f(1)}\left(\epsilon(n)-\sum_{i|n,i\neq1}f(i)g(\frac ni)\right)\]
这样的话:
\[\sum_{i|n}f(i)g(\frac ni) \=f(1)g(n) \sum_{i|n, i\neq 1} f(i)g(\frac ni) \=\epsilon(n) \\therefore f*g=\epsilon\]
定义一个函数\(\mu\)使得\(\mu*1=\epsilon\)
这样的话,如果\(g*1=f\),则\(f*\mu=g\)
即:如果\(f(n)=\sum_{d|n}g(d)\),则\(g(n)=\sum_{d|n}\mu(d)f(\frac nd)\)
是不是很简单
若\(n<m\)求:
\[\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=1]\]
设\(f=\epsilon\),\(\because \epsilon=1*\mu,\;\therefore g=\mu\)
于是原式变为:
\[\sum_{i=1}^n\sum_{j=1}^m\sum_{d|i,d|j}\mu(d) \=\sum_{d=1}^n \mu(d)\sum_{i=1}^n\sum_{j=1}^m[d|i][d|j] \=\sum_{d=1}^n \mu(d)\left[\frac nd\right]\left[\frac md\right]\]
预处理\(\mu\),数论分块就可以\(O(\sqrt n)\)求了
\[\sum_{i=1}^n\sum_{j=1}^m gcd(i,j)\]
设\(f=id\),\(\because id=1*\varphi,\;\therefore g=\varphi\)
于是原式变为:
\[\sum_{d=1}^n\varphi(d)\left[\frac nd\right]\left[\frac md\right]\]
\[\sum_{i=1}^n\sum_{j=1}^m\sigma(gcd(i,j))\]
设\(f=\sigma\),\(\because\sigma=1*(\mu*\sigma),\;\therefore g=\mu*\sigma\)
于是原式变为:
\[\sum_{d=1}^ng(d)\left[\frac nd\right]\left[\frac md\right]\]
因为积性函数可以线性筛,所以也可以预处理
\[\sum_{i=1}^n\sum_{j=1}^mf(gcd(i,j))\]
设\(g=\mu*f\),如果我们能预处理出\(f\),那么可以这样求\(g\)
(当然是蒯的啦)
void get_g_1(int N, const int *f, int *g){ for (int i = 1; i <= N; i ) g[i] = 0; for (int i = 1; i <= N; i ) for (int j = 1; i * j <= N; j ) g[i * j] = (g[i * j] mu[i] * f[j]) % mod;} // 依照定义,O(nlogn)void get_g_2(int N, const int *f, int *g){ for (int i = 1; i <= N; i ) g[i] = f[i]; for (int i = 1; i <= N; i ) for (int j = 2; i * j <= N; j ) g[i * j] = (g[i * j] - g[i]) % mod;} // 类似求狄利克雷卷积逆的方式,不需要线性筛 mu ,O(nlogn)void get_g_3(int N, const int *f, int *g){ for (int i = 1; i <= N; i ) g[i] = f[i]; for (int i = 0; i < prime_count; i ) for (int j = N / prime[i]; j >= 1; j--) g[j * prime[i]] = (g[j * prime[i]] - g[j]) % mod;} // Magic! O(nloglogn)
于是原式变为:
\[\sum_{d=1}^ng(d)\left[\frac nd\right]\left[\frac md\right]\]
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