莫比乌斯反演 学习笔记

Posted 2021-05-15 _Veritas

定义和常用公式

1

\\[h=f*g \\iff h(x)=\\sum_{d|x}f(d)g(\\frac{x}{d}) \\]

2

\\([p]=(p?1:0)\\)
\\(\\epsilon(x)=[x=1]\\)
\\(id(x)=x\\)
\\(I(x)=1\\)
\\(\\varphi(x)=\\sum_{i=1}^x [gcd(i,x)=1]\\)
\\(d(x)=\\sum_{d|x}1\\)

3

\\[\\mu * I = \\epsilon, \\varphi * I = id, \\mu * id = \\varphi \\]

4

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

\\(=\\sum_{i=1}^{n}\\sum_{j=1}^m\\epsilon(gcd(i,j))\\)

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

\\(=\\sum_{d=1}^{min(n,m)}\\sum_{i=1}^{n/d}\\sum_{j=1}^{m/d}\\mu(d)\\)

\\(=\\sum_{d=1}^{min(n,m)}\\mu(d)(n/d)(m/d)\\)

5

当公式4中\\(n=m\\)时，有
\\(\\sum_{i=1}^n\\sum_{j=1}^n[gcd(i,j)=1]\\)
\\(=\\sum_{1\\leq i,j\\leq n}[gcd(i,j)=1]\\)
\\(=\\sum_{1\\leq i\\leq j\\leq n}[gcd(i,j)=1]+\\sum_{1\\leq j\\leq i\\leq n}[gcd(i,j)=1]-\\sum_{1\\leq i=j\\leq n}[gcd(i,j)=1]\\)
\\(=2\\sum_{1\\leq i\\leq j\\leq n}[gcd(i,j)=1]-1\\)
\\(=2\\sum_{1\\leq j\\leq n}\\sum_{1\\leq i\\leq j}[gcd(i,j)=1]-1\\)
\\(=2\\sum_{1\\leq j\\leq n}\\varphi(j)-1\\)
\\(=2\\sum_{j=1}^n\\varphi(j)-1\\)

常用模板

线性筛

整除分块

例题

luoguP1390

\\(\\sum_{i=1}^{n}\\sum_{j=i+1}^{n}gcd(i,j)\\)
\\(=\\sum_{i}\\sum_{j}\\sum_{k}k[k=gcd(i,j)][1\\leq i<j\\leq n]\\)
\\(=\\sum_{k}k\\sum_{i}\\sum_{j}[k=gcd(i,j)][1\\leq i<j\\leq n]\\)
\\(=\\sum_{k}k\\sum_{i}\\sum_{j}[gcd(i,j)=1][1\\leq i<j\\leq n/k]\\)
\\(=\\sum_{k}k\\sum_{j=1}^{n/k}\\sum_{i=1}^{j-1}[gcd(i,j)=1]\\)
\\(=\\sum_{k}k\\sum_{j=1}^{n/k}(\\varphi(j)-[gcd(j,j)=1])\\)
\\(=\\sum_{k}k(-1+\\sum_{j=1}^{n/k}\\varphi(j))\\)
线性筛+前缀和
\\(O(n)\\)

luoguP2398

\\(\\sum_{i=1}^n\\sum_{j=1}^ngcd(i,j)\\)
\\(=\\sum_{i=1}^n\\sum_{j=1}^n\\sum_{k=1}^nk[k=gcd(i,j)]\\)
\\(=\\sum_{k=1}^nk\\sum_{i=1}^{n/k}\\sum_{j=1}^{n/k}[gcd(i,j)=1]\\)
\\(=\\sum_{k=1}^nk\\sum_{d=1}^{n/k}\\mu(d)(n/k/d)(n/k/d)\\)(由公式4)
\\(=\\sum_{k=1}^n\\sum_{d|k}(k/d)\\mu(d)(n/k)(n/k)\\)
\\(=\\sum_{k=1}^n(n/k)(n/k)\\sum_{d|k}(k/d)\\mu(d)\\)
\\(=\\sum_{k=1}^n(n/k)(n/k)\\varphi(k)\\)
或
\\(\\sum_{i=1}^n\\sum_{j=1}^ngcd(i,j)\\)
\\(=\\sum_{i=1}^n\\sum_{j=1}^n\\sum_{d|gcd(i,j)}\\varphi(d)\\)
\\(=\\sum_{d}\\sum_{i=1}^{n/d}\\sum_{j=1}^{n/d}\\varphi(d)\\)
\\(=\\sum_{d}(n/d)(n/d)\\varphi(d)\\)
线性筛即可
复杂度\\(O(n)\\)

