可以用 N 卢比购买的最大升水量

给定的$N$卢比。一升塑料瓶水的成本$A$卢比和一升玻璃瓶水的费用$B$卢比。但是购买后的空玻璃瓶可以换$C$卢比。找出可以购买的最大升水量$N$卢比。
例子：

Input: N = 10 , A = 11 , B = 9 , C = 8 
Output: 2 
One glass bottle can be bought and then can be returned to buy one more glass bottle
Input: N = 15 , A = 6 , B = 4 , C = 3 
Output: 12 
 

方法：如果我们至少有$b$钱那么一个玻璃瓶的成本是b – c 。这意味着如果a ≤ (b – c)那么我们不需要买玻璃瓶，只买塑料瓶，答案是floor(n / a) 。否则，我们需要尽可能购买玻璃瓶。
所以，如果我们至少有$b$钱，那么我们将购买floor((n – c) / (b – c))玻璃瓶，然后将剩余的钱花在塑料瓶上。
下面是上述方法的实现：

C++

// CPP implementation of the above approach
#include
using namespace std;
 
 
void maxLitres(int budget,int plastic,int glass,int refund)
{
 
    // if buying glass bottles is profitable
    if (glass - refund < plastic)
       {
           // Glass bottles that can be bought
        int ans = max((budget - refund) / (glass - refund), 0);
 
        // Change budget according the bought bottles
        budget -= ans * (glass - refund);
 
        // Plastic bottles that can be bought
        ans += budget / plastic;
        cout<

---

Java

// Java implementation of the above approach
class GFG
{
 
    static void maxLitres(int budget, int plastic,
                            int glass, int refund)
    {
 
        // if buying glass bottles is profitable
        if (glass - refund < plastic)
        {
             
            // Glass bottles that can be bought
            int ans = Math.max((budget - refund) / (glass - refund), 0);
 
            // Change budget according the bought bottles
            budget -= ans * (glass - refund);
 
            // Plastic bottles that can be bought
            ans += budget / plastic;
            System.out.println(ans);
        }
         
        // if only plastic bottles need to be bought
        else
        {
            System.out.println((budget / plastic));
        }
 
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int budget = 10, plastic = 11, glass = 9, refund = 8;
        maxLitres(budget, plastic, glass, refund);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the above approach
 
def maxLitres(budget, plastic, glass, refund):
 
    # if buying glass bottles is profitable
    if glass - refund < plastic:
 
        # Glass bottles that can be bought
        ans = max((budget - refund) // (glass - refund), 0)
 
        # Change budget according the bought bottles
        budget -= ans * (glass - refund)
 
        # Plastic bottles that can be bought
        ans += budget // plastic
        print(ans)
 
    # if only plastic bottles need to be bought
    else:
        print(budget // plastic)
 
# Driver Code
budget, plastic, glass, refund = 10, 11, 9, 8
maxLitres(budget, plastic, glass, refund)

C#

// C# implementation of the above approach
using System;   
 
class GFG
{
 
    static void maxLitres(int budget, int plastic,
                            int glass, int refund)
    {
 
        // if buying glass bottles is profitable
        if (glass - refund < plastic)
        {
             
            // Glass bottles that can be bought
            int ans = Math.Max((budget - refund) / (glass - refund), 0);
 
            // Change budget according the bought bottles
            budget -= ans * (glass - refund);
 
            // Plastic bottles that can be bought
            ans += budget / plastic;
            Console.WriteLine(ans);
        }
         
        // if only plastic bottles need to be bought
        else
        {
            Console.WriteLine((budget / plastic));
        }
 
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int budget = 10, plastic = 11, glass = 9, refund = 8;
        maxLitres(budget, plastic, glass, refund);
    }
}
 
// This code contributed by Rajput-Ji

Javascript

 
// Javascript implementation of the above approach
 
    function maxLitres(budget, plastic,
                            glass, refund)
    {
   
        // if buying glass bottles is profitable
        if (glass - refund < plastic)
        {
               
            // Glass bottles that can be bought
            let ans = Math.max((budget - refund) / (glass - refund), 0);
   
            // Change budget according the bought bottles
            budget -= ans * (glass - refund);
   
            // Plastic bottles that can be bought
            ans += Math.floor(budget / plastic);
            document.write(ans);
        }
           
        // if only plastic bottles need to be bought
        else
        {
            document.write(Math.floor(budget / plastic));
        }
   
    }
     
    // Driver code
 
    let budget = 10, plastic = 11, glass = 9, refund = 8;
    maxLitres(budget, plastic, glass, refund);
       

输出：

2

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