莫比乌斯反演

标签：limits   lse   函数   cas   algo   its   \n   aio   blank   

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\[{\rm{F(n)}} = \sum\limits_{{\rm{d|n}}}^{} {{\rm{f}}(d)}
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqiVu0Je9sqqr
% pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs
% 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai
% aabeqaamaabaabauaakeaacaqGgbGaaeikaiaab6gacaqGPaGaeyyp
% a0ZaaabCaeaacaqGMbGaaiikaiaadsgacaGGPaaaleaacaqGKbGaae
% iFaiaab6gaaeaaa0GaeyyeIuoaaaa!4B7C!
\]，那么我们得到结论\[f(n) = \sum\limits_{d|n}^{} {\mu (d)F(\frac{n}{d})}
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqiVu0Je9sqqr
% pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs
% 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai
% aabeqaamaabaabauaakeaacaWGMbGaaiikaiaad6gacaGGPaGaeyyp
% a0ZaaabCaeaacqaH8oqBcaGGOaGaamizaiaacMcacaWGgbGaaiikam
% aalaaabaGaamOBaaqaaiaadsgaaaGaaiykaaWcbaGaamizaiaacYha
% caWGUbaabaaaniabggHiLdaaaa!5084!
\]

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：\[F(n) = \sum\limits_{d|n}^{} {f(d)}  \Rightarrow f(n) = \sum\limits_{d|n}^{} {\mu (d)F(\frac{n}{d})}
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqiVu0Je9sqqr
% pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs
% 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai
% aabeqaamaabaabauaakeaacaWGgbGaaiikaiaad6gacaGGPaGaeyyp
% a0ZaaabCaeaacaWGMbGaaiikaiaadsgacaGGPaaaleaacaWGKbGaai
% iFaiaad6gaaeaaa0GaeyyeIuoakiabgkDiElaadAgacaGGOaGaamOB
% aiaacMcacqGH9aqpdaaeWbqaaiabeY7aTjaacIcacaWGKbGaaiykai
% aadAeacaGGOaWaaSaaaeaacaWGUbaabaGaamizaaaacaGGPaaaleaa
% caWGKbGaaiiFaiaad6gaaeaaa0GaeyyeIuoaaaa!5F53!
\]

其中μ(d)为莫比乌斯函数

μ(d)的性质：\[\mu (d) = \left\{ \begin{array}{l}1,d = 1\\{( - 1)^k},d = {p_1}*{p_2}*...{p_k}\\0,\end{array} \right.
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbWexLMBbXgBd9gzLbvyNv2CaeHbl7mZLdGeaGqiVu0Je9sqqr
% pepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs
% 0-yqaqpepae9pg0FirpepeKkFr0xfr-xfr-xb9adbaqaaeGaciGaai
% aabeqaamaabaabauaakeaacqaH8oqBcaGGOaGaamizaiaacMcacqGH
% 9aqpdaGabaabaeqabaGaaGymaiaacYcacaWGKbGaeyypa0JaaGymaa
% qaaiaacIcacqGHsislcaaIXaGaaiykamaaCaaaleqabaGaam4Aaaaa
% kiaacYcacaWGKbGaeyypa0JaamiCamaaBaaaleaacaaIXaaabeaaki
% aacQcacaWGWbWaaSbaaSqaaiaaikdaaeqaaOGaaiOkaiaac6cacaGG
% UaGaaiOlaiaadchadaWgaaWcbaGaam4AaaqabaaakeaacaaIWaGaai
% ilaaaacaGL7baaaaa!5AE1!
\]

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\n",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\n",i,res1-res2/2);
    }
}

莫比乌斯反演

标签：limits   lse   函数   cas   algo   its   \n   aio   blank   

原文地址：https://www.cnblogs.com/donke/p/10383087.html