luoguP2257

\\(\\sum_{p}\\sum_{i=1}^N\\sum_{j=1}^M[p=gcd(i,j)]\\)
\\(=\\sum_{p}\\sum_{d=1}^{min(n/p,m/p)}\\mu(d)(n/p/d)(m/p/d)\\)(由P2398)
\\(=\\sum_{k}\\sum_{p|k}\\mu(k/p)(n/k)(m/k)\\)
\\(=\\sum_{k}(n/k)(m/k)\\sum_{p|k}\\mu(k/p)\\)
\\(=\\sum_{k}(n/k)(m/k)f(k)\\)
其中\\(f(x)=\\sum_{p|x}\\mu(x/p)\\)
\\(f\\)可通过枚举所有素数，枚举它们的倍数做到\\(O(n\\log\\log n)\\)
也可由线性筛(\\(f(px)=mu(x)-f(x)\\))\\(O(n)\\)求出
然后进行整除分块
总复杂度\\(O(n+T\\sqrt{n})\\)

luoguP2568

P2257中取\\(N=M\\)即可,但此题时间限制为P2257的\\(\\frac{1}{4}\\)
或
\\(\\sum_{p}\\sum_{i=1}^n\\sum_{j=1}^n[p=gcd(i,j)]\\)
\\(=\\sum_{p}\\sum_{i=1}^{n/p}\\sum_{j=1}^{n/p}[gcd(i,j)=1]\\)
\\(=\\sum_{p}(2\\sum_{j=1}^{n/p}\\varphi(j)-1)\\)
线性筛+前缀和即可
复杂度\\(O(n)\\)

luoguP3327

先推导一个关于\\(d(ij)\\)的公式

设
\\(i=p_1^{a_1}p_2^{a_2}p_3^{a_3}...\\)
\\(j=p_1^{b_1}p_2^{b_2}p_3^{b_3}...\\)
又设\\(X\\)为\\(i\\)的因数，\\(Y\\)为\\(j\\)的因数
\\(X=p_1^{x_1}p_2^{x_2}p_3^{x_3}...\\)
\\(Y=p_1^{y_1}p_2^{y_2}p_3^{y_3}...\\)
则\\(\\frac{jX}{Y}=p_1^{x_1+b_1-y_1}p_2^{x_2+b_2-y_2}p_3^{x_3+b_3-y_3}...\\)

对于任意的\\(i\\)：
当\\(x_i=0,y_i>0\\)时，\\(x_i+b_i-y_i\\)取\\(0\\)到\\(b_i-1\\)的所有值；
当\\(x_i=0,y_i=0\\)时，\\(x_i+b_i-y_i\\)取\\(b_i\\)；
当\\(x_i>0,y_i=0\\)时，\\(x_i+b_i-y_i\\)取\\(1+b_i\\)到\\(a_i+b_i\\)的所有值；
所以当\\(\\forall i,\\min(x_i,y_i)=0\\)即\\(gcd(X,Y)=1\\)时，\\(\\{\\frac{jX}{Y}\\}\\)为\\(ij\\)的所有因数
所以得到下式：

\\[d(ij)=\\sum_{x|i}\\sum_{y|j}[gcd(x,y)=1] \\]

下面看题
\\(\\sum_{i=1}^{n}\\sum_{j=1}^{m}d(ij)\\)
\\(=\\sum_{i=1}^{n}\\sum_{j=1}^{m}\\sum_{x|i}\\sum_{y|j}[gcd(x,y)=1]\\)
\\(=\\sum_{x=1}^{n}\\sum_{y=1}^{m}\\sum_{i=1}^{n/x}\\sum_{j=1}^{m/y}[gcd(x,y)=1]\\)
\\(=\\sum_{x=1}^{n}\\sum_{y=1}^{m}(n/x)(m/y)[gcd(x,y)=1]\\)
\\(=\\sum_{x=1}^{n}\\sum_{y=1}^{m}(n/x)(m/y)\\sum_{d|gcd(x,y)}\\mu(d)\\)
\\(=\\sum_{d=1}^{\\min(m,n)} \\sum_{x=1}^{n/d}\\sum_{y=1}^{m/d}(n/(xd))(m/(yd))\\mu(d)\\)
\\(=\\sum_{d=1}^{\\min(m,n)} \\mu(d) (\\sum_{x=1}^{n/d}(n/d)/x) (\\sum_{y=1}^{m/d}(m/d)/y)\\)
\\(=\\sum_{d=1}^{\\min(m,n)} \\mu(d) (\\sum_{k=1}^{n/d}(n/d)/k) (\\sum_{k=1}^{m/d}(m/d)/k)\\)
\\(=\\sum_{d=1}^{\\min(m,n)} \\mu(d) f(n/d) f(m/d)\\)
其中\\(f(x)=\\sum_{k=1}^{x}x/k\\)，可\\(O(n)\\)预处理
再用线性筛\\(O(n)\\)求出\\(\\mu\\)的前缀和
整除分块即可
\\(O(n+T\\sqrt{n})\\)

