simple c++ solution::

#include<bits/stdc++.h>
using namespace std;

int main() 
{
    long n;
    cin>>n;
    long long tiles=n/4;
    long long  k=int(sqrt(tiles));
    long long sum=0;
    for(int i=1;i<=k;i++)
    {
        sum=sum+(tiles/i-i);
        
    }
    cout<<sum;
    return 0;
}

I was able to make it working using the below algorithm.

private static long findNums(long totalTiles){
    int maxLvel = (int)Math.sqrt(totalTiles-1)/2;
    long count = 0;
    for(int i = 1;i<=maxLvel;i++){
        //Finds number of laminates of the same level. For a a given hole size h and a level l (2*(l+h/2))^2 <= tiles + h^2
        count += (long) ((totalTiles/(4*i))-i);
    }

    return count;
}

Is there no way to find out the input of each test case?

This is my program. But it faces a time-out since the range extends to 10^12. My program's complexity is O(n^2). Any tips?

int main() {

long int k,n,i,f,count=0,temp;
int max;
scanf("%ld",&i);
k=1;
        f=(4*k)+4;
for(n=0;n<=i;n++)
        {

        if(n==0)
        f=f+(8*n);
        else
        f=f+8;
        temp=temp+f;
        if(temp<=i)
        max=n;
        else
        break;
}
for(k=1;k<i;k++)
    {
    f=(4*k)+4;
    temp=0;
    for(n=0;n<=max;n++)
        {
        if(n==0)
        f=f+(8*n);
        else
        f=f+8;
        temp=temp+f;
        if(temp<=i)
            {

            count++;
        }
    }
}
printf("%ld",count);

}

PLEASE IF SOMEONE COULD HELP ME WITH THE BELOW CODE::: double d=Math.sqrt((double)n); int y=(int)d; for(int g=(int)d;g>=3;g--) { int p=g*g; if(g%2==0) { int v=2; int fl=v*v; while(fl