In Java, maximize the total profit of all people X

We have 5 integer variables Num, P1, P2, profit_P1, profit_P2, and the task is to maximize the profit and choose from all natural numbers in the range [1, Num]. The approach here is that if a positive number is divisible by P1, the profit is increased by profit_P1, similarly, if a number in the range is divisible by P2, the profit is increased by profit_P2. Furthermore, profits from positive integers can only be added at most once.

Let us understand through examples:

Input - int num = 4, P1 = 6, P2 = 2, profit_P1 = 8, profit_P2 = 2;

Output - Maximize the total profit for all )])

No number in the series is divisible by P1 1 and 2 are divisible by P2

1 and 2 are divisible by P2, resulting in a profit of 2 * 2 =

Enter - num = 3, P1 = 1, P2 = 2, profit_P1 = 3, profit_P2 = 4

Output - Maximize the total profit for everyone X 10

Explanation - 1, 2 and 3 are all divisible by A.

2 is the only number in the given range that is divisible by B. 2 is divisible by A and B.

1 and 3 can be divisible by A, and the profit is 2 * 3 = 6

2 can be divisible by B, and the profit 1 * 4 = 4

2 can be divided by A is divisible by B, but to maximize profit, it is divisible by B instead of A.

The method used in the following program is as follows -

We have 6 integer variables, including the positive range (Num), P1 represents the first person, and P2 represents The second person, profit_P1 represents the profit of the first person (i.e. profit_P1 increases if a number in the given range of numbers is divisible by P1), and similarly profit_P2.

• A method (profitMaximisation) is called in the main function, which is a utility method for all calculations.

• As you can see inside the function, only when the number is a multiple of the least common multiple of P1 or P2, it can be divisible by both P1 and P2 at the same time. Furthermore, it should be divided by a number that provides more profit.

• So, the calculation method here is

  profit_P1 * (num / P1) profit_P2 * (num / P2) - min(profit_P1, profit_P2) * (num / lcm(P1 , P2))
.

• Introduced a method CalculateGcd() to calculate the least common multiple of a given number.

• The final output is captured in the main method and displayed to the user.

• Example

  public class testClass{
     static int CalculateGcd(int n1, int n2){
        if (n2 == 0)
           return n1;
        return CalculateGcd(n2, n1 % n2);
     }
     static int profitMaximisation(int n, int a, int b, int x, int y){
        int result = x * (n / a);
        result += y * (n / b);
        result -= Math.min(x, y) * (n / ((a * b) / CalculateGcd(a, b)));
        return result;
     }
     public static void main(String[] args){
        int num = 6, P1 = 6, P2 = 2, profit_P1 = 8, profit_P2 = 2;
        System.out.println("Maximize the total profit of all the persons X "+profitMaximisation(num, P1, P2, profit_P1, profit_P2));
     }
  }

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Output

If we run the above code, the following output will be generated

Maximize the total profit of all the persons X 12

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