luoguP2522

类似二维前缀和
\\(\\sum_{x=a}^{b}\\sum_{y=c}^{d}[gcd(x,y)=k]\\)
\\(=\\sum_{x=a}^{b}(\\sum_{y=1}^{d}[gcd(x,y)=k]-\\sum_{y=1}^{c-1}[gcd(x,y)=k])\\)
\\(=\\sum_{x=a}^{b}\\sum_{y=1}^{d}[gcd(x,y)=k]-\\sum_{x=a}^{b}\\sum_{y=1}^{c-1}[gcd(x,y)=k]\\)
\\(=\\sum_{y=1}^{d}\\sum_{x=a}^{b}[gcd(x,y)=k]-\\sum_{y=1}^{c-1}\\sum_{x=a}^{b}[gcd(x,y)=k]\\)
\\(=\\sum_{y=1}^{d}(\\sum_{x=1}^{b}[gcd(x,y)=k]-\\sum_{x=1}^{a-1}[gcd(x,y)=k])-\\sum_{y=1}^{c-1}(\\sum_{x=1}^{b}[gcd(x,y)=k]-\\sum_{x=1}^{a-1}[gcd(x,y)=k])\\)
\\(=\\sum_{y=1}^{d}\\sum_{x=1}^{b}[gcd(x,y)=k]-\\sum_{y=1}^{d}\\sum_{x=1}^{a-1}[gcd(x,y)=k]-\\sum_{y=1}^{c-1}\\sum_{x=1}^{b}[gcd(x,y)=k]+\\sum_{y=1}^{c-1}\\sum_{x=1}^{a-1}[gcd(x,y)=k])\\)
\\(=s[b][d]-s[a-1][d]-s[b][c-1]+s[a-1][c-1]\\)
其中
\\(s[n][m]\\)
\\(=\\sum_{x=1}^{n}\\sum_{y=1}^{m}[gcd(x,y)=k]\\)
\\(=\\sum_{x=1}^{n/k}\\sum_{y=1}^{m/k}[gcd(x,y)=1]\\)
\\(=\\sum_{d=1}^{min(n/k,m/k)}\\mu(d)(n/k/d)(m/k/d)\\)
线性筛+整除分块
复杂度\\(O(n+T\\sqrt n)\\)

luoguP3768

\\(\\sum_{i=1}^{n}\\sum_{j=1}^{n}ijgcd(i,j) \\pmod p\\)
\\(\\equiv \\sum_{i=1}^{n}\\sum_{j=1}^{n}\\sum_{k}ijk[k=gcd(i,j)] \\pmod p\\)
\\(\\equiv \\sum_{k}\\sum_{i=1}^{n/k}\\sum_{j=1}^{n/k}ijk^3[gcd(i,j)=1] \\pmod p\\)
\\(\\equiv \\sum_{k}k^3\\sum_{i=1}^{n/k}\\sum_{j=1}^{n/k}ij[gcd(i,j)=1] \\pmod p\\)
\\(\\equiv \\sum_{k}k^3\\sum_{i=1}^{n/k}\\sum_{j=1}^{n/k}ij\\sum_{d|gcd(i,j)}\\mu(d) \\pmod p\\)
\\(\\equiv \\sum_{k}k^3\\sum_{d}\\sum_{i=1}^{n/k/d}\\sum_{j=1}^{n/k/d}ijd^2\\mu(d) \\pmod p\\)
\\(\\equiv \\sum_{k}k^3\\sum_{d}d^2\\mu(d)\\sum_{i=1}^{n/k/d}i\\sum_{j=1}^{n/k/d}j \\pmod p\\)
\\(\\equiv \\sum_{k}k^3\\sum_{d}d^2\\mu(d)((\\sum_{i=1}^{n/k/d}i)(\\sum_{j=1}^{n/k/d}j)) \\pmod p\\)
\\(\\equiv \\sum_{k}k^3\\sum_{d}d^2\\mu(d)s[n/k/d]^2 \\pmod p\\)
后面要用杜教筛 等我学会了再接着写