莫比乌斯反演学习笔记

莫比乌斯反演

数论分块

用处：用于快速计算一些含有除法向下取整的和式

for example: \(\sum_{i=1}^n \left\lfloor\frac{n}{i}\right\rfloor\)

就是将 \(\left\lfloor\frac{n}{i}\right\rfloor\) 相等的数同时计算

直接说结论吧：
对于一个常数 \(n\) 使得 \(\left\lfloor\frac{n}{i}\right\rfloor = \left\lfloor\frac{n}{j}\right\rfloor\) 并且 \(i\le j\le n\) 的 \(j\) 的最大值为\(\left\lfloor\frac{n}{\left\lfloor\frac{n}{i}\right\rfloor}\right\rfloor\)

证明: 令 \(k=\left\lfloor\frac{n}{i}\right\rfloor\)，\(\therefore k\le \frac{n}{i}\)

\(\therefore \left\lfloor\frac{n}{k}\right\rfloor \ge \left\lfloor\frac{n}{\frac{n}{i}}\right\rfloor = \left\lfloor i\right\rfloor = i\)

\(j=\max\)满足条件的所有\(i = i_{max} = \left\lfloor\frac{n}{i}\right\rfloor = \left\lfloor\frac{n}{\left\lfloor\frac{n}{i}\right\rfloor}\right\rfloor\)

\(By\quad Oi-Wiki\)

模板题：余数求和

题目大意:给定正整数\(n,k(n,k \le 1e9)\),求\(\sum_{i=1}^n{k\mod i}\)

solution:

考虑推式子，将其转为数论分块的形式

\(\sum_{i=1}^n k\mod i\)

\(=\sum_{i=1}^n k-\left\lfloor\frac{k}{i}\right\rfloor*i\)

\(=n*k-\sum_{i=1}^n\left\lfloor\frac{k}{i}\right\rfloor*i\)

后面考虑数论分块，假设左端点为\(l\)，右端点为\(r\)，则该段的值\(=(k/l)*\sum_{i=l}^ri\)

\(done.\)

\(code:\)
```
ll n,k;read(n,k);
ll res = n*k;
for(int l = 1,r;l <= n;l = r+1){
    r = (k/l)?min(k/(k/l),n):n;
    res = res - k/l*(l+r)*(r-l+1)/2;
}
write(res);
```
狄利克雷卷积
对于两个数论函数\(f(x)\)和\(g(x)\),则它们的狄利克雷卷积得到的结果\(h(x)\)定义为

\[h(x)=\sum_{d|x}f(d)g(\frac{n}{d})=\sum_{ab=x}f(a)g(b) \]
简记为 \(h=f*g\)

简单性质：

1.\(交换律：f*g=g*f\)

2.\(结合律：(f*g)*h=f*(g*h)\)

3.\(分配律：(f+g)*h=f*h+g*h\)

两个结论：

1.两个积性函数的狄利克雷卷积也是积性函数

2.积性函数的逆元也是积性函数

详细证明请见\(oi-wiki\)

进入正题

莫比乌斯函数

\(\mu\)为莫比乌斯函数,定义为\(\mu=\begin{cases} 1,& n=1\\ 0,& n\text{含有平方因子}\\ (-1)^k,& k为n的本质不同质因子个数 \end{cases}\)

性质：
1.积性函数

2.\(\sum_{d|x}\mu(d)=\begin{cases} 1,& n=1\\ 0,& n\ne1 \end{cases}\)
\(\quad\)
即\(\mu*1=\epsilon\)

性质二证明：

设\(n=\prod_{i=1}^k p^{c_i}_i , n^{'}=\prod_{i=1}^k p_i\)
则有\(\sum_{d|n}\mu(d) = \sum_{d|n^{'}} = \sum_{i=0}^k\mathrm{C}_k^i*(-1)^i = (1+(-1))^k\)仅当\(k=0\)时为\(1\)

反演结论：\([\gcd(i,j)=1]=\sum_{d|\gcd(i,j)}\mu(d)\)

