(URI) Feynman - Solution

Once we know that the entry is between 1 and N (N <= 100), we can find the result for every number between 1 and N and store it in an array. Then, for every entry, we only need to call for the array using the entry as the position (e.g.: v[entry]).

import java.io.*;

class Main  {
    public static final int MAX = 100;
   
    public static int reader(BufferedReader br) throws NumberFormatException, IOException {     
        int n;
        int resp = 0;       
      
        while (true) {         
            n = br.read();         
            if (n >= '0' && n <= '9') {
                break;
            }   
        }
           
        while (true) {         
            resp = resp*10 + n-'0';         
            n = br.read();         
            if (n < '0' || n > '9') {
                break;     
            }
        }
      
        return resp;     
    }
   
    public static void process() throws NumberFormatException, IOException {   
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
       
        int[] v = new int[MAX+1];
        int result = 0;
        for (int i = 1; i <= MAX; i++) {
            result += i*i;
            v[i] = result;
        }
       
        int num = reader(br);
        while (num != 0) {
            bw.write(v[num] + "\n");
           
            num = reader(br);
        }
       
        bw.flush();
        bw.close();
       
        return;
    }
   
    public static void main(String[] args) throws NumberFormatException, IOException {
        Main m = new Main();
        m.process();

        System.exit(0);
    }
}