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++ (judge verdict: Accepted)

#include <cstdio>
#include <cstdlib>
#include <utility>
#include <typeinfo>
#include <cmath>
#include <numeric>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <iomanip>
#include <vector>
#include <string>
#define REP(i,n) for(int i=0;i<n;i++)
typedef long long ll;
using namespace std;
typedef vector<ll> vl;
typedef pair<ll,ll> pll;
typedef vector<pll> vpll;
typedef vector<string> vs;

int main() 
{
	int a[183]={};
	REP(i,183)a[i]=i*i;
			
	int n;
	while(cin>>n&&n!=0)
	{
		int ans(0);
		for(int i=182;i*i>=n/4&&i>=0;i--)
		{
			int tmp1=n;
			if(a[i]>tmp1){continue;}
			else
			{
				int tmp2 = tmp1 - a[i];
				for(int j=i;j*j>=tmp2/3&&j>=0;j--)
				{
					if(a[j]>tmp2){continue;}
					else
					{
						int tmp3 = tmp2 - a[j];
						for(int k=j;k*k>=tmp3/2&&k>=0;k--)
						{
							if(a[k]>tmp3){continue;}
							else
							{
								int tmp4 = tmp3 - a[k];
								for(int l=k;l*l>=tmp4&&l>=0;l--)
								{
									if(tmp4==a[l])
									{
										ans++;
										break;
									}
								}
							}
						}
					}
				}
			}
		}
		cout << ans << endl;
	}
	return 0;
}