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<iostream>
#include<algorithm>
using namespace std;
int main() {
	long long n, a = 0, b = 0, c = 0, ans = 0; cin >> n;
	int tt[2001], up[2001] = { 0 }, kl[2001] = { 0 };
	for (int h = 0; h < n; h++) {
		cin >> tt[h];
	}
	sort(tt, tt + n);
	for (int h = 0; h < n; h++) {
		if (h == 0) { up[a] = tt[h]; a++; }
		if (h >= 1 && tt[h - 1] == tt[h]) {
            if(kl[b-1]!=tt[h]){
			  kl[b] = tt[h]; b++;
            }
			if (h >= 2 && tt[h - 2] == tt[h - 1] && c != tt[h]) {
				ans++; c=tt[h];
			}
		}
		if (h >= 1 && tt[h - 1] != tt[h]) { up[a] = tt[h]; a++; }
	}
	for (int h = 0; h < b; h++) {
		for (int i = 0; i < a; i++) {
			if (kl[h] < kl[h] + up[i] && up[i] < kl[h] + kl[h]) {
                if(kl[h]!=up[i]){
				  ans++;
                }
			}
		}
	}
	for (int h = 0; h <= a; h++) {
		for (int i = h + 1; i <= a; i++) {
			for (int j = i + 1; j <= a; j++) {
				if (up[h] < up[i] + up[j] && up[i] < up[h] + up[j] && up[j] < up[h] + up[i]) {
					ans++;
				}
			}
		}
	}
	cout << ans << endl;
}