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: Accepted)


#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <numeric>
#include <unordered_map>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <memory>

using namespace std;

long long MOD = 1e9 + 7;

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n;
    cin >> n;
    vector<int> l(n, 0);
    for (int i = 0; i < n; i++)
    {
        cin >> l[i];
    }

    std::sort(l.begin(), l.end());
    
    int count = 0;
    // 短いほうから2つ棒を選ぶ
    for (int i = 0; i < n; i++)
    {
        for (int j = i + 1; j < n; j++)
        {
            // l[i]: 一番短い棒
            // l[j]: 二番目に短い棒
            // l[i] + l[j]以上で最も小さな値の入った箇所のイテレータを返す。無い場合大きくマイナスの値。
            auto itr = std::lower_bound(l.begin(), l.end(), l[i] + l[j]);
            // vector上でのインデックスに変換
            int lowerBoundIndex = itr - l.begin();
            // l[j] < num && num < l[i] + l[j](残り1辺の条件)を満たす数をカウント。
            count += std::max(lowerBoundIndex - (j + 1), 0);
        }
    }
    cout << count << endl;
    return 0;
}