数学

# 懵逼鸟斯繁衍

$F(n)=\sum_{d|n}f(d)$

$F(n)$与1的雷子卷积

$F=f*1 <=> f=F*\mu$

$$
\huge\sum_{d|n}\mu(x)=\{^{1(n=1)}_{0(n>1)}
$$

 

$$
\huge\sum_{i=1}^{n}\sum_{j=1}^{m}[\gcd(i,j)==1]
$$

等价于

$$
\huge\sum_{i=1}^{n}\sum_{j=1}^{m}\huge\sum_{d|gcd(i,j)}\mu(d)
$$

等价于

$$
\huge\sum_{d=1}^{n}[n/d][m/d]\mu(d)
$$

将n/d分块处理所有n/d相等的数
$$
\huge\sum_{d=1}^{n}[n/d][m/d]
$$

```c++
for(int i=1;i<=min(n,m);i=last+1){
last=min(n/(n/i),m/(m/i));
ans+=(n/i)*(m/i)*(last-i+1);
}

for(int i=1;i<=min(n,m);i=last+1){
last=min(n/(n/i),m/(m/i));
ans+=(n/i)*(m/i)*(sum[last]-sum[i-1]);
}
//sum为mu的前缀和
```

形如

$x==1$

这样的式子

用
$$
\huge\sum_{d|x}\mu(d)
$$
代替

### eg

求$1\le i \le n, 1 \le j \le m$
$$
\huge\sum_{i=1}^{n}\sum_{j=1}^{m}[\gcd(i,j)为质数]
$$

$$
\huge\sum_{p}\sum_{i=1}^{n/p}\sum_{j=1}^{m/p}[\gcd(i,j)==1]
$$

$$
\huge\sum_{p为质数}\sum_{i=1}^{n/p}\sum_{j=1}^{m/p}\sum_{d|gcd(i,j)}\mu(d)
$$

$$
\huge\sum_{p为质数}\sum_{d=1}^{\frac{n}{p}}[\frac{\frac{n}{p}}{d}][\frac{\frac{m}{p}}{d}]\mu(d)
$$

$$
\huge\sum_{dp\le n}\sum_{d=1}^{\frac{n}{p}}[\frac{n}{pd}][\frac{m}{pd}]\mu(d)
$$

令Q=dp
$$
\huge\sum_{Q=1}^{n}[\frac{n}{Q}][\frac{m}{Q}]\sum_{p|q}prime(p)\mu(\frac{Q}{p})
$$

$$
\huge\sum_{Q=1}^{n}[\frac{n}{Q}][\frac{m}{Q}]f(Q)
$$

$$
\huge f(n)=\sum_{p|n}prime(p)\mu(\frac{n}{p})
$$