线性筛筛莫比乌斯函数
运用定义式即可

code：

vector<int> prime;
int mu[N];bitset<N> pd;
inline void mobius(int n){
    mu[1]=1;
    for(int i=2;i<=n;++i){
        if(!pd[i])mu[i]=-1,prime.emplace_back(i);
        for(int j:prime){
            if(i*j>n) break;
            pd[i*j]=true;
            if(i%j == 0) break;
            else mu[i*j]=-mu[i];
        }
    }
}

莫比乌斯反演
1. 应用反演结论\([\gcd(i,j)==1] = \sum_{d|\gcd(i,j)}\mu(d)\)
设\(f(n),g(n)\)为两个数论函数
1. 形式一\(\qquad if\quad f(n)=\sum_{d|n}g(d),then\quad g(n)=\sum_{d|n}\mu(d)f(\frac{n}{d})\)
  
  证明：
  
  \(\begin{aligned} \sum_{d\mid n}\mu(d)f\left(\frac{n}{d}\right) &= \sum_{d\mid n}\mu(d)\sum_{k\mid \frac{n}{d}}g(k)\\ &= \sum_{k\mid n}g(k)\sum_{d\mid \frac{n}{k}}\mu(d)\\ &= \sum_{k\mid n}\left[\frac{n}{k} = 1\right]g(k)\\ &= g(n)\\ \end{aligned}\)
  
  3.形式二\(\qquad if \quad f(n)=\sum_{n|d}g(d),then\quad g(n)=\sum_{n|d}f(d)\mu(\frac{d}{n})\)
  
  证明：
  
  \(\begin{aligned} \sum_{n|d}\mu(\frac{d}{n})f(d) &= \sum_{k=1}^{+\infty}\mu(k)f(kn)\\ &= \sum_{k=1}^{+\infty}\mu(k)\sum_{kn|d}^{+\infty}g(d)\\ &= \sum_{n|d}g(d)\sum_{k|\frac{d}{n}}\mu(k)\\ &= \sum_{n|d}g(d)\epsilon(\frac{d}{n})\\ &= g(n) \end{aligned}\)
- 应用
  1. problem b
    
    题目大意:对于给出的\(n\)个询问，每次求有多少个数对\((x,y)\)，满足 \(a≤x≤b,c≤y≤d\)，且 \(gcd⁡(x,y)=k\)。
    
    solution:
    
    直接求\(\sum_{i=a}^b\sum_{j=c}^d[\gcd(i,j)=k]\)不好求，考虑用简单的容斥原理转化为\(\sum_{i=1}^b\sum_{j=1}^d[\gcd(i,j)=k]-\sum_{i=1}^{a-1}\sum_{j=1}^{d}[\gcd(i,j)=k]-\sum_{i=1}^b\sum_{j=1}^{c-1}[\gcd(i,j)=k]+\sum_{i=1}^{a-1}\sum_{j=1}^{c-1}[\gcd(i,j)=k]\)
    
    则只需要推\(\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=k]\) (假设n\(\le\)m)
    
    \(\begin{aligned} \sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=k] &= \sum_{i=1}^{\left\lfloor\frac{n}{k}\right\rfloor}\sum_{j=1}^{\left\lfloor\frac{m}{k}\right\rfloor}[\gcd(i,j)=1]\\ &= \sum_{i=1}^{\left\lfloor\frac{n}{k}\right\rfloor}\sum_{j=1}^{\left\lfloor\frac{m}{k}\right\rfloor}\sum_{d|\gcd(i,j)}\mu(d)\\ &=\sum_{d=1}^n\mu(d)\sum_{i=1}^{\left\lfloor\frac{n}{k}\right\rfloor}\sum_{j=1}^{\left\lfloor\frac{m}{k}\right\rfloor}[d|i][d|j]\\ &= \sum_{d=1}^n\mu(d)\left\lfloor\frac{n}{kd}\right\rfloor\left\lfloor\frac{m}{kd}\right\rfloor \end{aligned}\)
    
