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 (Clang 10.0.0) (judge verdict: Accepted)

#include<stdio.h>

int main(){
  long long n;
  scanf("%lld",&n);
  long long ans[n+1];
  for(long long i=1;i<=n;i++)ans[i]=0;
  
  for(long long a=1;a<=100;a++){
    for(long long b=a;b<=100;b++){
      for(long long c=b;c<=100;c++){
        if((a+b)*(a+b)+(c+b)*(c+b)+(a+c)*(a+c)<=2*n){
          if(a==c){
            ans[((a+b)*(a+b)+(c+b)*(c+b)+(a+c)*(a+c))/2]+=1;
          }else if(a==b || b==c){
            ans[((a+b)*(a+b)+(c+b)*(c+b)+(a+c)*(a+c))/2]+=3;
          }else{
            ans[((a+b)*(a+b)+(c+b)*(c+b)+(a+c)*(a+c))/2]+=6;
          }
        }
      }
    }
  }
  
  for(long long i=1;i<=n-1;i++){
    printf("%lld\n",ans[i]);
  }
  printf("%lld",ans[n]);
  return 0;
}