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);

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.

#include<stdio.h>

int power(int x, int y);

int main()

{

int n,a[30],i,j,count=0;

scanf("%d",&n);

i=0;

while(i<n)

{

scanf("%d",&a[i]);

i++;

}

for(i=0;i<n;i++)

{

for(j=i+1;j<n;j++)

{

if(power(a[i],a[j])>power(a[j],a[i]))

count++;

}

}

printf("%d",count);

return 0;

}

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;

}