传送门
解法：
[egin{align}
sum_{i=1}^nsum_{j=1}^nphi(gcd(i,j))&=sum_{d=1}phi(d)sum_{i=1}^nsum_{j=1}^n[d==gcd(i,j)]\
&=sum_{d=1}^nphi(d)sum_{i=1}^{lfloorfrac{n}{d}
floor}sum_{j=1}^{lfloorfrac{n}{d}
floor}sum_{k=1}^{lfloorfrac{n}{d}
floor}mu(k)[k|gcd(i,j)]\
&=sum_{T=1}^nlfloorfrac{n}{T}
floor^2mu*phi(T)\
end{align}
]
下面考虑转化这个式子
[sum_{i=1}^nf*g(i)=sum_{i=1}^nf(i)sum_{j=1}^{lfloorfrac{n}{i}
floor}g(j)\
egin{align}
sum_{i=1}^nphi(i)sum_{j=1}^{lfloorfrac{n}{i}
floor}phi(j)&=sum_{i=1}^nmu*phi(i)sum_{j=1}^{lfloorfrac{n}{i}
floor}id(j)\
&=sum_{i=1}^nmu*phi(i)frac{lfloorfrac{n}{i}
floor(lfloorfrac{n}{i}
floor+1)}{2}\
&=frac{sum_{i=1}^nlfloorfrac{n}{i}
floor^2mu*phi(i)+sum_{i=1}^nlfloorfrac{n}{i}
floormu*phi(i)}{2}\
&=frac{sum_{i=1}^nlfloorfrac{n}{i}
floor^2mu*phi(i)+sum_{i=1}^nphi(i)}{2}\
end{align}\
sum_{i=1}^nlfloorfrac{n}{i}
floor^2mu*phi(i)=2 imessum_{i=1}^nphi(i)sum_{j=1}^{lfloorfrac{n}{i}
floor}phi(j)-sum_{i=1}^nphi(i)
]
代码：
//https://www.lydsy.com/JudgeOnline/problem.php?id=4804
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<cmath>
#include<map>
#include<queue>
#include<bitset>
#include<set>
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define dwn(i,a,b) for(int i=(a);i>=(b);--i)
using namespace std;
typedef long long ll;
const int N=10000000;
int T,n,cnt,prime[1000010];
bool v[N+10];
ll phi[N+10];
inline void init(int n)
{
phi[1]=1;
rep(i,2,n)
{
if(!v[i])
{
prime[++cnt]=i;
phi[i]=i-1;
}
for(int j=1;j<=cnt&&prime[j]*i<=n;++j)
{
v[prime[j]*i]=1;
if(i%prime[j]==0)
{
phi[prime[j]*i]=phi[i]*prime[j];
break;
}
phi[prime[j]*i]=phi[i]*(prime[j]-1);
}
}
rep(i,2,n)
{
phi[i]+=phi[i-1];
}
}
int main()
{
init(N);
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
ll ans=0;
for(int l=1,r;l<=n;l=r+1)
{
r=n/(n/l);
ans+=phi[n/l]*(phi[r]-phi[l-1]);
}
printf("%lld
",2*ans-phi[n]);
}
return 0;
}