杜教篩

杜教篩

求積性函數f(x)$f(x)$ 的前綴和S(n)=∑ni=1f(i)$S(n)=\sum _{i=1}^{n}f(i)$

狄利克雷卷積：

(g∗f)(x)=∑d|xg(d)f(xd)$$(g\ast f)(x)=\sum _{d|x}g(d)f({\displaystyle \frac{x}{d}})$$

∑x=1n(g∗f)(x)=∑x=1n∑d|xg(d)f(xd)$$\sum _{x=1}^n(g\ast f)(x)=\sum _{x=1}^n\sum _{d|x}g(d)f({\displaystyle \frac{x}{d}})$$

=∑d=1n∑d|xng(d)f(xd)$$=\sum _{d=1}^n\sum _{d|x}^{n}g(d)f({\displaystyle \frac{x}{d}})$$

=∑d=1ng(d)∑x=1n/df(x)$$=\sum _{d=1}^{n}g(d)\sum _{x=1}^{n/d}f(x)$$

=∑d=1ng(d)S(⌊nd⌋)$$=\sum _{d=1}^{n}g(d)S(\lfloor {\displaystyle \frac{n}{d}}\rfloor )$$

即

∑x=1n(g∗f)(x)=∑d=1ng(d)S(⌊nd⌋)$$\sum _{x=1}^n(g\ast f)(x)=\sum _{d=1}^{n}g(d)S(\lfloor {\displaystyle \frac{n}{d}}\rfloor )$$

我們有

g(1)S(n)=∑i=1ng(i)S(⌊ni⌋)−∑i=2ng(i)S(⌊ni⌋)$$g(1)S(n)=\sum _{i=1}^{n}g(i)S(\lfloor {\displaystyle \frac{n}{i}}\rfloor )-\sum _{i=2}^{n}g(i)S(\lfloor {\displaystyle \frac{n}{i}}\rfloor )$$

g(1)S(n)=∑i=1n(g∗f)(i)−∑i=2ng(i)S(⌊ni⌋)$$g(1)S(n)=\sum _{i=1}^n(g\ast f)(i)-\sum _{i=2}^{n}g(i)S(\lfloor {\displaystyle \frac{n}{i}}\rfloor )$$

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假設我們要求S(n)=∑ni=1μ(i)$S(n)=\sum _{i=1}^n\mu (i)$

我們只需要找出g(x)$g(x)$ 使得(g∗f)(x)$(g\ast f)(x)$ 的前綴和好算就可以了
我們知道

∑d|nμ(d)=[n=1]$$\sum _{d|n}\mu (d)=[n=1]$$

那麼我們可以設g(x)=1$g(x)=1$
這樣的話

(g∗f)(x)=∑d|xf(d)g(dx)=[x=1]$$(g\ast f)(x)=\sum _{d|x}f(d)g({\displaystyle \frac{d}{x}})=[x=1]$$

∑i=1x(g∗f)(x)=1$$\sum _{i=1}^x(g\ast f)(x)=1$$

g(1)S(n)=1−∑i=2nS(⌊ni⌋)$$g(1)S(n)=1-\sum _{i=2}^{n}S(\lfloor {\displaystyle \frac{n}{i}}\rfloor )$$

後面的部分可以數論分塊
首先先線性篩出前面一部分（1000000$1000000$ ）μ(x)$\mu (x)$ 的前綴和，後面記憶化搜索即可

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再假設我們要求S(n)=∑ni=1ϕ(i)$S(n)=\sum _{i=1}^n\varphi (i)$

我們知道

∑d|nϕ(d)=n$$\sum _{d|n}\varphi (d)=n$$

(g∗f)(x)=∑d|xf(d)g(xd)$$(g\ast f)(x)=\sum _{d|x}f(d)g({\displaystyle \frac{x}{d}})$$

我們發現如果令g(x)=1$g(x)=1$

(g∗f)(x)=∑d|xf(d)=x$$(g\ast f)(x)=\sum _{d|x}f(d)=x$$

g(1)S(n)=∑i=1ni−∑i=2ng(i)S(⌊ni⌋)$$g(1)S(n)=\sum _{i=1}^{n}i-\sum _{i=2}^{n}g(i)S(\lfloor {\displaystyle \frac{n}{i}}\rfloor )$$

S(n)=x(x+1)2−∑i=2nS(⌊ni⌋)$$S(n)={\displaystyle \frac{x(x+1)}{2}}-\sum _{i=2}^{n}S(\lfloor {\displaystyle \frac{n}{i}}\rfloor )$$

