莫比乌斯反演的入门题，设 \(F(x): gcd(i,j)\%x=0\) 的对数，\(f(x): gcd(i,j)=x\)的对数。
易知$$F(p) = \lfloor \frac{n}{p} \rfloor * \lfloor \frac{n}{p} \rfloor$$
\(F(x) = \sum_{x|d} f(d)\)
根据莫比乌斯反演得，\(f(x) = \sum_{x|d}u(\frac{d}{x})F(d)\)
所求的是\(gcd(i,j)\)为素数的对数，所以\(ans = \sum_{isprime(p)}f(p)\)
\[ans = \sum_{p}^{N}\lfloor \frac{N}{d}\rfloor*\lfloor \frac{N}{d}\rfloor\sum_{p|d}u(\frac{d}{p})
\]
\]
其中\(d=i*p\)
\(\sum_{p|d}u(\frac{d}{p})\)打表求出
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=1e7+5;
bool vis[maxn];
int prime[maxn],mu[maxn];
int sum[maxn];
void init(){
memset(vis,false,sizeof(vis));
mu[1] = 1;
prime[0] = 0;
int cnt=0;
for(int i=2;i<maxn;++i){
if(!vis[i]){
mu[i] = -1;
sum[i] = 1;
prime[++cnt] = i;
}
for(int j=1;j<=cnt;++j){
if(i*prime[j] >= maxn) break;
vis[i*prime[j]] = true;
if(i % prime[j]){
mu[i*prime[j]] = -mu[i];
sum[i*prime[j]] = mu[i] - sum[i];
}
else{
mu[i*prime[j]] = 0;
sum[i*prime[j]] = mu[i];
break;
}
}
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
init();
LL N;
while(scanf("%lld",&N)==1){
LL res=0;
for(LL i=1;i<=N;++i){
res += sum[i]*(N/i)*(N/i);
}
printf("%lld\n",res);
}
return 0;
}