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 Rust (1.15.1) (judge verdict: Time Limit Exceeded)

extern crate core;

use std::fmt;
use std::i32::{MAX};
use std::cmp::{Ordering, min, max };
use std::ops::{Add, Sub, Mul, Div, Neg, Index, IndexMut, SubAssign, Range};
use std::collections::{BTreeMap, VecDeque, BinaryHeap, BTreeSet, HashMap, HashSet};
use std::fmt::{Display, Formatter, Error};

fn show<T: Display>(vec: &Vec<T>) {
    if vec.is_empty() {
        println!("[]");
    }else {
        print!("[{}", vec[0]);
        for i in 1 .. vec.len() {
            print!(", {}", vec[i]);
        }
        println!("]");
    }
}
fn show2<T: Display>(vec: &Vec<Vec<T>>) {
    if vec.is_empty() {
        println!("[]");
    }else {
        for l in vec {
            show(l);
        }
    }
}
macro_rules! read_lines{
    ($count: expr; $delimiter: expr; $ty: ty) => {{
        let line_count = $count;
        let mut vec: Vec<Vec<$ty>> = Vec::with_capacity(line_count);
        for _ in 0 .. line_count {
            vec.push(read_line!($delimiter; $ty));
        }
        vec
    }};
    ($count: expr; $ty: ty) => {{
        let line_count = $count;
        let mut vec: Vec<$ty> = Vec::with_capacity(line_count);
        for _ in 0 .. line_count {
            vec.push(read_value!());
        }
        vec
    }}
}
macro_rules! read_line{
    () => {{
        let mut line = String::new();
        std::io::stdin().read_line(&mut line).ok();
        line
    }};
    (delimiter: ' ') => {
        read_line!().split_whitespace().map(|x|x.to_string()).collect::<Vec<_>>()
    };
    (delimiter: $p:expr) => {
        read_line!().split($p).map(|x|x.to_string()).collect::<Vec<_>>()
    };
    (' ') => {
        read_line!(delimiter: ' ')
    };
    ($delimiter:expr) => {
        read_line!(delimiter: $delimiter)
    };
    (' '; $ty:ty) => {
        read_line!().split_whitespace().map(|x|x.parse::<$ty>().ok().unwrap()).collect::<Vec<$ty>>()
    };
    ($delimiter:expr; $ty:ty) => {
        read_line!($delimiter).into_iter().map(|x|x.parse::<$ty>().ok().unwrap()).collect::<Vec<$ty>>()
    };
}
macro_rules! read_value{
    () => {
        read_line!().trim().parse().ok().unwrap()
    }
}
macro_rules! let_all {
    ($($n:ident:$t:ty),*) => {
        let line = read_line!(delimiter: ' ');
        let mut iter = line.iter();
        $(let $n:$t = iter.next().unwrap().parse().ok().unwrap();)*
    };
}

macro_rules! let_mut_all {
    ($($n:ident:$t:ty),*) => {
        let line = read_line!(delimiter: ' ');
        let mut iter = line.iter();
        $(let mut $n:$t = iter.next().unwrap().parse().ok().unwrap();)*
    };
}

fn count_digit_one_just(digit: i32) -> i32 {
    match digit {
        0 => 0,
        _ => power(10, digit - 1) + count_digit_one_just(digit - 1) * 10
    }
}
fn count_digit(n: i32) -> i32 {
    match n {
        0 => 0,
        _ => 1 + count_digit(n / 10)
    }
}
fn power(base: i32, exp: i32) -> i32 {
    match exp {
        0 => 1,
        1 => base,
        _ => power(base, exp & 1) * power(base * base, exp >> 1)
    }
}
struct Solution;
impl Solution {
    pub fn count_digit_one(n: i32) -> i32 {
        match count_digit(n) {
            0 => 0,
            digit => {
                let lower = power(10, digit - 1);
                match n / lower {
                    1 => n % lower + Solution::count_digit_one(n % lower) + count_digit_one_just(digit - 1) + 1,
                    left => lower + Solution::count_digit_one(n % lower) + count_digit_one_just(digit - 1) * left
                }
            }
        }
    }
}
fn main(){
    let_all!(n: usize);
    let mut sticks = read_line!(' '; i32);
    sticks.sort();
    let sticks = sticks;
    let mut sum = 0;
    let mut i = 0;
    while i < n {
        let mut j = i + 1;
        while j < n {
            let mut k = j + 1;
            while k < n {
                if sticks[i] + sticks[j] > sticks[k] {
                    sum += 1;
                }
                k += 1;
            }
            j += 1;
        }
        i += 1;
    }
    println!("{}", sum);
}