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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define N 100000
#define L 2005
#define MAX(a,b) (a>b?a:b)
#define MIN(a,b) (a<b?a:b)
int length[L];
int s[L][L];
int A(int m,int n)
{
    int x=1,i;
    for(i=m; i>=m-n+1; i--) x*=i;
    for(i=2; i<=n; i++) x/=i;
    return x;
}
int main()
{
    int n,i,j,kt,l,head,tail;
    long long ans=0;
  scanf("%d",&n);

        memset(length,0,sizeof(length));
        memset(s,0,sizeof(s));
        for(i=0; i<n; i++)
        {
            scanf("%d",&l);
            length[l]++;
        }
        head=0;
        while(length[head]==0) head++;
        tail=L-1;
        while(length[tail]==0) tail--;
        for(i=head; i<=tail; i++) s[i][i]=length[i];
        for(i=head; i<tail; i++)
            for(j=i+1; j<=tail; j++)
                s[i][j]=s[i][j-1]+length[j];
        ans=0;
        for(i=head; i<=tail; i++)
            for(j=i; j<=tail; j++)
            {
                kt=MIN(tail,i+j-1);
                if(i==j)
                {
                    if(length[j]>2)
                    {
                        ans+=A(length[j],3);
                        ans+=(A(length[j],2)*s[j+1][kt]);
                    }
                    else if(length[i]>1) ans+=s[j+1][kt];
                }
                else
                {
                    if(length[j]>1)
                    {
                        ans+=(length[i]*A(length[j],2));
                        ans+=(length[i]*length[j]*s[j+1][kt]);
                    }
                    else if(length[j]>0)
                    {
                        ans+=(length[i]*s[j+1][kt]);
                    }
                }
            }
        printf("%lld\n",ans);

    return 0;
}