Sum of matrix element where each elements is integer division of row and column

Consider a N X N matrix where each element is row number divide by column number (integer division), i.e. mat[i][j] = floor((i+1)/(j+1)) where 0 <= i < n and 0 <= j < n. The task is to find the sum of all matrix element.

Examples :

Input  : N = 2
Output : 4
2 X 2 matrix with given constraint:
1 0
2 1
Sum of matrix element: 4

Input  : N = 3
Output : 9

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 (Brute Force):
Run two loops, one for row and another for column and find integer part of (i / j) and add to the answer.

Below is the implementation of this approach:

C++

// C++ program to find sum of matrix element 
// where each element is integer division of 
// row and column. 
#include<bits/stdc++.h> 
using namespace std; 
  
// Return sum of matrix element where each element 
// is division of its corresponding row and column. 
int findSum(int n) 
{ 
    int ans = 0; 
    for (int i = 1; i <= n; i++)   // for rows 
        for (int j = 1; j <= n; j++) // for columns 
            ans += (i/j); 
    return ans; 
} 
  
// Driven Program 
int main() 
{ 
    int N = 2; 
    cout << findSum(N) << endl; 
    return 0; 
} 

Java

// java program to find sum of matrix 
// element where each element is integer 
// division of row and column. 
  
import java.io.*; 
  
class GFG { 
      
    // Return sum of matrix element  
    // where each element is division 
    // of its corresponding row and 
    // column. 
    static int findSum(int n) 
    { 
        int ans = 0; 
          
        // for rows 
        for (int i = 1; i <= n; i++)  
          
            // for columns 
            for (int j = 1; j <= n; j++)  
                ans += (i/j); 
                  
        return ans; 
    } 
      
    // Driven Program 
    public static void main (String[] args)  
    { 
        int N = 2; 
        System.out.println( findSum(N)); 
    } 
} 
  
// This code is contributed by anuj_67. 

Python3

# Python 3 program to find sum of  
# matrix element where each element  
# is integer division of row and column.  
  
# Return sum of matrix element  
# where each element is division  
# of its corresponding row and column.  
def findSum(N): 
    ans = 0
    for i in range(1, N + 1): 
        for j in range(1, N + 1): 
            ans += i // j 
    return ans 
  
# Driver code 
N = 2
print(findSum(N)) 
  
# This code is contributed  
# by Shrikant13 

C#

// C# program to find sum of matrix 
// element where each element is integer 
// division of row and column. 
using System; 
  
class GFG { 
      
    // Return sum of matrix element  
    // where each element is division 
    // of its corresponding row and 
    // column. 
    static int findSum(int n) 
    { 
        int ans = 0; 
          
        // for rows 
        for (int i = 1; i <= n; i++)  
          
            // for columns 
            for (int j = 1; j <= n; j++)  
                ans += (i/j); 
                  
        return ans; 
    } 
      
    // Driven Program 
    public static void Main ()  
    { 
        int N = 2; 
        Console.WriteLine( findSum(N)); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// PHP program to find sum of matrix element 
// where each element is integer division of 
// row and column. 
  
// Return sum of matrix element 
// where each element is division  
// of its corresponding row and  
// column. 
function findSum( $n) 
{ 
    $ans = 0; 
      
    // for rows 
    for($i = 1; $i <= $n; $i++)  
      
        // for columns 
        for($j = 1; $j <= $n; $j++)  
            $ans += ($i / $j); 
    return floor($ans); 
} 
  
    // Driver Code 
    $N = 2; 
    echo findSum($N); 
      
// This code is contributed by anuj_67. 
?> 

Output:

Method 2 (Efficient):
Let N = 9, matrix will be

Observe, for each jth column
mat[i][k] = 0, for 1 <= k < j, 1 <= i <= N
mat[i][k] = 1, for j <= k < 2*j, 1 <= i <= N
mat[i][k] = 2, for 2*j <= k < 3*j, 1 <= i <= N
and so on.
So, in each column i, there are i – 1 zeroes, followed by i times 1, followed by i times 2 and so on.

We traverse matrix column by column and sum elements.