    后面的整除分块加\(\mu\)前缀和即可
    
    \(end\)
    
    code：
```
const int N = 1e7+10;
vector<int> prime;
bitset<N> pd;
int mu[N];
inline void mobius(int n){
    mu[1] = 1;
    for(int i = 2;i <= n; ++i){
        if(!pd[i]) prime.emplace_back(i),mu[i] = -1;
        for(auto j : prime){
            if(i*j > n) break;
            pd[i*j] = true;
            if(i%j == 0) break;
            mu[i*j] = -mu[i];
        }
    }
    for(int i = 1;i <= n; ++i) mu[i] += mu[i-1];
}
inline int solve(int n,int m,int k){
    if(n > m) swap(n,m);
    int res = 0;
    for(int l = 1,r = 0;l <= n;l = r+1){
        r = min(n/(n/l),m/(m/l));
        res = res+(mu[r]-mu[l-1])*(n/(l*k))*(m/(l*k));
    }
    return res;
}
//main
mobius(N-10);
int T;read(T);
while(T--){
    int a,b,c,d,k;read(a,b,c,d,k);
    write(solve(b,d,k)-solve(a-1,d,k)-solve(b,c-1,k)+solve(a-1,c-1,k),'\n');
}
```
  2. YY的gcd
    
    题目大意：\(给定 N,M，求 1≤x≤N，1≤y≤M 且 gcd⁡(x,y) 为质数的 (x,y) 有多少对。\)
    
    solution:
    
    先写出暴力式子\(\sum_{p\in\mathbb{{p}}}\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=p]\)
    
    推倒
    
    \(\begin{aligned} \sum_{p\in\mathbb{{p}}}\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=p] &= \sum_{p\in\mathbb{{p}}}\sum_{i=1}^{\left\lfloor\frac{n}{p}\right\rfloor}\sum_{j=1}^{\left\lfloor\frac{m}{p}\right\rfloor}[\gcd(i,j)=1]\\ &= \sum_{p\in\mathbb{{p}}}\sum_{i=1}^{\left\lfloor\frac{n}{p}\right\rfloor}\sum_{j=1}^{\left\lfloor\frac{m}{p}\right\rfloor}\sum_{d|\gcd(i,j)}\mu(d)\\ &= \sum_{p\in\mathbb{{p}}}\sum_{d=1}^n\mu(d)\left\lfloor\frac{n}{pd}\right\rfloor\left\lfloor\frac{m}{pd}\right\rfloor \end{aligned}\)
    
    (一个套路：)用\(T\)替换\(pd\)
    
    \(\begin{aligned} \sum_{p\in\mathbb{{p}}}\sum_{d=1}^n\mu(d)\left\lfloor\frac{n}{pd}\right\rfloor\left\lfloor\frac{m}{pd}\right\rfloor = \sum_{T=1}^n\left\lfloor\frac{n}{T}\right\rfloor\left\lfloor\frac{m}{T}\right\rfloor\sum_{d|T}\mu(d) \end{aligned}\)
    
    后面的\(\sum_{d|T}\mu(d)\)可以暴力预处理求
    
    \(end\)
    
    code:
```
const int N = 1e7+10;
vector<int> prime;
bitset<N> pd;
int mu[N];
ll f[N];
inline void mobius(int n){
    mu[1] = 1;
    for(int i = 2;i <= n; ++i){
        if(!pd[i]) prime.emplace_back(i),mu[i] = -1;
        for(auto j : prime){
            if(i*j > n) break;
            pd[i*j] = true;
            if(i%j == 0) break;
            mu[i*j] = -mu[i];
        }
    }
    for(auto i:prime){
        for(int j = 1;j*i <= n; ++j){
            f[i*j] += mu[j];
        }
    }
    for(int i = 1;i <= n; ++i) f[i] += f[i-1];
}
inline ll solve(int n,int m){
    if(n > m) swap(n,m);
    ll res = 0;
    for(int l = 1,r = 0;l <= n;l = r+1){
        r = min(n/(n/l),m/(m/l));
        res = res+1ll*(f[r]-f[l-1])*(n/l)*(m/l);
    }
    return res;
}
//main
mobius(N-10);
int T;read(T);
while(T--){
    int n,m;read(n,m);
    write(solve(n,m),'\n');
}
```
  进阶：
  
