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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef vector<ld> vd;
typedef vector<ll> vll;
typedef vector<string> vstr;
typedef vector<vector<ll>> vvll;
typedef vector<pair<ll, ll>> vpll;
typedef priority_queue<ll, vector<ll>, greater<ll>> spqll;  // 小さい順に取り出し
typedef priority_queue<ll, vector<ll>, less<ll>> bpqll;     // 大きい順に取り出し

#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define FOR(i, a, b) for (ll i = (ll)a; i < (ll)b; i++)
#define IREP(i, v) for (auto i = (v).begin(); i != (v).end(); i++)
#define FI first
#define SE second
#define MP make_pair
#define MT make_tuple
#define PB push_back
#define TS to_string
#define NP next_permutation
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) (ll) x.size()
#define SP(x) setprecision((ll)x)


ll INF = 1e9;
ll NIL = -1;
ll MOD = 1000000007;
ll LINF = 1e18;
ld EPS = 1e-9;
ld PI = M_PI;
vll dx = {1, 0, -1, 0, 1, -1, -1, 1};
vll dy = {0, 1, 0, -1, 1, 1, -1, -1};
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a;} //最大公約数
ll lcm(ll a, ll b) { return a / gcd(a, b) * b;}     //最小公倍数

void yes(){ cout << "Yes" << endl;}
void no(){ cout << "No" << endl;}

//-----------------------------------------

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

    ll n;
    cin >>n;
    vll l(n);
    REP(i,n){
        cin >> l[i];
    }
    sort(ALL(l));
    ll ans = 0;
    REP(i,n-2){
        FOR(j,i+1,n-1){
            ll lef = j;
            ll rig = n;
            while(rig-lef>1){
                ll mid = (lef+rig) /2;
                if(l[i]+l[j]>l[mid]){
                    lef = mid;
                }
                else{
                    rig = mid;
                }
            }
            ans += (lef-j);
        }
    }
    cout << ans <<endl;

    
    return 0;
}