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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
int main() {
  int N,a,b,c,count;
  count = 0;
  cin >> N;
  vector<int> L(N+1);
  for(int i = 0;i < N;i++){
    cin >> L.at(i);
  }
  L.at(N) = 0;
  for(int i = 0;i < N*2;i++){
    for(int j = 0;j < N;j++){
      if (L.at(j) > L.at(j+1)){
        a = L.at(j);
        L.at(j) = L.at(j + 1);
        L.at(j+1) = a;
      }
    }
  }
  for(int i = 1;i < N-1;i++){
    a = L.at(i);
    for(int j = i+1;j < N;j++){
      b = L.at(j);
      if (a == b){
        continue;
      }
      for(int k = j+1;k < N+1;k++){
        c = L.at(k);
        if (a == c){
          continue;
        }
        else if(b == c){
          continue;
        }
        else if(b+c > a and a+b > c and a+c > b){
          count += 1;
        }
      }
    }
  }
  cout << count << endl;
}