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，仅当数字是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

Javascript

 
// 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.
 

输出：

12

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。