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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoder.Abc
{
    class QuestionD
    {
        public static void Main(string[] args)
        {
            // 整数の入力
            int n = int.Parse(Console.ReadLine());

            var lArray = Console.ReadLine().Split(' ').Select(i => int.Parse(i)).ToList();
            lArray.Sort();

            //配列をユニークにする
            Dictionary<int, int> dic = new Dictionary<int, int>();
            foreach (var l in lArray)
            {
                if (dic.ContainsKey(l))
                {
                    dic[l]++;
                }
                else
                {
                    dic.Add(l, 1);
                }
            }

            int currentI = -1;
            int currentJ = -1;
            int currentK = -1;

            long result = 0;

            for (int i = 0; i < n - 2; i++)
            {
                if (lArray[i] == currentI)
                {
                    continue;
                }
                currentI = lArray[i];
                currentJ = -1;
                for (int j = i + 1; j < n - 1; j++)
                {
                    if (lArray[j] == currentJ)
                    {
                        continue;
                    }
                    currentJ = lArray[j];
                    currentK = -1;
                    for (int k = j + 1; k < n; k++)
                    {
                        if (lArray[k] == currentK)
                        {
                            continue;
                        }
                        currentK = lArray[k];
                        if (currentK < currentI + currentJ)
                        {
                            int sum;

                            //resultに全パターンをまとめて追加する
                            if (currentI == currentJ && currentI == currentK)
                            {
                                //全部同じ長さ
                                var count = dic[currentI];
                                sum = count * (count - 1) * (count - 2) / 6;
                            }
                            else if (currentI == currentJ)
                            {
                                //I,Jが同じ長さで、Kだけ違う
                                var count = dic[currentI];
                                sum = dic[currentK] * (count * (count - 1) / 2);
                            }
                            else if (currentJ == currentK)
                            {
                                //J,Kが同じ長さで、Iだけ違う
                                var count = dic[currentJ];
                                sum = dic[currentI] * (count * (count - 1) / 2);
                            }
                            else
                            {
                                //全部違う長さ
                                sum = dic[currentI] * dic[currentJ] * dic[currentK];
                            }
                            Console.WriteLine($"{currentI}-{currentJ}-{lArray[k]} => {sum}");
                            result += sum;
                        }
                        else
                        {
                            break;
                        }
                    }
                }
            }

            Console.WriteLine(result);
        }
    }
}