Sum of each element raised to (prime-1) % prime

Given an array arr[] and a positive integer P where P is prime and none of the elements of array are divisible by P. Find sum of all the elements of the array raised to the power P – 1 i.e. arr[0]^{P – 1} + arr[1]^{P – 1} + … + arr[n – 1]^{P – 1} and print the result modulo P.
Examples:

Input: arr[] = {2, 5}, P = 3 
Output: 2 
2² + 5² = 29 and 29 % 3 = 2
Input: arr[] = {5, 6, 8}, P = 7 
Output: 3 

Approach: This problem is a direct application of Fermats’s Little Theorem, a^(P-1) = 1 (mod p) where a is not divisible by P. Since, none of the elements of array arr[] are divisible by P, each element arr[i] will give the value 1 with the given operation. 
Therefore, our answer will be 1 + 1 + … (upto n(size of array)) = n.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <iostream>
#include <vector>

using namespace std;

// Function to return the required sum
int getSum(vector<int> arr, int p)
{
    return arr.size();
}

// Driver code
int main()
{
    vector<int> arr = { 5, 6, 8 };
    int p = 7;
    cout << getSum(arr, p) << endl;

    return 0;
}

// This code is contributed by Rituraj Jain

Java

// Java implementation of the approach
public class GFG {

    // Function to return the required sum
    public static int getSum(int arr[], int p)
    {
        return arr.length;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 5, 6, 8 };
        int p = 7;
        System.out.print(getSum(arr, p));
    }
}

Python3

# Python3 implementation of the approach
# Function to return the required sum
def getSum(arr, p) :

    return len(arr)

# Driver code
if __name__ == "__main__" :

    arr = [5, 6, 8]
    p = 7
    print(getSum(arr, p))

# This code is contributed by Ryuga

C#

// C# implementation of the approach

using System;

public class GFG{

    // Function to return the required sum
    public static int getSum(int []arr, int p)
    {
        return arr.Length;
    }

    // Driver code
    static public void Main (){
        int []arr = { 5, 6, 8 };
        int p = 7;
        Console.WriteLine(getSum(arr, p));
    }

//This Code is contributed by akt_mit
}

PHP

<?php
// PHP implementation of the approach

// Function to return the required sum
function getSum($arr, $p)
{
    return count($arr);
}

// Driver code
$arr = array( 5, 6, 8 );
$p = 7;
echo (getSum($arr, $p));

// This code is contributed
// by Sach_Code
?>

Javascript

<script>

    // Javascript implementation of the approach

    // Function to return the required sum
    function getSum(arr, p)
    {
        return arr.length;
    }

    let arr = [ 5, 6, 8 ];
    let p = 7;
    document.write(getSum(arr, p));

</script>

Output:

3

Time Complexity: O(1)

Auxiliary Space: O(1)