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
\(\sum_{i=1}^N \sum_{j=1}^M (i,j)\)
\(f(d)=\sum_{i=1}^N \sum_{j=1}^M d[(i,j)=d]=\sum_{i=1}^{\lfloor \frac{\min(N,M)}{d} \rfloor} d\mu(i) \lfloor \frac{N}{id} \rfloor \lfloor \frac{M}{id} \rfloor\)
\(Ans=\sum_{d=1}^{\min(N,M)} f(d)=\sum_{d=1}^{\min(N,M)} \sum_{i=1}^{\lfloor \frac{\min(N,M)}{d} \rfloor} d\mu(i) \lfloor \frac{N}{id} \rfloor \lfloor \frac{M}{id} \rfloor\)
设\(w=id\)
\(Ans=\sum_{w=1}^{\min(N,M)} \sum_{d|w} d\mu(\frac{w}{d}) \lfloor \frac{N}{w} \rfloor \lfloor \frac{M}{w} \rfloor\)
\(\sum_{d|w} d\mu(\frac{w}{d})=\phi(w)\)显然是积性函数，线性筛后做下前缀和，离线\(\Theta(\min(N,M))\)
\(\sum_{w=1}^{\min(N,M)} \lfloor \frac{N}{w} \rfloor \lfloor \frac{M}{w} \rfloor\) 整除分块可以做到在线\(\Theta(\sqrt{N}+\sqrt{M})\)
多组询问下总复杂度\(\Theta(\min(N,M)+T(\sqrt{N}+\sqrt{M}))\)
T3
\(\sum_{i=1}^N \sum_{j=1}^M \frac{ij}{(i,j)}\)
\(f(d)=\sum_{i=1}^{\lfloor \frac{N}{d} \rfloor} \sum_{j=1}^{\lfloor \frac{M}{d} \rfloor} ijd[(i,j)=1]=d \sum_{i=1}^{\lfloor \frac{N}{d} \rfloor} \sum_{j=1}^{\lfloor \frac{M}{d} \rfloor} ij[(i,j)=1]\)
\(Ans=\sum_{d=1}^{\min(N,M)} f(d)\)
\(Ans=\sum_{d=1}^{\min(N,M)} d \sum_{i=1}^{\lfloor \frac{N}{d} \rfloor} \sum_{j=1}^{\lfloor \frac{M}{d} \rfloor} ij\sum_{n|(i,j)} \mu(n)\)
\(Ans=\sum_{d=1}^{\min(N,M)} d \sum_{n=1}^{\lfloor \frac{\min(N,M)}{d} \rfloor} n (\sum_{i=1}^{\lfloor \frac{N}{dn} \rfloor} i)n(\sum_{j=1}^{\lfloor \frac{M}{dn} \rfloor} j)\mu(n)\)
设\(w=dn\)
\(Ans=\sum_{w=1}^{\min(N,M)} (\sum_{i=1}^{\lfloor \frac{N}{w} \rfloor} i)(\sum_{j=1}^{\lfloor \frac{M}{w} \rfloor} j) w\sum_{n|w} n \mu(n)\)
线筛前缀和+整除分块
复杂度与上题相同
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] $
设\(w=(i,j)\)
\(\sum_{w=1}^{\min(N,M)} \mu(w) \sum_{i=1}^{\lfloor \frac{N}{w} \rfloor} \sum_{j=1}^{\lfloor \frac{M}{w} \rfloor} \lfloor \frac{N}{iw} \rfloor \lfloor \frac{M}{jw} \rfloor\)
整除分块+线筛前缀和
复杂度仍然与上题相同