  3.数表
  
  \(有一张 n×m的数表，其第 i 行第 j 列（1≤i≤n,1≤j≤m）的数值为能同时整除 i 和 j 的所有自然数之和。给定 a，计算数表中不大于 a 的数之和。\)
  
  多出来的\(\le a\)的限制很难评，那就先不考虑了
  
  推式子
  
  \(\begin{aligned} &\sum_{d=1}^n\sigma(d)\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=d]\\ &=\sum_{d=1}^n\sigma(d)\sum_{k=1}^n\mu(k)\left\lfloor\frac{n}{kd}\right\rfloor\left\lfloor\frac{m}{kd}\right\rfloor \end{aligned}\)
  
  \(let\quad T = kd\)
  
  \(\begin{aligned} &\sum_{T=1}^n\left\lfloor\frac{n}{T}\right\rfloor\left\lfloor\frac{m}{T}\right\rfloor\sum_{d|T}\sigma(d)\mu(\frac{T}{d}) \end{aligned}\)
  
  后面明显的狄利克雷卷积，但因为那个\(\sigma(d)\le a\)的限制,所以就寄了。
  
  ~~那就暴力？~~ no，可以考虑离线，将询问按\(a\)升序排序，每次将符合条件的\(\sigma(d)\)插入一个BIT中，查询前缀和即可。
  
  \(end\)
  
  \(code:\)
```
const int N = 1e5+10;
vector<int> prime;
bitset<N> pd;
int mu[N];
int s[N],low[N];
struct node{int val,id;}a[N];
struct Query{int n,m,a,id;}Q[N];
uint ans[N];
class BIT{
private:
    int tree[N];
    inline int lowbit(int x){return (x&(-x));}
public:
    int n = 0;
    inline void update(int pos,int val){for(int i = pos;i <= n;i += lowbit(i)) tree[i] += val;}
    inline int query(int pos){int res = 0;for(int i = pos; i;i -= lowbit(i)) res += tree[i];return res;}
}T;
inline void mobius(int n){
    mu[1] = 1,s[1] = 1;
    for(int i = 2;i <= n; ++i){
        if(!pd[i]) prime.emplace_back(i),mu[i] = -1,s[i] = i+1,low[i] = i;
        for(auto j : prime){
            if(i*j > n) break;
            pd[i*j] = true;
            if(i%j == 0){
                low[i*j] = low[i]*j;
                if(low[i] == i) s[i*j] = s[i]+i*j;
                else s[i*j] = s[i/low[i]]*s[low[i]*j];
                break;
            }
            low[i*j] = j;
            s[i*j] = s[i]*s[j];
            mu[i*j] = -mu[i];
        }
    }
    for(int i = 1;i <= n; ++i) a[i] = {s[i],i};
    sort(a+1,a+1+n,[](node x,node y){return x.val < y.val;});
}
inline int solve(int n,int m){
    int res = 0;
    for(int l = 1,r;l <= n;l = r+1){
        r = min(n/(n/l),m/(m/l));
        res = res + (T.query(r)-T.query(l-1))*(n/l)*(m/l);
    } 
    return res;
}
//main
int t;read(t);
int mx = 0;
for(int i = 1;i <= t; ++i){
    int n,m,a;read(n,m,a);
    if(n > m) swap(n,m);
    T.n = max(T.n,n);
    mx = max(mx,n);
    Q[i]={n,m,a,i};
}
sort(Q+1,Q+1+t,[](Query x,Query y){return x.a<y.a;});
mobius(mx);
int now = 1;
for(int i = 1;i <= t; ++i){
    int n = Q[i].n,m = Q[i].m,lim = Q[i].a;
    while(now<=mx&&a[now].val<=lim){
        for(int j = a[now].id;j <= mx;j += a[now].id) T.update(j,mu[j/a[now].id]*a[now].val);
        ++now;
    }
    ans[Q[i].id] = solve(n,m);
}
for(int i = 1;i <= t; ++i){
    write(ans[i]&2147483647,'\n');
}
```
  \(to\quad be\quad continued\)

posted @ 2024-06-28 21:13 CuFeO4 阅读(13) 评论(0) 编辑收藏举报

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