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Count numbers up to C that can be reduced to 0 by adding or subtracting A or B
• Last Updated : 22 Apr, 2021

Given three non-negative integers A, B, and C, the task is to count the numbers in the range [1, C] that can be reduced to 0 by adding or subtracting A or B.

Examples:

Input: A = 2, B = 4, C = 7
Output: 3
Explanation: The numbers from the range [1, 7] that can be reduced to 0 by given operations are:

1. For element 2: The number can be modified as 2 – 2 = 0.
2. For element 4: The number can be modified as 4 – 2 – 2 = 0.
3. For element 6: The number can be modified as 6 – 4 – 2 = 0.

Therefore, the total count is 3.

Input: A = 2, B = 3, C = 5
Output: 5

Approach: The given problem can be solved based on the following observations:

• Consider X and Y number of addition or subtraction of A and B are performed respectively.
• After applying the operations on any number N, it becomes Ax + By. Therefore, by Extended Euclidean Algorithm, it can be said that there exist integer coefficients x and y such that Ax + By = GCD(A, B).
• Therefore, N must be a multiple of GCD(A, B), say G. Now the problem is reduced to finding the number of multiples of G which are in the range [1, C] which is floor (C / G).

Follow the below steps to solve the problem:

• Find the GCD of A and B and store it in a variable, say G.
• Now, the count of numbers over the range [1, C] is the multiples of G having values at most C which is given by floor(C/G).

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to calculate GCD of the``// two numbers a and b``long` `long` `gcd(``long` `long` `a, ``long` `long` `b)``{``    ``// Base Case``    ``if` `(b == 0)``        ``return` `a;` `    ``// Recursively find the GCD``    ``return` `gcd(b, a % b);``}` `// Function to count the numbers up``// to C that can be reduced to 0 by``// adding or subtracting A or B``void` `countDistinctNumbers(``long` `long` `A,``                          ``long` `long` `B,``                          ``long` `long` `C)``{``    ``// Stores GCD of A and B``    ``long` `long` `g = gcd(A, B);` `    ``// Stores the count of multiples``    ``// of g in the range (0, C]``    ``long` `long` `count = C / g;` `    ``// Print the result``    ``cout << count;``}`  `// Driver Code``int` `main()``{``    ``long` `long` `A = 2, B = 3, C = 5;``    ``countDistinctNumbers(A, B, C);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``class` `GFG{` `// Function to calculate GCD of the``// two numbers a and b``static` `long` `gcd(``long` `a, ``long` `b)``{``    ` `    ``// Base Case``    ``if` `(b == ``0``)``        ``return` `a;` `    ``// Recursively find the GCD``    ``return` `gcd(b, a % b);``}` `// Function to count the numbers up``// to C that can be reduced to 0 by``// adding or subtracting A or B``static` `void` `countDistinctNumbers(``long` `A, ``long` `B,``                                 ``long` `C)``{``    ` `    ``// Stores GCD of A and B``    ``long` `g = gcd(A, B);` `    ``// Stores the count of multiples``    ``// of g in the range (0, C]``    ``long` `count = C / g;` `    ``// Print the result``    ``System.out.println(count);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``long` `A = ``2``, B = ``3``, C = ``5``;``    ` `    ``countDistinctNumbers(A, B, C);``}``}` `// This code is contributed by abhinavjain194`

## Python3

 `# Python3 program for the above approach` `# Function to calculate GCD of the``# two numbers a and b``def` `gcd(a, b):``    ` `    ``# Base Case``    ``if` `(b ``=``=` `0``):``        ``return` `a``        ` `    ``# Recursively find the GCD``    ``return` `gcd(b, a ``%` `b)` `# Function to count the numbers up``# to C that can be reduced to 0 by``# adding or subtracting A or B``def` `countDistinctNumbers(A, B, C):``    ` `    ``# Stores GCD of A and B``    ``g ``=` `gcd(A, B)``    ` `    ``# Stores the count of multiples``    ``# of g in the range (0, C]``    ``count ``=` `C ``/``/` `g`` ` `    ``# Print the result``    ``print``(count)` `# Driver code``A ``=` `2``B ``=` `3``C ``=` `5` `countDistinctNumbers(A, B, C)` `# This code is contributed by abhinavjain194`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{``    ` `// Function to calculate GCD of the``// two numbers a and b``static` `long` `gcd(``long` `a, ``long` `b)``{``    ` `    ``// Base Case``    ``if` `(b == 0)``        ``return` `a;` `    ``// Recursively find the GCD``    ``return` `gcd(b, a % b);``}` `// Function to count the numbers up``// to C that can be reduced to 0 by``// adding or subtracting A or B``static` `void` `countDistinctNumbers(``long` `A, ``long` `B,``                                 ``long` `C)``{``    ` `    ``// Stores GCD of A and B``    ``long` `g = gcd(A, B);` `    ``// Stores the count of multiples``    ``// of g in the range (0, C]``    ``long` `count = C / g;` `    ``// Print the result``    ``Console.Write(count);``}` `// Driver Code``static` `void` `Main()``{``    ``long` `A = 2, B = 3, C = 5;``    ``countDistinctNumbers(A, B, C);``}``}` `// This code is contributed by abhinavjain194`

## Javascript

 ``
Output:
`5`

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

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