NPTEL Count pairs such that x^y > y^x

Given a  sequence A of N positive integers, write a program to find the number of pairs (A[i], A[j]) such that i < j and A[i]A[j] > A[j]A[i] (A[i] raised to the power A[j] > A[j] raised to the power A[i]).

You are provided with a function named power( ) that takes two positive integers x & y and returns xy. If y is 0, the function returns 1.

The prototype of this function is

int power(int x, int y);

You do not have to write the program for power ( ) function. This function is automatically added at the end of the code segment that you write.

In your program, use this function to count the number of such pairs.

INPUT: Line 1 contains N.

Line 2 contains a sequence of N integers, separated by whitespace.

OUTPUT: A single integer representing the number of required pairs.

CONSTRAINTS: The inputs will satisfy the following properties. It is not necessary to validate the inputs.

1 <= N <= 30

0 <= A[i] <= 8 The input sequence can have repetitions and the count must reflect that.
Solution:
#include<stdio.h>

int power(int x, int y);

int main()
{  
int n,arr[30]={0},i,count=0,j;
 scanf("%d",&n);
 for(i=0;i<n;i++ )
  scanf("%d",&arr[i]);

  for(i=0;i<n;i++)
  {
          for(j=i+1;j<n;j++)
          {
            if(power(arr[i],arr[j])>power(arr[j],arr[i]))
                    count++;
          }
  }
  printf("%d",count);

    return 0;

}

/* THE CODE BELOW WILL BE AUTOMATICALLY UNCOMMENTED DURING EXECUTION. PLEASE DO NOT MODIFY ANYTHING BELOW THIS LINE.*/

// int power(int x, int y)
// {
//   int result = x;
//
//   if(y == 0) return 1;
//   if(x < 0 || y < 0) return 0;
//
//   for(int i = 1; i < y; ++i)
//   result *= x;
//
//   return result;
//}