题解
显然可以O（nlogn）计算
代码
//by 减维
#include<set>
#include<map>
#include<queue>
#include<ctime>
#include<cmath>
#include<bitset>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define il inline
#define rg register
#define db double
#define mpr make_pair
#define maxn 2000005
#define inf (1<<30)
#define eps 1e-8
#define pi 3.1415926535897932384626L
using namespace std; inline int read()
{
int ret=;bool fla=;char ch=getchar();
while((ch<''||ch>'')&&ch!='-')ch=getchar();
if(ch=='-'){fla=;ch=getchar();}
while(ch>=''&&ch<=''){ret=ret*+ch-'';ch=getchar();}
return fla?-ret:ret;
} int t,n,num,mu[maxn],pri[maxn],phi[maxn];
ll cnt[maxn],sum[maxn];
bool pd[maxn]; void pre()
{
phi[]=;mu[]=;
for(int i=;i<=maxn-;i++)
{
if(!pd[i]) pri[++num]=i,phi[i]=i-,mu[i]=-;
for(int j=;j<=num&&i*pri[j]<=maxn-;++j)
{
pd[i*pri[j]]=;
if(i%pri[j]==)
{
phi[i*pri[j]]=phi[i]*pri[j];
break ;
}
phi[i*pri[j]]=phi[i]*phi[pri[j]];
mu[i*pri[j]]=-mu[i];
}
}
} il ll gcd(ll x,ll y){return y==?x:gcd(y,x%y);} ll solve(int x)
{
memset(cnt,,sizeof cnt);
memset(sum,,sizeof sum);
for(int i=;i<=x;++i) cnt[phi[i]]++;
for(int i=;i<=x;++i)
for(int j=;i*j<=x;++j) sum[i]+=cnt[i*j];
ll ret=;
for(int d=;d<=x;++d)
if(mu[d])
for(int dd=;dd*d<=x;++dd)
ret+=phi[dd]*mu[d]*sum[d*dd]*sum[d*dd];
return ret;
} int main()
{
t=read();
pre();
while(t--)
{
n=read();
printf("%lld\n",solve(n));
}
return ;
}