A

Consider a N*N*N lattice. One corner is at (0,0,0) and the opposite one is at (N,N,N). How many lattice points are visible from corner at (0,0,0) ? A point X is visible from point Y iff no other lattice point lies on the segment joining X and Y. 
 
Input : 
The first line contains the number of test cases T. The next T lines contain an interger N 
 
Output : 
Output T lines, one corresponding to each test case. 
 
Sample Input : 
3 
1 
2 
5 
 
Sample Output : 
7 
19 
175 
 
Constraints : 
T <= 50 
1 <= N <= 1000000

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e6+10;
const int INF=0x3f3f3f3f;
int cas=1,T;
int c[N];
void init()
{
    memset(c,0,sizeof(c));
    c[1]=1;
    for(int i=1;i<N;i++)
    {
        if(c[i]) for(int j=i<<1;j<N;j+=i) c[j]-=c[i];
        c[i]+=c[i-1];
    }
}
int p[N],vis[N],pn;
void getMu(int *mu)  //O(n)
{  
    memset(vis,0,sizeof(vis));  
    mu[1]=1;  
    pn=0;  
    for(int i=2; i<N; i++)  
    {  
        if(!vis[i]) { p[pn++]=i;mu[i]=-1; }  
        for(int j=0; j<pn&&i*p[j]<N; j++)  
        {  
            vis[i*p[j]]=1;  
            if(i%p[j]) mu[i*p[j]]=-mu[i];  
            else { mu[i*p[j]]=0;break;}  
        }  
        mu[i]+=mu[i-1];
    }  
}  
int main()
{
    //freopen("1.in","w",stdout);
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    //init();
    getMu(c);
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        LL ans=0;
        //for(int i=1;i<=n;i++) if(c[i]) ans+=c[i] * ((LL) (n/i+1)*(n/i+1)*(n/i+1)-1);
        for(int i=1;i<=n;)
        {
            int x=n/i;
            int y=n/x;
            ans+=(c[y]-c[i-1]) * ((LL) (n/i+1)*(n/i+1)*(n/i+1)-1);
            i=y+1;
        }
        printf("%lld
",ans);
    }
    //printf("time=%.3lf
",(double)clock()/CLOCKS_PER_SEC);
    return 0;
}