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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
constexpr int64_t mod=1e9+7;

int main()
{
  int n;
  cin>>n;
  map<int,int> map;
  for (int i = 0; i < n; ++i)
  {
    int tmp;
    cin>>tmp;
    map[tmp]++;
  }
  std::vector<int> l;
  for(auto&& e: map) {
    l.push_back(e.first);
  }
  int64_t ans=0;

  if(l.size()<3){
    for(auto&& e: map) {
      if(e.second==1) continue;
      else{
        for(auto&& f: map) {
          if(f.first==e.first) continue;
          if(f.first<2*e.first) ans++;
        }
        if(e.second>=3) ans++;
      }
    }
    cout<<ans;
    return 0;
  }
  n=l.size();
  sort(l.begin(), l.end());
  for (int i = n-1; i > 1; --i)
  {
    for (int j = i-1; j > 0 ; --j)
    {
      int tmp = l[i]-l[j];
      if(tmp>=l[j]) break;
      auto lb = lower_bound(l.begin(), l.end(),tmp);
      if(*lb==tmp) lb++;
      ans +=l.end()-lb-(n-i)-(i-j);
    }
  }
  for(auto&& e: map) {
    if(e.second==1) continue;
    else{
      for(auto&& f: map) {
        if(f.first==e.first) continue;
        if(f.first<2*e.first) ans++;
      }
      if(e.second>=3) ans++;
    }
  }
  cout<<ans;
  
}