Problem:
Score : 200 points

Problem Statement

We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.

In how many ways can we choose three of the sticks with different lengths that can form a triangle?

That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:

L_i, L_j, and L_k are all different.

There exists a triangle whose sides have lengths L_i, L_j, and L_k.

Constraints

1 \leq N \leq 100

1 \leq L_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
L_1 L_2 \cdots L_N

Output

Print the number of ways to choose three of the sticks with different lengths that can form a triangle.

Sample Input 1

5
4 4 9 7 5

Sample Output 1

5

The following five triples (i, j, k) satisfy the conditions: (1, 3, 4), (1, 4, 5), (2, 3, 4), (2, 4, 5), and (3, 4, 5).

Sample Input 2

6
4 5 4 3 3 5

Sample Output 2

8

We have two sticks for each of the lengths 3, 4, and 5. To satisfy the first condition, we have to choose one from each length.

There is a triangle whose sides have lengths 3, 4, and 5, so we have 2 ^ 3 = 8 triples (i, j, k) that satisfy the conditions.

Sample Input 3

10
9 4 6 1 9 6 10 6 6 8

Sample Output 3

39

Sample Input 4

2
1 1

Sample Output 4

0

No triple (i, j, k) satisfies 1 \leq i < j < k \leq N, so we should print 0.

Solution in Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
      	int n = sc.nextInt();
      	int[] a = new int[n];
      	for(int i=0;i<n;i++){
          a[i] = sc.nextInt();
        }
      	int ans = 0;
      	Arrays.sort(a);
        for(int i=0;i<a.length;i++){
          for(int j=i+1;j<a.length;j++){
            for(int k=j+1;k<a.length;k++){
              if( (a[i]!=a[j])&&(a[j]!=a[k])&&(a[i]!=a[k])&&(a[i]+a[j]>a[k])&&(a[i]+a[k]>a[j])&&(a[j]+a[k]>a[i])){
                ans++;
                System.out.println(i+":"+j+":"+k);
              }
            }
          }
        }
      /*
      	int count = 1;
      	for(int i=1;i<n;i++){
          if(a[i]!=a[i-1]){
            count++;
          }
        }
      	int[] num = new int[count];
      	num[0] = a[0];
      	count = 1;
      	for(int i=1;i<n;i++){
          if(a[i]!=a[i-1]){
            num[count] = a[i];
            //System.out.println(num[count-1]);
            count++;
            
          }
        }
      	for(int i=0;i<num.length;i++){
          for(int j=i+1;j<num.length;j++){
            for(int k=j+1;k<num.length;k++){
              if( (i!=j)&&(j!=k)&&(i!=k)&&(num[i]+num[j]>num[k])&&(num[i]+num[k]>num[j])&&(num[j]+num[k]>num[i])){
                ans++;
              }
            }
          }
        }
        */
      System.out.println(ans);
	}
}