Mobius反演的套路

T1

\(\sum_{i=1}^N \sum_{j=1}^M [(i,j)=1]\)

\(f(d)=\sum_{i=1}^N \sum_{j=1}^M [(i,j)=d]\)

\(g(d)=\sum_{i=1}^N \sum_{i=1}^M [d|(i,j)]=\lfloor \frac{N}{d} \rfloor \lfloor \frac{M}{d} \rfloor\)

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

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

\(f(1)=\sum_{i=1}^{\min(N,M)} \mu(i)\lfloor \frac{N}{i} \rfloor \lfloor \frac{M}{i} \rfloor\)

T2

\(\sum_{i=1}^N \sum_{j=1}^M (i,j)\)

\(f(d)=\sum_{i=1}^N \sum_{j=1}^M d[(i,j)=d]=\sum_{i=1}^{\lfloor \frac{\min(N,M)}{d} \rfloor} d\mu(i) \lfloor \frac{N}{id} \rfloor \lfloor \frac{M}{id} \rfloor\)

\(Ans=\sum_{d=1}^{\min(N,M)} f(d)=\sum_{d=1}^{\min(N,M)} \sum_{i=1}^{\lfloor \frac{\min(N,M)}{d} \rfloor} d\mu(i) \lfloor \frac{N}{id} \rfloor \lfloor \frac{M}{id} \rfloor\)

設\(w=id\)

\(Ans=\sum_{w=1}^{\min(N,M)} \sum_{d|w} d\mu(\frac{w}{d}) \lfloor \frac{N}{w} \rfloor \lfloor \frac{M}{w} \rfloor\)

\(\sum_{d|w} d\mu(\frac{w}{d})=\phi(w)\)顯然是積性函數，線性篩後作下前綴和，離線\(\Theta(\min(N,M))\)

\(\sum_{w=1}^{\min(N,M)} \lfloor \frac{N}{w} \rfloor \lfloor \frac{M}{w} \rfloor\) 整除分塊能夠作到在線\(\Theta(\sqrt{N}+\sqrt{M})\)

多組詢問下總複雜度\(\Theta(\min(N,M)+T(\sqrt{N}+\sqrt{M}))\)

T3

\(\sum_{i=1}^N \sum_{j=1}^M \frac{ij}{(i,j)}\)

\(f(d)=\sum_{i=1}^{\lfloor \frac{N}{d} \rfloor} \sum_{j=1}^{\lfloor \frac{M}{d} \rfloor} ijd[(i,j)=1]=d \sum_{i=1}^{\lfloor \frac{N}{d} \rfloor} \sum_{j=1}^{\lfloor \frac{M}{d} \rfloor} ij[(i,j)=1]\)

\(Ans=\sum_{d=1}^{\min(N,M)} f(d)\)

\(Ans=\sum_{d=1}^{\min(N,M)} d \sum_{i=1}^{\lfloor \frac{N}{d} \rfloor} \sum_{j=1}^{\lfloor \frac{M}{d} \rfloor} ij\sum_{n|(i,j)} \mu(n)\)

\(Ans=\sum_{d=1}^{\min(N,M)} d \sum_{n=1}^{\lfloor \frac{\min(N,M)}{d} \rfloor} n (\sum_{i=1}^{\lfloor \frac{N}{dn} \rfloor} i)n(\sum_{j=1}^{\lfloor \frac{M}{dn} \rfloor} j)\mu(n)\)

設\(w=dn\)

\(Ans=\sum_{w=1}^{\min(N,M)} (\sum_{i=1}^{\lfloor \frac{N}{w} \rfloor} i)(\sum_{j=1}^{\lfloor \frac{M}{w} \rfloor} j) w\sum_{n|w} n \mu(n)\)

線篩前綴和+整除分塊

複雜度與上題相同

T4

\(\sum_{i=1}^N \sum_{j=1}^M d(ij)\)

\(\sum_{i=1}^N \sum_{j=1}^M \sum_{a|i} \sum_{b|j} [(a,b)=1]\)

$\sum_{i=1}^N \sum_{j=1}^M \lfloor \frac{N}{i} \rfloor \lfloor \frac{M}{j} \rfloor[(i,j)=1] $

設\(w=(i,j)\)

\(\sum_{w=1}^{\min(N,M)} \mu(w) \sum_{i=1}^{\lfloor \frac{N}{w} \rfloor} \sum_{j=1}^{\lfloor \frac{M}{w} \rfloor} \lfloor \frac{N}{iw} \rfloor \lfloor \frac{M}{jw} \rfloor\)

整除分塊+線篩前綴和

複雜度仍然與上題相同