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: Accepted)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#define PI 3.141592653589793
#define MOD 1000000007
//#define DEBUG

//qsort用、昇順
int compare_int(const void *a, const void *b)
{
	return *(int*)a-*(int*)b;
}
//qsort(array, 10, sizeof(int), compare_int)

int main(){
	int n, lim;
	scanf("%d", &n);
	for(int i=1; i<=5; i++){
		if(i>n) break;
		printf("0\n");
	}
	
	lim = pow(n, 0.5) + 1;
	for(int i=6; i<=n; i++){
		int count=0;
		for(int x=1; x<=lim; x++){
			for(int y=1; y<=lim; y++){
				if(x*x+y*y+x*y>i) break;
				for(int z=1; z<=lim; z++){
					int tmp1 = pow(x+y+z, 2)-(x*y+y*z+z*x);
					if(tmp1==i) count++;
					else if(tmp1>i) break;
				}
			}
		}
		//printf("i:%d, count:%d\n", i, count);
		printf("%d\n", count);
	}
	
	
	return 0;
}