积性函数$$forall p,q wedge gcd(p,q)=1 , f(pq)=f(p)*f(q)$$
(mu)函数定义
$$
n=a_1{p_1}*a_2{p_2}cdots*a_k^{p_k}
mu(n)=left{
egin{array}{lr}
1,n = 1
(-1)^k,a_i=1
0,otherwise
end{array}
ight.
[
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##**$mu$函数性质**
###1.]
sum_{d|n}mu(d)=[n=1]
[([n=1]表示仅当n=1时返回值为1,其余为0)
###2.]
sum_{d|n}{mu(d)over{d}}={varphi(n)over{n}}
[
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##**线性筛莫比乌斯函数**
```cpp
void init()
{
memset(vis,0,sizeof vis);
mu[1]=1;
for (int i=2;i<=n;++i)
{
if (!vis[i])
{
mu[i]=-1;
prime[++tot]=i;
}
for (int j=1;j<=tot&&i*prime[j]<=n;++j)
{
vis[i*prime[j]]=1;
if (i%prime[j]==0)
{
mu[i*prime[j]]=0;
break;
}
else
{
mu[i*prime[j]]=-mu[i];
}
}
}
}
```
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##**莫比乌斯反演**
### 若$$F(n)=sum_{d|n}f(d)]
则有 $$f(n)=sum_{d|n}mu(d)*F({{n}over{d}})$$
证明
$$sum_{d|n}mu(d)F(frac{n}{d})=sum_{d|n}mu(d)sum_{k|frac{n}{d}}f(k)
=sum_{p|n}f(p)sum_{k|frac{n}{p}}mu(k) (p=dk)\
=sum_{p|nwedge p
eq n}f(p)sum_{k|frac{n}{p}}mu(k)+f(n)sum_{k|1}mu(k)\=0+f(n)
=f(n)
[
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###**另一种形式**
### 若$$F(n)=sum_{n|d}f(d)]
则有 $$f(n)=sum_{n|d}mu(frac {d} {n})*F({d})$$
另一种形式证明
$$sum_{n|d}mu(frac{d}{n})F(d)=sum_{n|d}mu(frac{d}{n})sum_{d|p}f(p)
=sum_{q=1}^{+infty}mu(q)sum_{nq|p}f(p) (q=frac{d}{n})
=sum_{n|p}f(p)sum_{q|frac{p}{n}}mu(q)
=sum_{n|p wedge n
eq p}f(p)sum_{q|frac{p}{n}}mu(q)+f(n)sum_{q|1}mu(q)
=0+f(n)=f(n)
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###$$f(d)=sum_{i=1}^{n} sum_{j=1}^{m} [gcd(i,j)=d]]