I think that I have got right logic, but it reports compile time out. Can anyone guide me?

include

include

include

define max 1000000

using namespace std; int num,gsn=0; int *fact = new int[max];

int factorize_real(int); void square_sum(int,int);

int main(){ int loop=2,count; cin>>num; if(num!=1){ while(loop<=num){ count = factorize_real(loop); /* now fact[] has all real factors of the input number from 0 to count-1 for all x+iy a complex number to be a guassian divisor ((x^2)+(y^2)) should belong to fact[] for this we are using square_sum(n) */ square_sum(loop,count); loop++; } gsn++;//to add for one cout<

} else{ gsn=1; cout<

} delete [] fact; return 0; }

int factorize_real(int num){ //gives real factors of an integer number int count=2; int i=2,j=1,end; int *front = new int[max]; int *back = new int[max]; end=num; front[0]=1; back[0]=num;

//end marks the last number in factor comparison as given in while loop
while(i<end){
    if(num%i==0){
        front[j]=i;
        back[j]=end=(num/i);
        (front[j]!=back[j])?(count+=2):count++;
        j++;
    }
    i++;
}
//count determines the number of real factors
for(i=0; i<count/2; i++){
    fact[i]=front[i];
    gsn+=fact[i];
}
j=i;
if(count%2 != 0){
    //this means num is a square and is not 1
    fact[j]=front[j];
    gsn+=fact[j];
    j++;
}
i--;//because now i is used to read from behind in back[]
for(;i>=0;i--){
    fact[j]=back[i];
    gsn+=fact[j];
    j++;
}

delete [] front;
delete [] back;
return count;

}

void square_sum(int n,int count){

int x,y,x2,y2,root_n,multiple;
root_n=(int) (sqrt(n));
for(x=1;x<=root_n;x++){
    x2=x*x;
    y=(int) (sqrt(n-x2));
    y2=y*y;
    if( n%(x2+y2) ==0 && y2!=0){
        gsn+=(2*(x));//multiply it with 2 as conjugate pairs will exist and add only real part x
        multiple=n/(x2+y2);
        for(int i=1;i<count;i++){
            if(multiple%fact[i] ==0){
                //this is for cases where the conjugate pair multiplied a real factor is also a Gaussian intger divisor
                gsn+=(2*(fact[i]*x));//cases like 2(1+i) and 2(1-i) that divide 8 and 4 are satisfied here
            }
        }
    }
}

}

Can anyone tell me the correct solution for input = 1000?

i think my logic is right..just it is taking more time...please someone help :( .............................. import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
    long n=sc.nextInt();
    long sum=0;
    for(long ni=1;ni<=n;ni++)
        {
        for(long a=1;a<=ni;a++)
            {
            for(long b=0;b<ni;b++)
                {
                long den1=a*a+b*b,num_a=ni*a,num_b=ni*b;
                if((num_a%den1==0)&(num_b%den1==0)&(b!=0))
                    sum=sum+2*a;
                if((num_a%den1==0)&(num_b%den1==0)&(b==0))
                    sum=sum+a;

            }
        }
    }
    System.out.println(sum);
}

}

hi can anybody tell me the application for this question in real live projects.

Hi guys,

would anybody be so nice to explain to me, how 2+2i is a divisor of 4. My result is

(4*(2-2i)) / 8

Which shouldn't be a divisor if I understood everything correctly.

Thanks in advance.