Problem:
Score : 400 points

Problem Statement

Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.

He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:

a < b + c

b < c + a

c < a + b

How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Constraints

All values in input are integers.

3 \leq N \leq 2 \times 10^3

1 \leq L_i \leq 10^3

Input

Input is given from Standard Input in the following format:

N
L_1 L_2 ... L_N

Constraints

Print the number of different triangles that can be formed.

Sample Input 1

4
3 4 2 1

Sample Output 1

1

Only one triangle can be formed: the triangle formed by the first, second, and third sticks.

Sample Input 2

3
1 1000 1

Sample Output 2

0

No triangles can be formed.

Sample Input 3

7
218 786 704 233 645 728 389

Sample Output 3

23

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
//登録しよう！→ https://www.youtube.com/channel/UCRXsI3FL_kvaVL9zoolBfbQ
using namespace std;
using ll = long long;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}//a,bの最大公約数(gcd)を求める
ll lcm(ll a,ll b){return abs(a*b)/gcd(a,b);}//a,bの最小公倍数(lcm)を求める
vector<ll> enum_div(ll n){vector<ll> ret;for(int i=1 ; i*i<=n ; i++){if(n%i == 0){ret.push_back(i);if(i!=1 && i*i!=n)ret.push_back(n/i);}}ret.push_back(n);return ret;}
//↑nの約数を求める
vector<bool> IsPrime; void sieve(size_t max){if(max+1 > IsPrime.size())IsPrime.resize(max+1,true);IsPrime[0] = false;IsPrime[1] = false;for(size_t i=2; i*i<=max; ++i)if(IsPrime[i])for(size_t j=2; i*j<=max; ++j)IsPrime[i*j] = false;}
//↑エラトステネスの篩で素数を求める
#define roundup(divisor,dividend) (divisor + (dividend - 1)) / dividend //切り上げ割り算
#define all(x) (x).begin(),(x).end() //xの初めから終わりまでのポインタ
#define size_t ll //size_tは自動でllに変換される
#define pb(x) push_back(x)
#define gre greater //sortを降順にする
#define pri_queue priority_queue //優先度付きキュー
#define syo(x) fixed << setprecision(x) //iostreamで小数をx桁表示

void solve(long long N, std::vector<long long> L){
  sort(all(L));
  int ans = 0;
  for (int i = 0; i < N-1; i++) {
    for (int l = i+1; l < N; l++) {
      sort(all(L));
      auto it = lower_bound(all(L),L[i]+L[l]);
      ans += (int)(it - L.begin() - 1) - l;
    }
  }
  cout << ans << endl;
}

// Generated by 1.1.6 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
int main(){
    long long N;
    scanf("%lld",&N);
    std::vector<long long> L(N);
    for(int i = 0 ; i < N ; i++){
        scanf("%lld",&L[i]);
    }
    solve(N, std::move(L));
}