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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <iomanip>
#include <cmath>
#include <cstdio>
#include <utility>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#define NDEBUG
#include <cassert>
#include <bitset>
using namespace std;

#define prv(a)            \
    for (auto x : a)      \
    {                     \
        cout << x << " "; \
    }                     \
    cout << '\n';
#define prvp(a)                                     \
    for (auto x : a)                                \
    {                                               \
        cout << x.first << " " << x.second << '\n'; \
    }                                               \
    cout << '\n';
#define ll long long
#define ld long double
#define INF pow(10, 10)
#define all(v) v.begin(), v.end()
#define bg begin()
#define ed end()
#define tr(v) for (auto it = v.bg; it != v.ed; it++)
#define f(a, b) for (ll i = a; i < b; i++)
#define fj(a, b) for (ll j = a; j < b; j++)
#define fb(b, a) for (ll i = b; i > a; i--)
#define out(x) cout << x << '\n';
#define pr1(x) cout << (x) << " ";
#define pr(x) cout << #x << " " << (x) << endl;
#define pr2(x, y) cout << #x << " " << (x) << "  " << #y << " " << (y) << endl;
#define pr3(x, y, z) cout << #x << " " << (x) << "  " << #y << " " << (y) << "  " << #z << " " << (z) << endl;
#define pp(x) cout << x.first << " " << x.second << endl;
ll min(ll a, ll b) {if(a < b) return a; else return b;} 
ll max(ll a, ll b) {if(a < b) return b; else return a;} 
ll min3(ll a, ll b, ll c) {return min(a,min(b,c));} 
ll max3(ll a, ll b, ll c) {return max(a,max(b,c));} 

const ll N = 1e9+7;
const ll S = 1e5 + 7;

ll mod(ll n);
ll gcd(ll a, ll b); 
ll modM(ll n, ll m);
ll modA(ll n, ll m);
ll modS(ll n, ll m);
ll power(ll a, ll b);
void ipgraph(ll n, ll m);
//==============================================================================================

vector<vector<ll>> v;
vector<bool> vis;

void solve()
{    
    ll n;
    cin >> n;
    vector<ll> a(n);
    f(0,n) cin >> a[i]; 
    sort(all(a)); 
    ll ans=0; 
    f(0,n)
    {
        auto it=lower_bound(all(a),a[i]+1); 
        ll i2=it-a.begin(); 
        fj(i2, n)
        {
            auto it2=lower_bound(all(a),a[i]+a[j]); 
            ll cnt=upper_bound(a.bg+j,a.ed,a[j])-lower_bound(a.bg+j,a.ed,a[j]); 
            ll i3=it2-a.begin(); 
            //pr3(i3,j,cnt)
            ans+=(i3-j-cnt); 
            //pr3(i,j,ans)
        }
    }
    out(ans)
}   

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

#ifndef ONLINE_JUDGE
    freopen("/Users/aturgupta/Desktop/CP/input.txt", "r", stdin);
    freopen("/Users/aturgupta/Desktop/CP/output.txt", "w", stdout);
#endif

    ll t;
    //cin >> t;
    t=1; 
    while (t--) solve();
    cerr << endl << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
    return 0;
}

void ipgraph(ll n, ll m)
{
    v.assign(n, {});
    vis.assign(n, 0);
    ll x, y;
    f(0, m)
    {
        cin >> x >> y;
        x--, y--;
        v[y].push_back(x);
    }
}

ll gcd(ll a, ll b)
{
    if (b > a) return gcd(b, a);
    if (b == 0) return a;
    return gcd(b, a % b);
}

ll power(ll a, ll b)
{
    if (b == 0) return 1;
    ll c = power(a, b / 2);
    if (b % 2 == 0) return modM(c,c);
    else return modM(modM(c,c),a);
}

ll mod(ll n) {return (n % N + N) % N; }
ll modM(ll n, ll m) { return ((n % N * m % N) + N) % N; }
ll modA(ll n, ll m) { return ((n % N + m % N) + N) % N; }
ll modS(ll n, ll m) { return ((n % N - m % N) + N) % N; }