Given a number n, write a program to calculate number of ways in which number can be expressed as product of two different factors.
Examples :
Input : 12 Output : 3 12 can be expressed as 1 * 12, 2 * 6 and 3*4. Input : 36 Output : 4 36 can be expressed as 1 * 36, 2 * 18, 3 * 12 and 4 * 9.
All factors of 12 are = 1, 2, 3, 4, 6, 12 We can observe that factors always exist in pair which is equal to number. Here (1, 12), (2, 6) and (3, 4) are such pairs.
As a number can expressed as product of two factors we only need to find number of factors of number upto square root of number. And we only need to find only different pairs so in case of perfect square we don’t include that factor.
C++
// CPP program to find number of ways // in which number expressed as // product of two different factors #include <bits/stdc++.h> using namespace std; // To count number of ways in which // number expressed as product // of two different numbers int countWays(int n) { // To store count of such pairs int count = 0; // Counting number of pairs // upto sqrt(n) - 1 for (int i = 1; i * i < n; i++) if (n % i == 0) count++; // To return count of pairs return count; } // Driver program to test countWays() int main() { int n = 12; cout << countWays(n) << endl; return 0; } |
Java
// Java program to find number of ways // in which number expressed as // product of two different factors public class Main { // To count number of ways in which // number expressed as product // of two different numbers static int countWays(int n) { // To store count of such pairs int count = 0; // Counting number of pairs // upto sqrt(n) - 1 for (int i = 1; i * i < n; i++) if (n % i == 0) count++; // To return count of pairs return count; } // Driver program to test countWays() public static void main(String[] args) { int n = 12; System.out.println(countWays(n)); } } |
Python 3
# Python 3 program to find number of ways # in which number expressed as # product of two different factors # To count number of ways in which # number expressed as product # of two different numbers def countWays(n): # To store count of such pairs count = 0 i = 1 # Counting number of pairs # upto sqrt(n) - 1 while ((i * i)<n) : if(n % i == 0): count += 1 i += 1 # To return count of pairs return count # Driver program to test countWays() n = 12print (countWays(n)) # This code is contributed # by Azkia Anam. |
C#
// C# program to find number of ways // in which number expressed as // product of two different factors using System; public class main { // To count number of ways in which // number expressed as product // of two different numbers static int countWays(int n) { // To store count of such pairs int count = 0; // Counting number of pairs // upto sqrt(n) - 1 for (int i = 1; i * i < n; i++) if (n % i == 0) count++; // To return count of pairs return count; } // Driver program to test countWays() public static void Main() { int n = 12; Console.WriteLine(countWays(n)); } } // This code is contributed by Anant Agarwal. |
PHP
<?php // PHP program to find number of ways // in which number expressed as // product of two different factors // To count number of ways in which // number expressed as product // of two different numbers function countWays($n) { // To store count of such pairs $count = 0; // Counting number of pairs // upto sqrt(n) - 1 for ($i = 1; $i * $i < $n; $i++) if ($n % $i == 0) $count++; // To return count of pairs return $count; } // Driver Code $n = 12; echo countWays($n), "\n"; // This code is contributed by ajit ?> |
Output :
3
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