Problem:
Problem B: Lagrange's Four-Square Theorem

The fact that any positive integer has a representation as the sum of at most four positive
squares (i.e. squares of positive integers) is known as Lagrange’s Four-Square Theorem. The
first published proof of the theorem was given by Joseph-Louis Lagrange in 1770. Your mission
however is not to explain the original proof nor to discover a new proof but to show that the
theorem holds for some specific numbers by counting how many such possible representations
there are.

For a given positive integer n, you should report the number of all representations of n as the
sum of at most four positive squares. The order of addition does not matter, e.g. you should
consider 42 + 32 and 32 + 42 are the same representation.

For example, let’s check the case of 25. This integer has just three representations 12 +22 +22 +42 ,
32 + 42 , and 52 . Thus you should report 3 in this case. Be careful not to count 42 + 32 and
32 + 42 separately.

Input

The input is composed of at most 255 lines, each containing a single positive integer less than
215 , followed by a line containing a single zero. The last line is not a part of the input data.

Output

The output should be composed of lines, each containing a single integer. No other characters
should appear in the output.

The output integer corresponding to the input integer n is the number of all representations
of n as the sum of at most four positive squares.

Sample Input

1
25
2003
211
20007
0

Output for the Sample Input

1
3
48
7
738

Solution in C++14 (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
int main(){
  vector<vector<vector<int>>> dp(182, vector<vector<int>>(5, vector<int>(1 << 15, 0)));
  for (int i = 0; i <= 4; i++){
    dp[0][i][0] = 1;
  }
  for (int i = 0; i < 181; i++){
    for (int j = 0; j <= 4; j++){
      for (int k = 0; k < (1 << 15); k++){
        for (int l = 0; l <= 4; l++){
          if (j + l <= 4 && k + (i + 1) * (i + 1) * l < (1 << 15)){
            dp[i + 1][j + l][k + (i + 1) * (i + 1) * l] += dp[i][j][k];
          }
        }
      }
    }
  }
  while (1){
    int N;
    cin >> N;
    if (N == 0){
      break;
    }
    cout << dp[181][4][N] << endl;
  }
}