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<math.h>

int triangle(long int a, long int b, long int c){
  int maxlen, maxint, yn;
    if(a==b) return(0);
    if(a==c) return(0);
    if(b==c) return(0);
    if(a>b) {
      maxlen=a;
      maxint=0;
    }else{
      maxlen=b;
      maxint=1;
    }
    if(maxlen<c){
      maxlen=c;
      maxint=2;
    }
    if(maxint==0){
      if(a<(b+c)) yn=1;
      else yn=0;}
    if(maxint==1){
      if(b<(a+c)) yn=1;
      else yn=0;}
    if(maxint==2){
      if(c<(a+b)) yn=1;
      else yn=0;}
//    printf("%ld, %ld, %ld, %d\n", a, b, c, yn);
    return (yn);

}

int main(void){
  int n, i, j, k, ans, yn;
  long int L[100];

  ans=0;
  scanf("%d", &n);
  for (i=0; i<n; i++){
    scanf("%ld", &L[i]);
  }
  
  if(n<=2){
    ans=0;
    printf("%d",ans);
    return 0;
  }
  for(i=0; i<n-2; i++){
    for(j=i+1; j<n-1; j++){
      for(k=j+1; k<n; k++){
        yn=triangle(L[i],L[j],L[k]);
        ans+=yn;
      }
    }
  }



  printf("%d", ans);

  return 0;
}