Count of all possible values of X such that A % X = B

Given two integers A and B. The task is to find the count of all possible values X such that A % X = B. If there are infinite number of possible values then print -1.

Examples:

Input: A = 21, B = 5
Output: 2
8 and 16 are the only valid values for X.



Input: A = 5, B = 5
Output: -1
X can have any value > 5

Approach: There are three possible cases:

  1. If A < B then no value of X can satisfy the given condition.
  2. If A = B then infinite solutions are possible. So, print -1 as X can be any value greater than A.
  3. If A > B then the number of divisors of (A – B) which are greater than B is the required count.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count
// of all possible values for x
// such that (A % x) = B
int countX(int a, int b)
{
    // Case 1
    if (b > a)
        return 0;
  
    // Case 2
    else if (a == b)
        return -1;
  
    // Case 3
    else {
        int x = a - b, ans = 0;
  
        // Find the number of divisors of x
        // which are greater than b
        for (int i = 1; i * i <= x; i++) {
            if (x % i == 0) {
                int d1 = i, d2 = b - 1;
                if (i * i != x)
                    d2 = x / i;
                if (d1 > b)
                    ans++;
                if (d2 > b)
                    ans++;
            }
        }
        return ans;
    }
}
  
// Driver code
int main()
{
    int a = 21, b = 5;
  
    cout << countX(a, b);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
class GFG
{
      
    // Function to return the count 
    // of all possible values for x 
    // such that (A % x) = B 
    static int countX(int a, int b) 
    
        // Case 1 
        if (b > a) 
            return 0
      
        // Case 2 
        else if (a == b) 
            return -1
      
        // Case 3 
        else
        
            int x = a - b, ans = 0
      
            // Find the number of divisors of x 
            // which are greater than b 
            for (int i = 1; i * i <= x; i++)
            
                if (x % i == 0)
                
                    int d1 = i, d2 = b - 1
                    if (i * i != x) 
                        d2 = x / i; 
                    if (d1 > b) 
                        ans++; 
                    if (d2 > b) 
                        ans++; 
                
            
            return ans; 
        
    
  
    // Driver code 
    static public void main (String args[]) 
    
        int a = 21, b = 5
      
        System.out.println(countX(a, b)); 
      
    
}
  
// This code is contributed by AnkitRai01

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 implementation of the approach
  
# Function to return the count
# of all possible values for x
# such that (A % x) = B
def countX( a, b):
    # Case 1
    if (b > a):
        return 0
  
    # Case 2
    elif (a == b):
        return -1
  
    # Case 3
    else:
        x = a - b
        ans = 0
  
        # Find the number of divisors of x
        # which are greater than b
        i = 1
        while i * i <= x:
            if (x % i == 0):
                d1 = i
                d2 = b - 1
                if (i * i != x):
                    d2 = x // i
                if (d1 > b):
                    ans+=1
                if (d2 > b):
                    ans+=1
            i+=1
        return ans
  
# Driver code
if __name__ == "__main__":
    a = 21
    b = 5
  
    print(countX(a, b))
      
    # This code is contributed by ChitraNayal

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG
{
      
    // Function to return the count 
    // of all possible values for x 
    // such that (A % x) = B 
    static int countX(int a, int b) 
    
        // Case 1 
        if (b > a) 
            return 0; 
      
        // Case 2 
        else if (a == b) 
            return -1; 
      
        // Case 3 
        else
        
            int x = a - b, ans = 0; 
      
            // Find the number of divisors of x 
            // which are greater than b 
            for (int i = 1; i * i <= x; i++)
            
                if (x % i == 0)
                
                    int d1 = i, d2 = b - 1; 
                    if (i * i != x) 
                        d2 = x / i; 
                    if (d1 > b) 
                        ans++; 
                    if (d2 > b) 
                        ans++; 
                
            
            return ans; 
        
    
  
    // Driver code 
    static public void Main () 
    
        int a = 21, b = 5; 
      
        Console.WriteLine(countX(a, b)); 
      
    
}
  
// This code is contributed by anuj_67..

chevron_right


Output:

2


My Personal Notes arrow_drop_up

pawanasipugmailcom

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : AnkitRai01, vt_m, chitranayal