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

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  • Last Updated : 21 Dec, 2022
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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:
22 + 52 = 29 and 29 % 3 = 2
Input: arr[] = {5, 6, 8}, P = 7 
Output:
 

 

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)


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