We need to again come up with diff algo after studying factors ( ..sum) under 10000 or 100000. It seems this problem require advance level of mathematics

hamad

I got this optimal solution in c++ but it still needs to be optimized can anyone help?

#include <cstdio>
#include<cmath>
#include <iostream>
using namespace std;
int main() 
{ 
    int N;
    cin>>N;
    long int n[N],count[N],max=0;
    for(int i=0;i<N;i++)
    {
        cin>>n[i];
        count[i]=0;
        if(max<n[i])
        {
            max=n[i];
        }
    }
    
    for(long int i=0;i<=max/7;i++)
    {
        long int b2c=(i+1)*(i+1)*(8*i+5);
        long int a=3*i+2;
        
        for(long int j=sqrt(b2c);j>=1;j--)
        {
            if(b2c%(j*j)==0&&b2c/(j*j)<=max)
            {
                long int b=j;
                long int c=b2c/(j*j);
                for(long int k=0;k<N;k++)    
                    if(8*a*a*a+15*a*a+6*a-27*b*b*c==1&&(a+b+c<=n[k]))
                    {
                        count[k]++;
                    }
            }
        }
    }
    for(int i=0;i<N;i++)
    {
        cout<<count[i]<<endl;
    }
    return 0;
}

is there any way to optimize it?

import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;

public class Solution {

    int check(long a,long b, long c)
    {
 double d=Math.pow(c,0.5);
   if(d>a)
     {
         double r=b*Math.pow(c,0.5);
         double h=(-1)*(a-r);
         double i=100*Math.pow(h,1.0/3.0) ;
         int y=(int)i;
         double o=(double)y/100;


         double r1=b*Math.pow(c,0.5);
         double h1=(a+r);
         double i1=100*Math.pow(h1,1.0/3.0) ;
         int y1=(int)i1;
         double o1=(double)y1/100;

         if(o1-o==1.0)
         return 1;
         else
         return 0;

     }
   else
     {
         double r=b*Math.pow(c,0.5);
         double h=(a-r);
         double i=100*Math.pow(h,1.0/3.0) ;
         int y=(int)i;
         double o=(double)y/100;


         double r1=b*Math.pow(c,0.5);
         double h1=(a+r);
         double i1=100*Math.pow(h1,1.0/3.0) ;
         int y1=(int)i1;
         double o1=(double)y1/100;

         if(o1+o==1.0)
         return 1;
         else
         return 0;

     }          

}




public static void main (String[] args) throws java.lang.Exception
{
    Scanner sc=new Scanner(System.in);
    int c=0;
    int u=sc.nextInt();
    long[] f=new long[u];
    for(int o=0;o<u;o++)
    {
        f[o]=sc.nextInt();
    }
    long n;

    for(int y=0;y<u;y++)
    {
        n=f[y];

    for(long i=1;i<n-3;i++)
    {
        for(long j=1;j<=n-i;j++)
        {
            for(long k=1;k<=n-i-j;k++)
            {
                Solution st=new Solution();
                if(i+j+k<=n)
                {
                if(st.check(i,j,k)==1)
                {
                //System.out.println(i+" "+j+" "+k);
                c++;
                }
                }
            }
        }
    }
    System.out.println(c);
    }

}

}

There is a nice way to do it in O(n^(2/3)) time, which is less than 1s for n=2*10^10.