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(void){
	int n,x=0,y=0,z=0,count=0;
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		for(x=1;;x++){
			for(y=x;;y++){
				for(z=y;;z++){
					if((x+y)*(x+y)+(y+z)*(y+z)+(z+x)*(z+x)==2*i){
						//printf("(%d,%d,%d),",x,y,z);
						if(x==y&&y==z){
							count+=1;
						}else if(x==y||y==z){
							count+=3;
						}else{
							count+=6;
						}
					}
					if((x+y)*(x+y)+(y+z)*(y+z)+(z+x)*(z+x)>2*i){
						break;
					}
				}
				if(z==y){
					break;
				}
			}
			if(y==x){
				break;
			}
		}
		printf("%d\n",count);
		count=0;
	}
}