随笔分类 - 莫比乌斯反演|杜教筛
摘要：\[ ans=\sum_{i=1}^n\sum_{j=1}^m(n\;mod\;i)\times(m\;mod\;j),i\neq j \] 我们假设 \[ n\leq m \] \[ \begin{aligned} ans & =\sum_{i=1}^n\sum_{j=1}^m(n\;mod\;i
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摘要：\[ ans=\sum_{i=1}^N\sum_{j=1}^N\sum_{p=1}^{\left\lfloor\dfrac{N}{j}\right\rfloor}\sum_{q=1}^{\left\lfloor\dfrac{N}{j}\right\rfloor}[gcd(i,j)=1][gcd(p,
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摘要：YY的GCD \(\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)\in prime]\) \(\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\sum_{p\in prime}\sum_{i=1}^{\left\lfloo
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摘要：约数个数和 \[ \begin{aligned} \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_
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摘要：数表 \[ ans=\sum_{i=1}^{n}\sum_{j=1}^{m}[a\geq\sum_{d|gcd(i,j)}d]\times\sum_{d|gcd(i,j)}d \] 我们假设 \[ G(n,m)=\sum_{d|gcd(n,m)}d \] \[ n \leq m \] 则 \[ \b
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摘要：\[ ans=\sum_{i=1}^{a}\sum_{j=1}^b[gcd(a,b)=d] \] 我们假设 \[ a\leq b \] \[ \begin{aligned} ans & = \sum_{i=1}^{a}\sum_{j=1}^b[gcd(a,b)=d] \\ & = \sum_{i=1
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摘要：\[ ans=\sum_{i=1}^n\sum_{j=1}^nlcm(A_i,A_j) \] \[ \begin{aligned} ans & =\sum_{i=1}^n\sum_{j=1}^nlcm(A_i,A_j) \\ & =\sum_{i=1}^n\sum_{j=1}^n\frac{A_i\
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摘要：\[ ans=\sum_{i=1}^n\sum_{j=1}^n(i+j)^kf(gcd(i,j))gcd(i,j) \] \[ \begin{aligned} ans & =\sum_{i=1}^n\sum_{j=1}^n(i+j)^kf(gcd(i,j))gcd(i,j) \\ & =\sum_{
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摘要：\[ ans=\sum_{i=1}^nlcm(i,n) \] \[ \begin{aligned} ans & =\sum_{i=1}^nlcm(i,n) \\ & =\sum_{i=1}^n\frac{i\cdot n}{gcd(i,n)} \\ & =n\times \sum_{i=1}^n\f
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