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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <math.h>
int main(void){
    
    int n;
    int l[2001];
    int ans=0;
    int tmp;
    int loop=0;
    
    scanf("%d", &n);
    
    for(int i=1; i<=n; i++){
        
        scanf("%d", &l[i]);
        
    }
    
    for(int i=1; i<=n; i++){
        
        for(int j=i+1; j<=n; j++){
            
            if(l[i]>l[j]){
                
                tmp = l[i];
                l[i] = l[j];
                l[j] = tmp;
                
            }
            
        }
        
    }
    
    for(int p=1; p<=n-2; p++){
        
        for(int q=p+1; q<=n-1; q++){
            
            loop = 0;
            
            for(int r=q+1; loop<999999; r++){
                
                loop += 1;
                
                if(l[p]+l[q]<=l[r]){ // kokokara-muri
                    
                    ans += r - q - 1;
                    
                    break;
                    
                } else if(loop>=n-q){ // owari
                    
                    ans += r - q;
                    
                    break;
                    
                } else if(l[p]+l[q]>l[n-r+q+1]){ // kokomade-dekiru
                    
                    ans += n - r + 1;
                    
                    break;
                    
                }
                
                loop += 1;
                
            }
            
        }
        
    }
    
    printf("%d", ans);
    
    return 0;
    
}