Problem:
Score : 200 points

Problem Statement

We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.

In how many ways can we choose three of the sticks with different lengths that can form a triangle?

That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:

L_i, L_j, and L_k are all different.

There exists a triangle whose sides have lengths L_i, L_j, and L_k.

Constraints

1 \leq N \leq 100

1 \leq L_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
L_1 L_2 \cdots L_N

Output

Print the number of ways to choose three of the sticks with different lengths that can form a triangle.

Sample Input 1

5
4 4 9 7 5

Sample Output 1

5

The following five triples (i, j, k) satisfy the conditions: (1, 3, 4), (1, 4, 5), (2, 3, 4), (2, 4, 5), and (3, 4, 5).

Sample Input 2

6
4 5 4 3 3 5

Sample Output 2

8

We have two sticks for each of the lengths 3, 4, and 5. To satisfy the first condition, we have to choose one from each length.

There is a triangle whose sides have lengths 3, 4, and 5, so we have 2 ^ 3 = 8 triples (i, j, k) that satisfy the conditions.

Sample Input 3

10
9 4 6 1 9 6 10 6 6 8

Sample Output 3

39

Sample Input 4

2
1 1

Sample Output 4

0

No triple (i, j, k) satisfies 1 \leq i < j < k \leq N, so we should print 0.

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define true 1
#define false 0
typedef long long ll;
typedef unsigned char Uchar; 

int main(void){
  
  int i,j,k;
  int n;
  int a[1000001];
  scanf("%d",&n);
  for(i=1;i<n+1;i++){
   scanf("%d",&a[i]); 
  }
  
  int tmp;
  for(i=1;i<n+1;i++){
    for(j=i+1;j<n+1;j++){
      if(a[i] > a[j]){
      tmp = a[i];
      a[i] = a[j];
      a[j] = tmp;
      }
    }
  }
  
  int s,t,u;
  int ans=0;
  for(i=1;i<n+1;i++){
    s = a[i];
    for(j=i+1;j<n+1;j++){
      if(s == a[j]) continue;
         //printf("[]%d,%d,%d\n",i,j,k);
      	t = a[j];
        for(k=j+1;k<n+1;k++){
          //printf("[]%d,%d,%d\n",i,j,k);
         if(t == a[k])continue;
          u = a[k]; 
           if(s+t > u){
             ans++;
             //printf("[]%d,%d,%d\n",i,j,k);
             //printf("--%d,%d,%d\n",s,t,u);
           }
        }
    }//j
    
  }//i
  
  printf("%d",ans);
  
  return 0;
}