Problem:
Score : 300 points

Problem Statement

Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions:

1 \leq x,y,z

x^2 + y^2 + z^2 + xy + yz + zx = n

Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).

Constraints

All values in input are integers.

1 \leq N \leq 10^4

Input

Input is given from Standard Input in the following format:

N

Output

Print N lines. The i-th line should contain the value f(i).

Sample Input 1

20

Sample Output 1

0
0
0
0
0
1
0
0
0
0
3
0
0
0
0
0
3
3
0
0

For n=6, only (1,1,1) satisfies both of the conditions. Thus, f(6) = 1.

For n=11, three triples, (1,1,2), (1,2,1), and (2,1,1), satisfy both of the conditions. Thus, f(6) = 3.

For n=17, three triples, (1,2,2), (2,1,2), and (2,2,1), satisfy both of the conditions. Thus, f(17) = 3.

For n=18, three triples, (1,1,3), (1,3,1), and (3,1,1), satisfy both of the conditions. Thus, f(18) = 3.

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<stdbool.h>

typedef  long long ll;

ll int_cmp(const void *a, const void *b) 
{ 
    const int *ia = (const int *)a;
    const int *ib = (const int *)b;
    return *ia  - *ib; 

} 



int max(int a,int b){
	if(a<b)
		a=b;
	return a;	
}

int min(int a,int b){
	if(a>b)
		a=b;
	return a;	
}

int absolu(int n){
	if(n<0)
		n*=-1;
	
	return n;	
}

void solve(){
	int i,j,k,n,a[10000]={0},res;
	scanf("%d",&n);
	
	for(i=1 ; i<=100 ; i++){
		for(j=1 ; j<=100 ; j++){
			for(k=1 ; k<=100 ; k++){
				res=i*i+j*j+k*k+i*j+j*k+k*i;
				if(res<=10000)
					a[res]++;	
			}
		}
	}
	
	for(i=1 ; i<=n ; i++)
		printf("%d\n",a[i]);

	
}



int main(){
/*int t;
	scanf("%d",&t);
	while(t--)*/
		solve();
	
	return 0;
}