给定元素数量的(不同顺序的)矩阵计数

给定数字N表示矩阵中元素的总数，任务是打印矩阵的所有可能顺序。顺序是一对(m，n)整数，其中m是行数，n是列数。例如，如果元素数为8，则所有可能的顺序为：
(1、8)，(2、4)，(4、2)，(8、1)。
例子：

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)

方法：
如果矩阵具有m行和n列，则称其为mxn阶。矩阵中元素的总数等于(m * n)。因此，我们从1开始并逐一检查是否除以N(元素总数)。如果它分裂，它将是一种可能的顺序。
下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function to print all possible order
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
            cout << i << " " << n / i << endl;
        }
    }
}
 
// Driver code
int main()
{
    int n = 10;
    printAllOrder(n);
    return 0;
}

Java

// Java implementation of the above approach
 
 
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
                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 order
def 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 code
n = 10
printAllOrder(n)
 
 
# This code is contributed by ihritik

C#

// C# implementation of the above approach
 
using 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

Javascript

// 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

输出：

1 10
2 5
5 2
10 1

时间复杂度： O(n)

辅助空间： O(1)