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Check if N can be obtained by repetitive addition or subtraction of two given numbers
• Last Updated : 06 Apr, 2021

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++

 `// C++ Program to check if``// a number can be obtained``// by repetitive addition``// or subtraction of two numbers` `#include ``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"``;``}`

## Java

 `// 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`

## Python3

 `# 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` `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``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`

## C#

 `// 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`

## Javascript

 ``
Output:
`YES`

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

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