[BZOJ3994] [SDOI2015]約數個數和

首先有一個結論$d(i*j)=\sum \limits _{x\mid i} \sum \limits _{y\mid j}\left [ gcd(i,j)=1 \right ] $函數

結論有點難想,不過以後就比較簡單了。spa

$\sum \limits _{i=1}^{N}\sum \limits _{j=1}^{M} \sum \limits _{x\mid i} \sum \limits _{y\mid j}\left [ gcd(x,y)=1 \right ]$blog

$\sum \limits _{i=1}^{N} \sum \limits _{j=1}^{M} \sum \limits _{x\mid i}\sum \limits _{y\mid j}\sum \limits _{g\mid gcd(i,j)}u(g)$it

$\sum \limits _{x=1}^{N}\sum \limits _{{i}'=1}^\left \lfloor \frac{N}{x} \right \rfloor \sum \limits _{y=1}^{M}\sum \limits _{{j}'=1}^{\left \lfloor \frac{M}{y} \right \rfloor}\sum \limits _{g\mid gcd(x,y)}u(g)$gc

$\sum \limits _{x=1}^{N}\sum \limits _{y=1}^{M} \sum \limits _{g\mid gcd(x,y)}u(g) \left \lfloor \frac{N}{xg} \right \rfloor \left \lfloor \frac{M}{yg} \right \rfloor$im

$\sum \limits _{g=1}^{min(N,M)} u(g)\sum \limits _{x=1}^{\left \lfloor \frac{N}{g} \right \rfloor}\sum \limits _{y=1}^{\left \lfloor \frac{M}{g} \right \rfloor}\left \lfloor \frac{N}{xg} \right \rfloor \left \lfloor \frac{M}{yg} \right \rfloor$d3

設$f(T)=\sum \limits _{T=1}^{N}\left \lfloor \frac{N}{T} \right \rfloor$img

那麼原式=$\sum \limits _{g=1}^{min(N,M)} u(g) f(\left \lfloor \frac{N}{g} \right \rfloor) f(\left \lfloor \frac{M}{g} \right \rfloor)$co

整除分塊預處理f函數就能夠了。d3

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