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Maximize profit when divisibility by two numbers have associated profits

  • Last Updated : 27 Apr, 2021

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++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

PHP




<?php
// PHP implementation of the approach
function __gcd($a, $b)
{
    if ($b == 0)
        return $a;
    return __gcd($b, $a % $b);
     
}
 
// Function to return the maximum profit
function 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
$n = 6;
$a = 6;
$b = 2;
$x = 8;
$y = 2;
print(maxProfit($n, $a, $b, $x, $y));
 
// This code is contributed by mits
?>

Javascript




<script>
 
// JavaScript implementation of the approach
function __gcd(a, b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);
       
}
 
// Function to return the maximum profit
 
function maxProfit(n, a, b, x, y)
{
    let res = x * Math.floor(n / a);
    res += y * Math.floor(n / b);
 
    // min(x, y) * n / lcm(a, b)
    res -= Math.min(x, y) * (n / ((a * b) / __gcd(a, b)));
    return res;
}
 
// Driver code
    let n = 6, a = 6, b = 2, x = 8, y = 2;
    document.write(maxProfit(n, a, b, x, y));
 
 
// This article is contributed by Surbhi Tyagi.
 
</script>
Output: 
12

 




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