洛谷 P3911 最小公倍数之和

最小公倍数之和

Here

仅作笔记和理解

题意

对于$A_1,A_2,\cdot \cdot \cdot ,A_N$求$\sum _{i=1}^N\sum _{j=1}^{N}lcm(A_i,A_j)$的值
数据范围：$1\le N\le 50000,1\le A_i\le 50000$

解

乍一看好像搞不了，实际上又搞得了

这道题是让我们求任意两个数的$lcm$的值

转换思路，我们不枚举下标，我们枚举每一个值，统计每种值的数的出现次数

莫比乌斯反演，开始化式子= =

$n=50000$

$\sum _{i=1}^n\sum _{j=1}^n\frac{i\cdot j\cdot C_i\cdot C_j}{gcd(i,j)}$

开始套路枚举$gcd$

$\sum _{g=1}^n\sum _{i=1}^{\frac{n}{g}}\sum _{j=1}^{\frac{n}{g}}\frac{i\cdot g\cdot j\cdot g\cdot C_i\cdot C_j}{g}[gcd(i,j)=1]$

套路替换莫比乌斯函数

$\sum _{g=1}^n\sum _{i=1}^{\frac{n}{g}}\sum _{j=1}^{\frac{n}{g}}i\cdot j\cdot g\cdot C_{ig}\cdot C_{jg}\cdot \sum _{x|gcd(i,j)}\mu (x)$

套路枚举$x$的值

$\sum _{g=1}^n\sum _{x=1}^{\frac{n}{g}}\sum _{i=1}^{\frac{n}{xg}}\sum _{j=1}^{\frac{n}{xg}}i\cdot x\cdot j\cdot x\cdot g\cdot C_{ixg}\cdot C_{jxg}\cdot \mu (x)$

变换一下$x$

$\sum _{g=1}^n\sum _{g|x}^n\sum _{i=1}^{\frac{n}{x}}\sum _{j=1}^{\frac{n}{x}}i\cdot \frac{x^2}{g}\cdot j\cdot C_{ix}\cdot C_{jx}\cdot \mu (\frac{x}{g})$

演他一手就得到了

$\sum _{x=1}^n\sum _{g|x}\sum _{i=1}^{\frac{n}{x}}\sum _{j=1}^{\frac{n}{x}}i\cdot \frac{x^2}{g}\cdot j\cdot C_{ix}\cdot C_{jx}\cdot \mu (\frac{x}{g})$

整理一下就变成了

$\sum _{x=1}^n\sum _{g|x}\frac{x^2}{g}\cdot \mu (\frac{x}{g})\sum _{i=1}^{\frac{n}{x}}i\cdot C_{ix}\sum _{j=1}^{\frac{n}{x}}j\cdot C_{jx}$

=>$\sum _{x=1}^n\sum _{g|x}\frac{x^2}{g}\cdot \mu (\frac{x}{g})(\sum _{i=1}^{\frac{n}{x}}i\cdot C_{ix}{)}^2$

所以接下来就可以预处理$\sum _{g|x}\frac{x^2}{g}\cdot \mu (\frac{x}{g})$，然后枚举$x$的值，$O(nlogn)$就可以算出式子的值了。

其实也可以预处理$(\sum _{i=1}^{\frac{n}{x}}i\cdot C_{ix}{)}^2$，然后枚举x的值

其实也可以两个都预处理（但没必要

Code 预处理$(\sum _{i=1}^{\frac{n}{x}}i\cdot C_{ix}{)}^2$

/****************************
* Author : 水娃             *
* Date : 2020-08-07-18.38.19*
****************************/
#pragma GCC optimize(3,"Ofast","inline")
#include
using namespace std;
#define mst(a,b) memset(a,b,sizeof(a))
#define ll long long
#define debug(a) cout<<#a<<" is "<
#define ull unsigned long long
#define fi first
#define se second
typedef pair<int,int>pii;
typedef pair<ll,int>pli;
typedef pair<int,ll>pil;
typedef pair<ll,ll>pll;
const ll mo=(ll)1e9+7;
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
const ll MAXN=5e4+100;
int cnt;
ll p[MAXN],mu[MAXN],vis[MAXN];
void init()
{
    ll num;
    mu[1]=1;
    for(ll i=2;i<MAXN;++i)
    {
        if(!vis[i])
        {
            mu[i]=-1;
            p[++cnt]=i;
        }
        for(int j=1;j<=cnt&&(num=i*p[j])<MAXN;++j)
        {
            vis[num]=1;
            if(i%p[j]==0)break;
            mu[num]=-mu[i];
        }
    }
}
int n;
ll a[51000],tot[51000],sum[51000];
void work()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        tot[a[i]]++;
    }
    for(ll x=1;x<=50000;x++)
    {
        ll top=50000/x;
        ll tmp=0;
        for(ll i=1;i<=top;i++)
        {
            tmp=tmp+i*tot[i*x];;
        }
        sum[x]=tmp*tmp;
    }
    ll ans=0;
    for(ll g=1;g<=50000;g++)
    {
        ll top=50000/g;
        ll tmp=0;
        for(ll j=1;j<=top;j++)
        {
            tmp=tmp+j*g*j*mu[j]*sum[g*j];
        }
        ans=ans+tmp;
    }
    cout<<ans<<"\n";
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    init();
    ///int T;cin>>T;while(T--)
        work();
    return 0;
}

Code 预处理$\sum _{g|x}\frac{x^2}{g}\cdot \mu (\frac{x}{g})$

/****************************
* Author : 水娃             *
* Date : 2020-08-07-18.38.19*
****************************/
#pragma GCC optimize(3,"Ofast","inline")
#include
using namespace std;
#define mst(a,b) memset(a,b,sizeof(a))
#define ll long long
#define debug(a) cout<<#a<<" is "<
#define ull unsigned long long
#define fi first
#define se second
typedef pair<int,int>pii;
typedef pair<ll,int>pli;
typedef pair<int,ll>pil;
typedef pair<ll,ll>pll;
const ll mo=(ll)1e9+7;
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
const ll MAXN=5e4+100;
int cnt;
ll p[MAXN],mu[MAXN],vis[MAXN];
void init()
{
    ll num;
    mu[1]=1;
    for(ll i=2;i<MAXN;++i)
    {
        if(!vis[i])
        {
            mu[i]=-1;
            p[++cnt]=i;
        }
        for(int j=1;j<=cnt&&(num=i*p[j])<MAXN;++j)
        {
            vis[num]=1;
            if(i%p[j]==0)break;
            mu[num]=-mu[i];
        }
    }
}
int n;
ll a[51000],tot[51000],sum[51000];
void work()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        tot[a[i]]++;
    }
    int num;
    for(int i=1;i<=50000;i++)
    {
        for(int j=1;(num=i*j)<=50000;j++)
        {
            sum[num]+=i*mu[j]*j*j;
        }
    }
    ll ans=0;
    for(int i=1;i<=50000;i++)
    {
        int top=50000/i;
        ll tmp=0;
        for(int j=1;j<=top;j++)
        {
            tmp=tmp+j*tot[i*j];
        }
        ans=ans+sum[i]*tmp*tmp;
    }
    cout<<ans<<"\n";
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    init();
    ///int T;cin>>T;while(T--)
        work();
    return 0;
}