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: Accepted)

#include <stdio.h>
#include <string.h>

int max(int a, int b){
    return(a>b)?a:b;
}

int nibutan(int* a,int i, int j, int n){
    int low=j;
    int high=n;
    int mid;
    while(low<=high){
        mid=(high+low)/2;
        if((a[mid]>(a[i]-a[j]))&&(a[mid+1]<=(a[i]-a[j])))break;
        else if(a[mid]>(a[i]-a[j]))low=mid+1;
        else high=mid-1;
    }
    return mid;
}

void swap(int *a,int *b){
    int c;
    c=*a;
    *a=*b;
    *b=c;
}
 
void add_num(int*a,int i){
    while(i>1){
        if(a[i]<=a[i/2])break;
        swap(&a[i],&a[i/2]);
        i/=2;
    }
}

void pushdown(int* a,int n){
    int i=1;
    while(1){
        if(a[i]>=max(a[2*i],a[2*i+1]))break;
        if(a[2*i]>a[2*i+1]){
            swap(&a[i],&a[2*i]);
            i*=2;
        }else{
            swap(&a[i],&a[2*i+1]);
            i=2*i+1;
        }
        if(i*2>n)break;
    }
}

int solve(int* a,int n){
    int res=0;
    for(int i=1;i<n-1;i++){
        for(int j=i+1;j<n;j++){
            res+=nibutan(a,i,j,n)-j;
        }
    }
    return res;
}

int main(void){
    int n;
    scanf("%d",&n);
    int a[n+2],b[n+1];
    for(int i=1;i<n+1;i++){
        scanf("%d",&a[i]);
        add_num(a,i);
    }
    a[n+1]=0;
    for(int i=1;i<n+1;i++){
        b[i]=a[1];
        a[1]=a[n-i+1];
        a[n-i+1]=0;
        pushdown(a,n-i);
    }
    printf("%d\n",solve(b,n));
    return 0;
}