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>

int main(){
	int n;
	int temp = 0;
	int ans[10010] = {0};
	
	scanf("%d", &n);
	
	for(int x = 1; x <= 100; x++){
		for(int y = 1; y <= 100; y++){
			for(int z = 1; z <= 100; z++){
				temp = x*x + y*y + z*z + x*y + y*z + x*z;
				if(temp < 10000){
					ans[temp]++;
				}
			}
		}
	}
	for(int i = 0; i < n; i++){
		printf("%d\n", ans[i]);
	}
}