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: Runtime Error)

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


int main() {

  int n;
  cin>>n;

  vector<int> L(n);
  for(int i=0;i<n;i++){
    cin>>L.at(i);
  }

  sort(L.begin(),L.end());

  vector<vector<vector<char>>> data(n,vector<vector<char>>(n, vector<char>(n,0)));
  int cnt=0;
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      for(int k=max(0,j-i);k<min(n,i+j);k++){
        if(i!=j&&j!=k&&i!=k&&L.at(i)<L.at(j)+L.at(k)&&L.at(j)<L.at(k)+L.at(i)&&L.at(k)<L.at(i)+L.at(j))
        {
          if(data.at(i).at(j).at(k)==0&&data.at(i).at(k).at(j)==0&&data.at(k).at(i).at(j)==0&&data.at(k).at(j).at(i)==0&&data.at(j).at(i).at(k)==0&&data.at(j).at(k).at(i)==0){
            cnt++;
            data.at(i).at(j).at(k)=1;
            data.at(i).at(k).at(j)=1;
            data.at(k).at(i).at(j)=1;
            data.at(k).at(j).at(i)=1;
            data.at(j).at(i).at(k)=1;
            data.at(j).at(k).at(i)=1;
          }
        }
      }
    }
  }

  cout<<cnt<<endl;

  return 0;
}

//全探索したら1000C3なので、1.67*10^8なのでTLEになる。
//(1)a<b+c -> a-b<c
//(2)b<c+a -> b-a<c
//(3)c<a+b -> c<a+b

// max(a-b,b-a)<c<a+b //cをこの範囲で探せばよい