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<bits/stdc++.h>
using namespace std;

#define F                   first
#define S                   second
#define int                 long long int
#define pb                  push_back
#define mp                  make_pair
#define pii                 pair<int,int>
#define vi                  vector<int>
#define mii                 map<int,int>
#define pqb                 priority_queue<int>
#define pqs                 priority_queue<int,vi,greater<int> >
#define mod                 1000000007
#define endl                '\n'
#define rep(i,a,b)            for(int i=a;i<b;i++)
void solve() {
    int n;
    cin >> n;
    int a[n];
    rep(i, 0, n) {
        cin >> a[i];
    }
    int ans = 0;
    rep(i, 0, n) {
        rep(j, i + 1, n) {
            rep(k, j + 1, n) {
                if (a[i] == a[j] or a[i] == a[k]) {
                    continue;
                }
                else {
                    if (a[i] + a[j] > a[k] and a[j] + a[k] > a[i] and a[i] + a[k] > a[j]) {
                        //cout << a[i] << " " << a[j] << " " << a[k] << endl;
                        ans++;
                    }
                }
            }
        }
    }
    cout << ans << endl;
}

void c_p_c()
{
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
}

int32_t main()
{
   // c_p_c();
    int t;
    //cin >> t;
    t = 1;
    while (t--) {
        solve();
    }


    return 0;
}