Below is the implementation of this approach
:

C++

// C++ program to find sum of matrix element 
// where each element is integer divison of 
// row and column. 
#include<bits/stdc++.h> 
using namespace std; 
  
// Return sum of matrix element where each 
// element is division of its corresponding 
// row and column. 
int findSum(int n) 
{ 
    int ans = 0, temp = 0, num; 
  
    // For each column. 
    for (int i = 1; i <= n && temp < n; i++) 
    { 
        // count the number of elements of 
        // each column. Initialize to i -1 
        // because number of zeroes are i - 1. 
        temp = i - 1; 
  
        // For multiply 
        num = 1; 
  
        while (temp < n) 
        { 
            if (temp + i <= n) 
                ans += (i * num); 
            else
                ans += ((n - temp) * num); 
  
            temp += i; 
            num ++; 
        } 
    } 
  
    return ans; 
} 
  
// Driven Program 
int main() 
{ 
    int N = 2; 
    cout << findSum(N) << endl; 
    return 0; 
} 

Java

// java program to find sum of matrix element 
// where each element is integer divison of 
// row and column. 
  
import java.io.*; 
  
class GFG { 
      
    // Return sum of matrix element where each 
    // element is division of its corresponding 
    // row and column. 
    static int findSum(int n) 
    { 
        int ans = 0, temp = 0, num; 
      
        // For each column. 
        for (int i = 1; i <= n && temp < n; i++) 
        { 
              
            // count the number of elements of 
            // each column. Initialize to i -1 
            // because number of zeroes are i - 1. 
            temp = i - 1; 
      
            // For multiply 
            num = 1; 
      
            while (temp < n) 
            { 
                if (temp + i <= n) 
                    ans += (i * num); 
                else
                    ans += ((n - temp) * num); 
      
                temp += i; 
                num ++; 
            } 
        } 
      
        return ans; 
    } 
      
    // Driven Program 
    public static void main (String[] args)  
    { 
        int N = 2; 
        System.out.println(findSum(N)); 
    } 
} 
  
// This code is contributed by anuj_67. 

Python3

# Program to find sum of matrix element  
# where each element is integer divison   
# of row and column.  
  
# Return sum of matrix element where each  
# element is division of its corresponding  
# row and column.  
def findSum(n): 
    ans = 0; temp = 0; 
  
    for i in range(1, n + 1): 
  
        # count the number of elements of  
        # each column. Initialize to i -1  
        # because number of zeroes are i - 1.  
        if temp < n: 
            temp = i - 1
  
            # For multiply 
            num = 1
            while temp < n: 
                if temp + i <= n: 
                    ans += i * num 
                else: 
                    ans += (n - temp) * num 
                temp += i 
                num += 1
    return ans 
  
# Driver Code 
N = 2
print(findSum(N)) 
  
# This code is contributed by Shrikant13 

C#

// C# program to find sum of matrix  
// element where each element is  
// integer divison of row and column. 
using System; 
  
class GFG  
{ 
      
    // Return sum of matrix element  
    // where each element is division  
    // of its corresponding row and column. 
    static int findSum(int n) 
    { 
        int ans = 0, temp = 0, num; 
      
        // For each column. 
        for (int i = 1; i <= n && temp < n; i++) 
        { 
              
            // count the number of elements  
            // of each column. Initialize  
            // to i -1 because number of  
            // zeroes are i - 1. 
            temp = i - 1; 
      
            // For multiply 
            num = 1; 
      
            while (temp < n) 
            { 
                if (temp + i <= n) 
                    ans += (i * num); 
                else
                    ans += ((n - temp) * num); 
      
                temp += i; 
                num ++; 
            } 
        } 
      
        return ans; 
    } 
      
    // Driver Code 
    public static void Main ()  
    { 
        int N = 2; 
        Console.WriteLine(findSum(N)); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// PHP program to find sum of  
// matrix element where each  
// element is integer divison  
// of row and column. 
  
// Return sum of matrix element  
// where each element is division  
// of its corresponding row and column. 
function findSum( $n) 
{ 
    $ans = 0; $temp = 0; $num; 
  
    // For each column. 
    for ($i = 1; $i <= $n and 
         $temp < $n; $i++) 
    { 
        // count the number of elements  
        // of each column. Initialize  
        // to i -1 because number of  
        // zeroes are i - 1. 
        $temp = $i - 1; 
  
        // For multiply 
        $num = 1; 
  
        while ($temp < $n) 
        { 
            if ($temp + $i <= $n) 
                $ans += ($i * $num); 
            else
                $ans += (($n - $temp) *      
                                $num); 
  
            $temp += $i; 
            $num ++; 
        } 
    } 
  
    return $ans; 
} 
  
// Driver Code 
$N = 2; 
echo findSum($N); 
  
// This code is contributed by anuj_67. 
?> 

Output :

Source: http://stackoverflow.com/questions/41094769/sum-of-integer-division-matrix?rq=1

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Sum of matrix element where each elements is integer division of row and column

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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