洛谷

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

求 $sumlimits_{i=1}^{n}sumlimits_{j=1}^{m} gcd(i,j) $

不会,看题解:

类似求gcd为p的求法:

$ f(n) = sumlimits_{i=1}^{n}sumlimits_{j=1}^{m} gcd(i,j) =sumlimits_{i=1}^{N} d sumlimits_{i=1}^{n}sumlimits_{j=1}^{m} [gcd(i,j)==d] $

提出 (d) :
$ f(n) =sumlimits_{i=1}^{N} d sumlimits_{i=1}^{lfloor frac{n}{d} floor }sumlimits_{j=1}^{ lfloorfrac{m}{d} floor } [gcd(i,j)==1] $

用 $sumlimits_{d|n}mu(d)=[n==1] $ 替换,反演:
$ f(n) = sumlimits_{i=1}^{N} d sumlimits_{k=1}^{N} mu(k) lfloorfrac{n}{kd} floor lfloorfrac{m}{kd} floor $

记 (T=kd) :
$ f(n) = sumlimits_{T=1}^{N} sumlimits_{d|T} d mu(frac{T}{d}) lfloorfrac{n}{T} floor lfloorfrac{m}{T} floor $

提出 (T)
$ f(n) = sumlimits_{T=1}^{N} lfloorfrac{n}{T} floor lfloorfrac{m}{T} floor sumlimits_{d|T} d mu(frac{T}{d}) $

因为:
$sumlimits_{d|n}frac{mu(d)}{d}=frac{varphi(n)}{n} $

$ f(n) = sumlimits_{T=1}^{N} lfloorfrac{n}{T} floor lfloorfrac{m}{T} floor 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) = sumlimits_{d=1}^{n} d sumlimits_{i=1}^{n}sumlimits_{j=1}^{i-1} [gcd(i,j)==d] $

提d:
$ sumlimits_{d=1}^{n} d sumlimits_{i=1}^{frac{n}{d}}sumlimits_{j=1}^{i-1} [gcd(i,j)==1] $

后面是欧拉函数的定义:
$ sumlimits_{d=1}^{n} d sumlimits_{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;
    }
}