当可除数为两个的数有关联的利润时，最大化利润

给定五个整数N ， A ， B ， X和Y。任务是找到从[1，N]范围内的数字获得的最大利润。如果正数可被A整除，则利润增加X ；如果正数可被B整除，则利润增加Y。
注意：从正数获利最多只能相加一次。

例子：

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

方法：容易看出，只有数字为lcm(A，B)的倍数时，我们才能将数字除以A和B。显然，该数字应除以可带来更多利润的数字。
因此，答案等于X *(N / A)+ Y *(N / B)– min(X，Y)*(N / lcm(A，B)) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

输出：

12