Since \(N \leq 100\), it is possible try all tuples \((i, j, k)\) and check if it forms a triangle. Three sides of length \(a\), \(b\) and \(c\) forms a triangle if and only if

• \(a + b > c\),
• \(b + c > a\) and
• \(c + a > b\),

so you can solve this problem by checking the following two conditions: - \(L_i\), \(L_j\) and \(L_k\) are all different, and - \(L_i\), \(L_j\) and \(L_k\) satisfies the condition above.

The total time complexity is \(O(N^3)\).

Just an aside, if you sort \(L\) firsthand and assume that they satisfy \(L_1 \leq L_2, \cdots, \leq L_N\), the only triangular inequality that has to be check is that \(L_i + L_j > L_k\).

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int N;
    cin >> N;
    vector<int> vec(N);
    for (int i = 0; i < N; ++i) cin >> vec[i];
    sort(vec.begin(), vec.end());

    int ans = 0;

    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < i; ++j) {
            for (int k = 0; k < j; ++k) {
                if (vec[k] != vec[j] and vec[i] != vec[j] and
                    vec[k] + vec[j] > vec[i]) {
                    ans++;
                }
            }
        }
    }

    cout << ans << endl;

    return 0;
}