Given a number N denotes the total number of elements in a matrix, the task is to print all possible order of matrix. An order is a pair (m, n) of integers where m is number of rows and n is number of columns. For example, if the number of elements is 8 then all possible orders are:
(1, 8), (2, 4), (4, 2), (8, 1).
Examples:
Input: N = 8
Output: (1, 2) (2, 4) (4, 2) (8, 1)
Input: N = 100
Output:
(1, 100) (2, 50) (4, 25) (5, 20) (10, 10) (20, 5) (25, 4) (50, 2) (100, 1)
Approach:
A matrix is said to be of order m x n if it has m rows and n columns. The total number of elements in a matrix is equal to (m*n). So we start from 1 and check one by one if it divides N(the total number of elements). If it divides, it will be one possible order.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <iostream>using namespace std;// Function to print all possible ordervoid printAllOrder(int n){ // total number of elements in a matrix // of order m * n is equal (m*n) // where m is number of rows and n is // number of columns for (int i = 1; i <= n; i++) { // if n is divisible by i then i // and n/i will be the one // possible order of the matrix if (n % i == 0) { // print the given format cout << i << " " << n / i << endl; } }}// Driver codeint main(){ int n = 10; printAllOrder(n); return 0;} |
Java
// Java implementation of the above approachclass GFG { // Function to print all possible order static void printAllOrder(int n) { // total number of elements in a matrix // of order m * n is equal (m*n) // where m is number of rows and n is // number of columns for (int i = 1; i <= n; i++) { // if n is divisible by i then i // and n/i will be the one // possible order of the matrix if (n % i == 0) { // print the given format System.out.println( i + " " + n / i ); } } } // Driver code public static void main(String []args) { int n = 10; printAllOrder(n); }}// This code is contributed by ihritik |
Python
# Python implementation of the above approach# Function to print all possible orderdef printAllOrder(n): # total number of elements in a matrix # of order m * n is equal (m*n) # where m is number of rows and n is # number of columns for i in range(1,n+1): # if n is divisible by i then i # and n/i will be the one # possible order of the matrix if (n % i == 0) : # print the given format print( i ,n // i ) # Driver coden = 10printAllOrder(n)# This code is contributed by ihritik |
C#
// C# implementation of the above approachusing System;class GFG { // Function to print all possible order static void printAllOrder(int n) { // total number of elements in a matrix // of order m * n is equal (m*n) // where m is number of rows and n is // number of columns for (int i = 1; i <= n; i++) { // if n is divisible by i then i // and n/i will be the one // possible order of the matrix if (n % i == 0) { // print the given format Console.WriteLine( i + " " + n / i ); } } } // Driver code public static void Main() { int n = 10; printAllOrder(n); }}// This code is contributed by ihritik |
PHP
<?php// PHP implementation of the above approach // Function to print all possible order function printAllOrder($n) { // total number of elements in a matrix // of order m * n is equal (m*n) // where m is number of rows and n is // number of columns for ($i = 1; $i <= $n; $i++) { // if n is divisible by i then i // and n/i will be the one // possible order of the matrix if ($n % $i == 0) { // print the given format echo $i, " ", ($n / $i), "\n"; } } } // Driver code $n = 10; printAllOrder($n); // This code is contributed by Ryuga?> |
Javascript
<script>// Java Script implementation of the above approach // Function to print all possible order function printAllOrder( n) { // total number of elements in a matrix // of order m * n is equal (m*n) // where m is number of rows and n is // number of columns for (let i = 1; i <= n; i++) { // if n is divisible by i then i // and n/i will be the one // possible order of the matrix if (n % i == 0) { // print the given format document.write( i + " " + n / i+"<br>" ); } } } // Driver code let n = 10; printAllOrder(n); // This code is contributed by sravan</script> |
Output:
1 10 2 5 5 2 10 1
Time Complexity: O(n)
Auxiliary Space: O(1)
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