莫比乌斯反演

Posted 2021-02-09 donke

定理：F(n)和f(n)是定义在非负整数集合上的两个函数，并且满足条件[{ m{F(n)}} = sumlimits_{{ m{d|n}}}^{} {{ m{f}}(d)}
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]，那么我们得到结论[f(n) = sumlimits_{d|n}^{} {mu (d)F(frac{n}{d})}
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]

 

 

根据F(n)的定义我们可以得出：

• F(1)=f(1)
• F(2)=f(1)+f(2)
• F(3)=f(1)+f(3)
• F(4)=f(1)+f(2)+f(4)
• F(5)=f(1)+f(5)
• F(6)=f(1)+f(2)+f(3)+f(6)
• F(7)=f(1)+f(7)
• F(8)=f(1)+f(2)+f(4)+f(8)

于是可以推导出f(n):

• f(1)=F(1)
• f(2)=F(2)-F(1)
• f(3)=F(3)-F(1)
• f(4)=F(4)-F(2)
• f(5)=F(5)-F(1)
• f(6)=F(6)-F(3)-F(2)+F(1)
• f(7)=F(7)-F(1)
• f(8)=F(8)-F(4)

可以得到公式：[F(n) = sumlimits_{d|n}^{} {f(d)}  Rightarrow f(n) = sumlimits_{d|n}^{} {mu (d)F(frac{n}{d})}
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]

其中μ(d)为莫比乌斯函数

μ(d)的性质：[mu (d) = left{ egin{array}{l}1,d = 1\{( - 1)^k},d = {p_1}*{p_2}*...{p_k}\0,end{array} ight.
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]

用线性筛求莫比乌斯函数值：

const int maxn=1e5+7;
bool vis[maxn];
int prime[maxn],mu[maxn];
int cnt;
void Init(int N)///线性筛求莫比乌斯函数的值
{
    //int N=maxn;
    memset(vis,0,sizeof(vis));
    mu[1] = 1;
    cnt = 0;
    for(int i=2; i<N; i++)
    {
        if(!vis[i])
        {
            prime[cnt++] = i;
            mu[i] = -1;
        }
        for(int j=0; j<cnt&&i*prime[j]<N; j++)
        {
            vis[i*prime[j]] = 1;
            if(i%prime[j]) mu[i*prime[j]] = -mu[i];
            else
            {
                mu[i*prime[j]] = 0;
                break;
            }
        }
    }
}

 例题：hdu-1695

代码：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
const int mod=10001;

int gcd(int a,int b){return (b==0)?a:gcd(b,a%b);}

const int maxn=1e5+7;
bool vis[maxn];
int prime[maxn],mu[maxn];
int cnt;
void Init(int N)///线性筛求莫比乌斯函数的值
{
    memset(vis,0,sizeof(vis));
    mu[1] = 1;
    cnt = 0;
    for(int i=2; i<N; i++)
    {
        if(!vis[i])
        {
            prime[cnt++] = i;
            mu[i] = -1;
        }
        for(int j=0; j<cnt&&i*prime[j]<N; j++)
        {
            vis[i*prime[j]] = 1;
            if(i%prime[j]) mu[i*prime[j]] = -mu[i];
            else
            {
                mu[i*prime[j]] = 0;
                break;
            }
        }
    }
}

int main()
{
    int t;
    cin>>t;//int T=t;
    Init(100000);
    for(int i=1;i<=t;i++){
        ll res1=0,res2=0;
        ll a,b,c,d,k;
        cin>>a>>b>>c>>d>>k;
        if(b>d)swap(b,d);
        if(k==0){
            printf("Case %d: 0
",i);continue;
        }
        b=b/k;d=d/k;
        for(int j=1;j<=b;j++){
            res1+=mu[j]*(b/j)*(d/j);
        }
        for(int j=1;j<=b;j++){
            res2+=mu[j]*(b/j)*(b/j);
        }

        printf("Case %d: %lld
",i,res1-res2/2);
    }
}