Comparing array members

i cant solve the 2nd part of this prob (found in this forum):
// Generate lottery numbers (from 1-49)
// Print number of occurrences of each number
public class Lotaria {
     public static void main(String[] args) {
          int x = 1;
          int p = 1;
          int s = 1;
          int n = 1;
          int[] loto = new int[49];    // n?s do loto
          // Generating lottery numbers (from 1-49)
          for ( n = 0; n < 49; n++ ) {
               loto[n] = (int)(Math.random() * 49 + 1);
               System.out.println(loto[n]);
          for ( p = 0; p < loto.length; p++ ) {
               for ( s = 0; s < loto.length; s++ ) {
                    if ( loto[p] == loto[s] ) {
                         x++;
                         System.out.println(loto[p]
                                          + " ocorre " + x + " vezes");// prints nonsenses
}looks like (inside the if) i'm comparing every array element to all the others and then counting each repetition...
can someone sort me out of this, or at least give me an hint? (pls dont ask for interfaces)

I did it!
// Generate lottery numbers (from 1-49)
// Print number of occurrences of each number
public class Lotaria {
     public static void main(String[] args) {
          int x = 0;
          int p = 1;
          int s = 1;
          int n = 1;
          int[] loto = new int[49];
          // Generating lottery numbers (from 1-49)
          for ( n = 0; n < 49; n++ ) {
               loto[n] = (int)(Math.random() * 49 + 1);
               System.out.println(loto[n]);
          for ( p = 0; p < loto.length; p++ ) {
               for ( s = 0; s < loto.length; s++ ) {
                    if ( loto[p] == loto[s] ) {
                    x++;
          System.out.println(loto[p] + " ocorre " + x + " vezes");
          x = 0;
}i think the 3rd loop (my original idea) can be more elegant but right now i cant cope with it
the secret was in resetting "x"!...
thanks to both for helping me

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Hello. I've been trying to solve this problem for hours, and I would be most grateful for any help you can give me.
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