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: Wrong Answer)

#include <iostream>
#include <fstream>
#include <vector>
#include <set>
#include <map>
#include <cstring>
#include <string>
#include <cmath>
#include <cassert>
#include <ctime>
#include <algorithm>
#include <sstream>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <cstdlib>
#include <cstdio>
#include <iterator>
#include <functional>
#include <bitset>
#include <iterator>
typedef long long ll;
typedef long double ld;
using namespace std;
typedef pair<int,int> pii;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int MAXN = 2e5+50;
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define f first
#define s second
#define rep(i,l,r) for(int i=l;i<=r;++i)
#define per(i,r,l) for(int i=r;i>=l;--i)
#define rep0(i,l,r) for(int i=l;i<r;++i
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) { return a * (b / gcd(a, b)); }

 
int findNumberOfTriangles(int arr[], int n) 
{ 
    // Count of triangles 
    int count = 0; 
  
    // The three loops select three 
    // different values from array 
    for (int i = 0; i < n; i++) { 
        for (int j = i + 1; j < n; j++) { 
  
            // The innermost loop checks for 
            // the triangle property 
            for (int k = j + 1; k < n; k++) 
  
                // Sum of two sides is greater 
                // than the third 
                if ( 
                    arr[i] + arr[j] > arr[k] 
                    && arr[i] + arr[k] > arr[j] 
                    && arr[k] + arr[j] > arr[i]) 
                    count++; 
        } 
    } 
    return count; 
} 
  
// Driver code 
int main() 
{ int size;
  cin>>size;
  int arr[size];
  for(int i=0;i<size;++i)
     cin>>arr[i];
  
    cout<<findNumberOfTriangles(arr, size); 
  
    return 0; 
}