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51nod1238 最小公倍數之和V3

求

∑i=1n∑j=1nlcm(i,j)$$\sum _{i=1}^n\sum _{j=1}^{n}lcm(i,j)$$

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ans=∑i=1n∑j=1nijgcd(i,j)$$ans=\sum _{i=1}^n\sum _{j=1}^n{\displaystyle \frac{ij}{gcd(i,j)}}$$

=∑d=1n1d∑i=1n∑j=1nij [gcd(i,j)=d]$$=\sum _{d=1}^n{\displaystyle \frac{1}{d}}\sum _{i=1}^n\sum _{j=1}^{n}ij[gcd(i,j)=d]$$

=∑d=1n1d∑i=1n/d∑j=1n/did jd [gcd(i,j)=1]$$=\sum _{d=1}^n{\displaystyle \frac{1}{d}}\sum _{i=1}^{n/d}\sum _{j=1}^{n/d}idjd[gcd(i,j)=1]$$

=∑d=1nd∑i=1n/d∑j=1n/dij [gcd(i,j)=1]$$=\sum _{d=1}^{n}d\sum _{i=1}^{n/d}\sum _{j=1}^{n/d}ij[gcd(i,j)=1]$$

設

f(x)=∑i=1x∑j=1xij[gcd(i,j)=1]$$f(x)=\sum _{i=1}^x\sum _{j=1}^{x}ij[gcd(i,j)=1]$$

=2(∑i=1xi∑j=1ij[gcd(i,j)=1])−1$$=2(\sum _{i=1}^{x}i\sum _{j=1}^{i}j[gcd(i,j)=1])-1$$

=2(∑i=1xi iϕ(i)+[i=1]2)−1$$=2(\sum _{i=1}^{x}i{\displaystyle \frac{i\varphi (i)+[i=1]}{2}})-1$$

=(∑i=1xi2 ϕ(i)+[i=1])−1$$=(\sum _{i=1}^x{i}^2\varphi (i)+[i=1])-1$$

=∑i=1xi2 ϕ(i)$$=\sum _{i=1}^x{i}^2\varphi (i)$$

代回原式子

ans=∑d=1nd∑i=1n/di2 ϕ(i)$$ans=\sum _{d=1}^{n}d\sum _{i=1}^{n/d}{i}^2\varphi (i)$$

設

S(n)=∑i=1ni2 ϕ(i)$$S(n)=\sum _{i=1}^n{i}^2\varphi (i)$$

S(n)=∑i=1n(g∗f)(i)−∑i=2ng(i)S(ni)$$S(n)=\sum _{i=1}^n(g\ast f)(i)-\sum _{i=2}^{n}g(i)S({\displaystyle \frac{n}{i}})$$

推一下卷積

∑i=1n(g∗f)(x)=∑i=1n∑d|if(d)g(id)$$\sum _{i=1}^n(g\ast f)(x)=\sum _{i=1}^n\sum _{d|i}f(d)g({\displaystyle \frac{i}{d}})$$

=∑i=1n∑d|id2ϕ(d)g(id)$$=\sum _{i=1}^n\sum _{d|i}{d}^2\varphi (d)g({\displaystyle \frac{i}{d}})$$

我們發現設g(x)=x2$g(x)=x^2$ 就可以把d2$d^2$ 約掉

=∑i=1n∑d|id2ϕ(d)i2d2$$=\sum _{i=1}^n\sum _{d|i}{d}^2\varphi (d){\displaystyle \frac{i^2}{d^2}}$$

=∑i=1ni2∑d|iϕ(d)$$=\sum _{i=1}^n{i}^2\sum _{d|i}\varphi (d)$$

=∑i=1ni3$$=\sum _{i=1}^n{i}^3$$

=(n(n+1)2)2$$=({\displaystyle \frac{n(n+1)}{2}}{)}^2$$

g(1)S(n)=∑i=1n(g∗f)(x)−∑i=2ng(i)S(ni)$$g(1)S(n)=\sum _{i=1}^n(g\ast f)(x)-\sum _{i=2}^{n}g(i)S({\displaystyle \frac{n}{i}})$$

S(n)=(n(n+1)2)2−∑i=2ni2S(ni)$$S(n)=({\displaystyle \frac{n(n+1)}{2}}{)}^2-\sum _{i=2}^n{i}^{2}S({\displaystyle \frac{n}{i}})$$

我們知道

∑i=1ni2=n(n+1)(2n+1)6$$\sum _{i=1}^n{i}^2={\displaystyle \frac{n(n+1)(2n+1)}{6}}$$

我們可以杜教篩S(n)$S(n)$

ans=∑d=1nd S(nd)$$ans=\sum _{d=1}^{n}dS({\displaystyle \frac{n}{d}})$$