Check if N can be obtained by repetitive addition or subtraction of two given numbers

Given three positive integers N, A, and B, the task is to check if it is possible to obtain N by adding or subtracting A and B multiple times. 
Examples:

Input: N = 11, A = 2, B = 5 
Output: YES 
Explanation:  11 = 5 + 5 + 5 – 2 -2
Input: N = 11, A = 2, B = 4 
Output: NO 
Explanation: Not possible to obtain 11 from 2 and 4 only. 
 

Approach: 
Follow the steps below to solve the problem:

  • The task is to check if it is possible to add or subtract A and B multiple times and obtain N as the end result.
  • Hence, in terms of linear equation, it can be written as: 
    Ax + By = N,  
    where x and y represent the number of times A and B are added or subtracted. Negative x represents A is subtracted x times and similarly, negative y represents B is subtracted y times
  • Now, the aim is to find the integral solutions for the above equation. Here, it is quite valid to use Extended Euclid Algorithm which says that the solutions exist if and only if N % gcd(a, b) is 0.

Below is the implementation of the above approach: 
 

C++

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// C++ Program to check if 
// a number can be obtained 
// by repetitive addition 
// or subtraction of two numbers 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to check and return if 
// it is possible to obtain N by 
// repetitive addition or subtraction 
// of A and B 
bool isPossible(int N, int a, int b) 
    // Calculate GCD of A and B 
    int g = __gcd(a, b); 
  
    // Condition to check 
    // if N can be obtained 
    if (N % g == 0) 
        return true
    else
        return false
  
// Driver Code 
int main() 
    int N = 11, a = 2; 
    int b = 5; 
  
    if (isPossible(N, a, b)) 
        cout << "YES"
    else
        cout << "NO"

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Java

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// Java program to check if 
// a number can be obtained 
// by repetitive addition 
// or subtraction of two numbers 
class GFG{ 
      
// Recursive function to return 
// gcd of a and b 
public static int gcd(int a, int b) 
    if (b == 0
        return a; 
    return gcd(b, a % b); 
      
// Function to check and return if 
// it is possible to obtain N by 
// repetitive addition or subtraction 
// of A and B 
public static boolean isPossible(int N, int a, 
                                 int b) 
      
    // Calculate GCD of A and B 
    int g = gcd(a, b); 
      
    // Condition to check 
    // if N can be obtained 
    if (N % g == 0
        return true
    else
        return false
  
// Driver code 
public static void main(String[] args) 
    int N = 11, a = 2
    int b = 5
      
    if (isPossible(N, a, b)) 
        System.out.print("YES"); 
    else
        System.out.print("NO"); 
  
// This code is contributed by divyeshrabadiya07 

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Python3

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# Python3 program to check if 
# a number can be obtained 
# by repetitive addition 
# or subtraction of two numbers 
  
# Recursive function to return 
# gcd of a and b 
def gcd(a, b): 
      
    if (b == 0): 
        return
    return gcd(b, a % b) 
      
# Function to check and return if 
# it is possible to obtain N by 
# repetitive addition or subtraction 
# of A and B 
def isPossible(N, a, b): 
      
    # Calculate GCD of A and B 
    g = gcd(a, b) 
      
    # Condition to check 
    # if N can be obtained 
    if (N % g == 0): 
        return True
    else
        return False
          
# Driver code 
N = 11
a = 2
b = 5
  
if (isPossible(N, a, b) != False): 
    print("YES"
else
    print("NO"
  
# This code is contributed by code_hunt 

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C#

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// C# program to check if 
// a number can be obtained 
// by repetitive addition 
// or subtraction of two numbers 
using System;
  
class GFG{
      
// Recursive function to return 
// gcd of a and b 
static int gcd(int a, int b) 
    if (b == 0) 
        return a; 
    return gcd(b, a % b); 
      
// Function to check and return if 
// it is possible to obtain N by 
// repetitive addition or subtraction 
// of A and B 
static bool isPossible(int N, int a,
                              int b) 
      
    // Calculate GCD of A and B 
    int g = gcd(a, b); 
      
    // Condition to check 
    // if N can be obtained 
    if (N % g == 0) 
        return true
    else
        return false
  
// Driver code
public static void Main()
{
    int N = 11, a = 2; 
    int b = 5; 
      
    if (isPossible(N, a, b)) 
        Console.Write("YES"); 
    else
        Console.Write("NO"); 
}
}
  
// This code is contributed by chitranayal

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Output: 

YES

Time Complexity: O(log(min(A, B)) 
Auxiliary Space: O(1)
 

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