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Count of distinct values till C formed by adding or substracting A, B, or 0 any number of times

  • Last Updated : 01 Jul, 2021

Given three integers A, B and C. You can add or substract A, B, or 0 any number of times to form new values in range 0 < final_value ≤ C. The task is to find the count of such distinct final values possible.

Examples :

Input : A = 2, B = 3, C = 10
Output:10
Possible values are : 
0 + 3 – 2 =1
0 + 2 = 2
0 + 3 = 3
2 + 2 = 4
2 + 3 = 5
3 + 3 = 6
3+2+2=7
2+2+2+2=8
2+2+2+3=9
3+3+2+2=10

Input : A = 10, B = 2, C = 10
Output: 5 

Approach: The idea is to use the GCD g of A and B.



The above approach works Because every distinct possible value is xA+yB

  • If A can be written as g×a, B can  be written as g×b
  • Then, A required final value can be written as xAg+yBg = (x*g*a + y*g*b) = g*(xa+yb)
  • Since maximum value possible is C, therefore C = g*(xa+yb).
  • Hence count of possible such values = C/g, which is the required answer.

Below is the implementation of the above approach :

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate gcd
int gcd(int A, int B)
{
    if (B == 0)
        return A;
    else
        return gcd(B, A % B);
}
 
// Function to find number of possible final values
int getDistinctValues(int A, int B, int C)
{
     
    // Find the gcd of two numbers
    int g = gcd(A, B);
 
    // Claculate number of distinct values
    int num_values = C / g;
 
    // Return values
    return num_values;
}
 
// Driver Code
int main()
{
    int A = 2;
    int B = 3;
    int C = 10;
 
    cout << (getDistinctValues(A, B, C));
    return 0;
}
 
// This code is contributed by subhammahato348

Java




// Java program for the above approach
 
import java.util.*;
 
class GFG {
    // Function to calculate gcd
    static int gcd(int A, int B)
    {
        if (B == 0)
            return A;
        else
            return gcd(B, A % B);
    }
 
    // Function to find number of possible final values
    static int getDistinctValues(int A, int B, int C)
    {
 
        // Find the gcd of two numbers
        int g = gcd(A, B);
 
        // Claculate number of distinct values
        int num_values = C / g;
 
        // Return values
        return num_values;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int A = 2;
        int B = 3;
        int C = 10;
 
        System.out.println(getDistinctValues(A, B, C));
    }
}

C#




// C# program for the above approach
using System;
class GFG
{
 
  // Function to calculate gcd
  static int gcd(int A, int B)
  {
    if (B == 0)
      return A;
    else
      return gcd(B, A % B);
  }
 
  // Function to find number of possible final values
  static int getDistinctValues(int A, int B, int C)
  {
 
    // Find the gcd of two numbers
    int g = gcd(A, B);
 
    // Claculate number of distinct values
    int num_values = C / g;
 
    // Return values
    return num_values;
  }
 
 
  // Driver code
  static void Main()
  {
    int A = 2;
    int B = 3;
    int C = 10;
 
    Console.Write(getDistinctValues(A, B, C));
  }
}
 
// This code is contributed by sanjoy_62.

Javascript




<script>
 
// JavaScript program for the above approach
 
    // Function to calculate gcd
    function gcd(A, B)
    {
        if (B == 0)
            return A;
        else
            return gcd(B, A % B);
    }
 
    // Function to find number of possible final values
    function getDistinctValues(A, B, C)
    {
 
        // Find the gcd of two numbers
        let g = gcd(A, B);
 
        // Claculate number of distinct values
        let num_values = C / g;
 
        // Return values
        return num_values;
    }
 
    // Driver Code
 
        let A = 2;
        let B = 3;
        let C = 10;
 
        document.write(getDistinctValues(A, B, C));
 
</script>
Output
10

Time Complexity : O(log(max(A, B))

Space Complexity : O(1)

 

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