Maximize profit when divisibility by two numbers have associated profits

Given five integers N, A, B, X and Y. The task is to find the maximum profit obtained from the numbers from the range [1, N]. If a positive number is divisible by A then the profit increases by X and if a positive number is divisible by B then the profit increases by Y.
Note: Profit from a positive number can be added at most once.

Examples:

Input: N = 3, A = 1, B = 2, X = 3, Y = 4
Output: 10
1, 2 and 3 are divisible by A.
2 is the only number in the given range which is divisible by B.
2 is divisible by both A and B.
1 and 3 can be divided by A to get the profit of 2 * 3 = 6
2 can be divided by B to get the profit of 1 * 4 = 4
2 is divisible by both but in order to maximise the profit it is divided by B instead of A.

Input: N = 6, A = 6, B = 2, X = 8, Y = 2
Output: 12

Approach: Easy to see that we can divide a number with both A and B only if the number is a multiple of lcm(A, B). Obviously, that number should be divided with the number that gives more profit.
So the answer equals to X * (N / A) + Y * (N / B) – min(X, Y) * (N / lcm(A, B)).

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the maximum profit
int maxProfit(int n, int a, int b, int x, int y)
{
    int res = x * (n / a);
    res += y * (n / b);
  
    // min(x, y) * n / lcm(a, b)
    res -= min(x, y) * (n / ((a * b) / __gcd(a, b)));
    return res;
}
  
// Driver code
int main()
{
    int n = 6, a = 6, b = 2, x = 8, y = 2;
    cout << maxProfit(n, a, b, x, y);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG
{
  
static int __gcd(int a, int b) 
    if (b == 0
        return a; 
    return __gcd(b, a % b); 
      
  
// Function to return the maximum profit
static int maxProfit(int n, int a, int b, 
                            int x, int y)
{
    int res = x * (n / a);
    res += y * (n / b);
  
    // min(x, y) * n / lcm(a, b)
    res -= Math.min(x, y) * (n / ((a * b) / __gcd(a, b)));
    return res;
}
  
// Driver code
public static void main (String[] args)
{
    int n = 6, a = 6, b = 2, x = 8, y = 2;
    System.out.println(maxProfit(n, a, b, x, y));
}
}
  
// This code is contributed by mits

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Python3

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# Python3 implementation of the approach 
from math import gcd
  
# Function to return the maximum profit 
def maxProfit(n, a, b, x, y) : 
      
    res = x * (n // a); 
    res += y * (n // b); 
  
    # min(x, y) * n / lcm(a, b) 
    res -= min(x, y) * (n // ((a * b) // 
                           gcd(a, b))); 
    return res; 
  
# Driver code 
if __name__ == "__main__"
  
    n = 6 ;a = 6; b = 2; x = 8; y = 2
      
    print(maxProfit(n, a, b, x, y)); 
      
# This code is contributed by Ryuga

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
  
static int __gcd(int a, int b) 
    if (b == 0) 
        return a; 
    return __gcd(b, a % b); 
      
  
// Function to return the maximum profit
static int maxProfit(int n, int a, int b, int x, int y)
{
    int res = x * (n / a);
    res += y * (n / b);
  
    // min(x, y) * n / lcm(a, b)
    res -= Math.Min(x, y) * (n / ((a * b) / __gcd(a, b)));
    return res;
}
  
// Driver code
static void Main()
{
    int n = 6, a = 6, b = 2, x = 8, y = 2;
    Console.WriteLine(maxProfit(n, a, b, x, y));
}
}
  
// This code is contributed by mits

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PHP

Output:

12


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