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<bits/stdc++.h>
using namespace std;

int main(){
	cin.tie(0),ios::sync_with_stdio(false);
	int n; cin>>n;
	unordered_map<int,int>m;
	for(int x=1;x<100;++x){
		for(int y=x;y<100;++y){
			for(int z=y;z<100;++z){
				int sum=x*x+y*y+z*z+x*y+y*z+z*x;
				if(sum>n)break;
				if(x!=y&&y!=z)m[sum]+=6;
				else if(x==y&&y==z)m[sum]+=1;
				else m[sum]+=3;
			}
		}
	}
	for(int i=1;i<=n;++i){
		cout<<m[i]<<"\n"s;
	}
}