洛谷 - P1390 - 公约数的和 - 莫比乌斯反演 - 欧拉函数

https://www.luogu.org/problemnew/show/P1390

求 $\sum\limits_{i=1}^{{n}\sum\limits_{j=1}} gcd(i,j) $

不会,看题解:

类似求gcd为p的求法:

$ f(n) = \sum\limits_{i=1}^{{n}\sum\limits_{j=1}} gcd(i,j) =\sum\limits_{i=1}^{N} d \sum\limits_{i=1}^{{n}\sum\limits_{j=1}} [gcd(i,j)==d] $

提出 \(d\) :
$ f(n) =\sum\limits_{i=1}^{N} d \sum\limits_{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum\limits_{j=1}^{ \lfloor\frac{m}{d}\rfloor } [gcd(i,j)==1] $

用 $\sum\limits_{d|n}\mu(d)=[n==1] $ 替换,反演:
$ f(n) = \sum\limits_{i=1}^{N} d \sum\limits_{k=1}^{N} \mu(k) \lfloor\frac{n}{kd}\rfloor \lfloor\frac{m}{kd}\rfloor $

记 \(T=kd\) :
$ f(n) = \sum\limits_{T=1}^{N} \sum\limits_{d|T} d \mu(\frac{T}{d}) \lfloor\frac{n}{T}\rfloor \lfloor\frac{m}{T}\rfloor $

提出 \(T\)
$ f(n) = \sum\limits_{T=1}^{N} \lfloor\frac{n}{T}\rfloor \lfloor\frac{m}{T}\rfloor \sum\limits_{d|T} d \mu(\frac{T}{d}) $

因为:
$\sum\limits_{d|n}\frac{\mu(d)}{d}=\frac{\varphi(n)}{n} $

$ f(n) = \sum\limits_{T=1}^{N} \lfloor\frac{n}{T}\rfloor \lfloor\frac{m}{T}\rfloor \varphi(T) $

#include<bits/stdc++.h>
using namespace std;
#define ll long long

#define N 2000005
int phi[N],pri[N],cntpri=0;
bool notpri[N];

void sieve_phi(int n)
{
    notpri[1]=phi[1]=1;
    for (int i=2;i<=n;i++)
    {
        if (!notpri[i]) pri[++cntpri]=i,phi[i]=i-1;
        for (int j=1;j<=cntpri&&i*pri[j]<=n;j++)
        {
            notpri[i*pri[j]]=1;
            if (i%pri[j]) phi[i*pri[j]]=phi[i]*phi[pri[j]];
            else {phi[i*pri[j]]=phi[i]*pri[j];break;}
        }
    }
}

int main(){
    int n;
    cin>>n;
    sieve_phi(n);
    ll ans=0;
    for(int i=1;i<=n;i++){
        ans+=1ll*phi[i]*(n/i)*(n/i);
    }
    cout<<(ans-(1ll*(1+n)*n)/2)/2<<endl;
}

---

另一种奇怪的做法:
$ f(n) = \sum\limits_{d=1}^{n} d \sum\limits_{i=1}^{{n}\sum\limits_{j=1}} [gcd(i,j)==d] $

提d:
$ \sum\limits_{d=1}^{n} d \sum\limits_{i=1}^{{\frac{n}{d}}\sum\limits_{j=1}} [gcd(i,j)==1] $

后面是欧拉函数的定义:
$ \sum\limits_{d=1}^{n} d \sum\limits_{i=2}^{\frac{n}{d}} \varphi(i) $

这里有个bug是因为1和1互质但是1和1相同，所以要去掉 \(\varphi(1)\)

#include<bits/stdc++.h>
using namespace std;
#define ll long long

#define N 2000000+5

int phi[N],pri[N],cntpri=0;
bool notpri[N];

ll prefix[N];

void sieve_phi(int n) {
    notpri[1]=phi[1]=1;
    prefix[0]=0;
    prefix[1]=1;
    for(int i=2; i<=n; i++) {
        if(!notpri[i])
            pri[++cntpri]=i,phi[i]=i-1;
        for(int j=1; j<=cntpri&&i*pri[j]<=n; j++) {
            notpri[i*pri[j]]=1;
            if(i%pri[j])
                phi[i*pri[j]]=phi[i]*phi[pri[j]];
            else {
                phi[i*pri[j]]=phi[i]*pri[j];
                break;
            }
        }
        prefix[i]=prefix[i-1]+phi[i];
    }
}

ll solve(ll n){
    ll ans=0;
    for(int d=1;d<=n;d++){
        ans+=d*((prefix[n/d])-1);
    }
    return ans;
}

int main() {
    sieve_phi(2000000+1);
    int n;
    while(cin>>n) {
        ll ans=solve(n);
        cout<<ans<<endl;
    }
}