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Sum of GCDs of each row of the given matrix
  • Last Updated : 07 Jun, 2019
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Given a matrix mat[][] of size N * N, the task is to find the sum of GCDs of all the rows of the given matrix.

Examples:

Input: arr[][] = {
{2, 4, 6, 8},
{3, 6, 9, 12},
{4, 8, 12, 16},
{5, 10, 15, 20}};
Output: 14
gcd(2, 4, 6, 8) = 2
gcd(3, 6, 9, 12) = 3
gcd(4, 8, 12, 16) = 4
gcd(5, 10, 15, 20) = 5
2 + 3 + 4 + 5 = 14

Input: arr[][] = {
{1, 1, 1, 1},
{2, 2, 2, 2},
{3, 3, 3, 3},
{4, 4, 4, 4}};
Output: 10

Approach: Find the GCD of each row of the matrix and add it to the total sum. The total sum will be the sum of GCD of all the rows.



Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Function to return the sum of gcd
// of each row of the given matrix
int sumGcd(int arr[][4], int n)
{
  
    // To store the required sum
    int sum = 0;
  
    for (int i = 0; i < n; i++) {
  
        // To store the gcd of the current row
        int gcdRow = arr[i][0];
        for (int j = 1; j < n; j++)
            gcdRow = gcd(arr[i][j], gcdRow);
  
        // Add gcd of the current row to the sum
        sum += gcdRow;
    }
  
    // Return the required sum
    return sum;
}
  
// Driver code
int main()
{
    int arr[][4] = { { 2, 4, 6, 8 },
                     { 3, 6, 9, 12 },
                     { 4, 8, 12, 16 },
                     { 5, 10, 15, 20 } };
    int size = sizeof(arr) / sizeof(arr[0]);
  
    cout << sumGcd(arr, size);
  
    return 0;
}

Java




// Java implementation of the approach 
  
class GFG 
{
  
    // Function to return gcd of a and b 
    static int gcd(int a, int b) 
    
        if (a == 0
            return b; 
        return gcd(b % a, a); 
    
      
    // Function to return the sum of gcd 
    // of each row of the given matrix 
    static int sumGcd(int arr[][], int n) 
    
      
        // To store the required sum 
        int sum = 0
      
        for (int i = 0; i < n; i++)
        
      
            // To store the gcd of the current row 
            int gcdRow = arr[i][0]; 
            for (int j = 1; j < n; j++) 
                gcdRow = gcd(arr[i][j], gcdRow); 
      
            // Add gcd of the current row to the sum 
            sum += gcdRow; 
        
      
        // Return the required sum 
        return sum; 
    
      
    // Driver code 
    public static void main (String[] args) 
    
        int arr[][] = { { 2, 4, 6, 8 }, 
                        { 3, 6, 9, 12 }, 
                        { 4, 8, 12, 16 }, 
                        { 5, 10, 15, 20 } }; 
                          
        int size = arr.length ;
      
        System.out.println(sumGcd(arr, size)); 
      
    }
}
  
// This code is contributed by AnkitRai01

Python3




# Python3 implementation of the approach
  
# Function to return gcd of a and b
def gcd(a, b):
    if (a == 0):
        return b
    return gcd(b % a, a)
  
  
# Function to return the Sum of gcd
# of each row of the given matrix
def SumGcd(arr, n):
  
    # To store the required Sum
    Sum = 0
  
    for i in range(n):
  
        # To store the gcd of the current row
        gcdRow = arr[i][0]
  
        for j in range(1,n):
            gcdRow = gcd(arr[i][j], gcdRow)
  
        # Add gcd of the current row to the Sum
        Sum += gcdRow
  
    # Return the required Sum
    return Sum
  
  
# Driver code
  
arr= [ [ 2, 4, 6, 8 ],
    [ 3, 6, 9, 12 ],
    [ 4, 8, 12, 16 ],
    [ 5, 10, 15, 20 ] ]
  
size = len(arr)
  
print(SumGcd(arr, size))
  
# This code is contributed by mohit kumar 29

C#




// C# implementation of the approach 
using System;
class GFG 
{
  
    // Function to return gcd of a and b 
    static int gcd(int a, int b) 
    
        if (a == 0) 
            return b; 
        return gcd(b % a, a); 
    
      
    // Function to return the sum of gcd 
    // of each row of the given matrix 
    static int sumGcd(int [ , ]arr, int n) 
    
      
        // To store the required sum 
        int sum = 0; 
      
        for (int i = 0; i < n; i++)
        
      
            // To store the gcd of the current row 
            int gcdRow = arr[i, 0]; 
            for (int j = 1; j < n; j++) 
                gcdRow = gcd(arr[i, j], gcdRow); 
      
            // Add gcd of the current row to the sum 
            sum += gcdRow; 
        
      
        // Return the required sum 
        return sum; 
    
      
    // Driver code 
    public static void Main () 
    
        int [ , ] arr = new int[ 4, 4 ]{{ 2, 4, 6, 8 }, 
                                        { 3, 6, 9, 12 }, 
                                        { 4, 8, 12, 16 }, 
                                        { 5, 10, 15, 20 }}; 
                          
        int size = 4;
        Console.WriteLine(sumGcd(arr, size)); 
    }
}
  
// This code is contributed by ihritik

PHP




<?php
  
// PHP implementation of the approach
  
// Function to return gcd of a and b
function gcd($a, $b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}
  
// Function to return the sum of gcd
// of each row of the given matrix
function sumGcd(&$arr, $n)
{
  
    // To store the required sum
    $sum = 0;
  
    for ($i = 0; $i < $n; $i++) 
    {
  
        // To store the gcd of the current row
        $gcdRow = $arr[$i][0];
        for ($j = 1; $j < $n; $j++)
            $gcdRow = gcd($arr[$i][$j], $gcdRow);
  
        // Add gcd of the current row to the sum
        $sum += $gcdRow;
    }
  
    // Return the required sum
    return $sum;
}
  
// Driver code
  
    $arr = array(array( 2, 4, 6, 8 ),
                array( 3, 6, 9, 12 ),
                array(4, 8, 12, 16 ),
                array( 5, 10, 15, 20));
    $size = sizeof($arr);
  
    echo sumGcd($arr, $size);
  
    return 0;
      
// This code is contributed by ChitraNayal    
?>
Output:
14

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