从 N 个装满的瓶子中最大化空水瓶的数量

给定两个整数N和E，其中N表示装满的水瓶的数量， E表示可以换成装满的水瓶的空瓶的数量。任务是找到可以清空的最大水瓶数量。

例子：

Input: N = 9, E = 3
Output: 13
Explanation:
Initially, there are 9 fully filled water bottles. 
All of them are emptied to obtain 9 empty bottles. Therefore, count = 9
Then exchange 3 bottles at a time for 1 fully filled bottle. 
Therefore, 3 fully filled bottles are obtained. 
Then those 3 bottles are emptied to get 3 empty bottles. Therefore, count = 9 + 3 = 12
Then those 3 bottles are exchanged for 1 full bottle which is emptied. 
Therefore, count = 9 + 3 + 1 = 13.

Input: N = 7, E = 5
Output: 8
Explanation:
Empty the 7 fully filled water bottles. Therefore, count = 7
Then exchange 5 bottles to obtain 1 fully filled bottle. Then empty that bottle. 
Therefore, count = 7 + 1 = 8.

方法：必须按照以下步骤来解决问题：

• 将N的值存储在一个临时变量中，比如a。
• 初始化一个变量，比如s，用来存储空瓶子的数量，另一个变量，比如b，用来存储更换后剩余瓶子的数量。
• 现在，迭代直到a不等于0并将s增加a ，因为它将是已清空的最小瓶子数。
• 更新以下值：
  • a到(a + b) / e
  • b到N – (a * e)
  • N到(a+b)
• 返回s作为必需的答案。

下面是上述方法的实现：

C++

// C++ program for the
// above approach
#include 
using namespace std;
 
// Function to find the maximum
// bottles that can be emptied
int maxBottles(int n, int e)
{
    int s = 0, b = 0;
    int a = n;
 
    // Iterate until a
  // is non-zero
    while (a != 0) {
       
        // Add the number of
      // bottles that are emptied
        s = s + a;
 
        // Update a after
        // exchanging empty bottles
        a = (a + b) / e;
 
        // Stores the number of bottles
        // left after the exchange
        b = n - (a * e);
        n = a + b;
    }
 
    // Return the answer
    return s;
}
 
// Driver Code
int main()
{
    int n = 9, e = 3;
 
    // Function call
    int s = maxBottles(n, e);
    cout << s << endl;
}

Java

// Java program for the
// above approach
import java.util.*;
 
class GFG{
 
// Function to find the maximum
// bottles that can be emptied
static int maxBottles(int n, int e)
{
    int s = 0, b = 0;
    int a = n;
 
    // Iterate until a
    // is non-zero
    while (a != 0)
    {
     
        // Add the number of
        // bottles that are emptied
        s = s + a;
 
        // Update a after
        // exchanging empty bottles
        a = (a + b) / e;
 
        // Stores the number of bottles
        // left after the exchange
        b = n - (a * e);
        n = a + b;
    }
 
    // Return the answer
    return s;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 9, e = 3;
 
    // Function call
    int s = maxBottles(n, e);
     
    System.out.print(s + "\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the
# above approach
 
# Function to find the maximum
# bottles that can be emptied
def maxBottles(n, e):
     
    s = 0
    b = 0
    a = n
 
    # Iterate until a
    # is non-zero
    while (a != 0):
 
        # Add the number of
        # bottles that are emptied
        s = s + a
 
        # Update a after
        # exchanging empty bottles
        a = (a + b) // e
 
        # Stores the number of bottles
        # left after the exchange
        b = n - (a * e)
        n = a + b
 
    # Return the answer
    return s
 
# Driver Code
if __name__ == '__main__':
     
    n = 9
    e = 3
 
    # Function call
    s = maxBottles(n, e)
    print(s)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the maximum
// bottles that can be emptied
static int maxBottles(int n, int e)
{
    int s = 0, b = 0;
    int a = n;
 
    // Iterate until a
    // is non-zero
    while (a != 0)
    {
     
        // Add the number of
        // bottles that are emptied
        s = s + a;
 
        // Update a after
        // exchanging empty bottles
        a = (a + b) / e;
 
        // Stores the number of bottles
        // left after the exchange
        b = n - (a * e);
        n = a + b;
    }
 
    // Return the answer
    return s;
}
 
// Driver Code
public static void Main()
{
    int n = 9, e = 3;
 
    // Function call
    int s = maxBottles(n, e);
     
    Console.Write(s + "\n");
}
}
 
// This code is contributed by code_hunt

Javascript

 
// javascript program for the
// above approach
 
// Function to find the maximum
// bottles that can be emptied
function maxBottles(n, e)
{
    var s = 0, b = 0;
    var a = n;
 
    // Iterate until a
    // is non-zero
    while (a != 0)
    {
     
        // Add the number of
        // bottles that are emptied
        s = s + a;
 
        // Update a after
        // exchanging empty bottles
        a = (a + b) / e;
 
        // Stores the number of bottles
        // left after the exchange
        b = n - (a * e);
        n = a + b;
    }
 
    // Return the answer
    return s;
}
 
// Driver Code
 
var n = 9, e = 3;
 
// Function call
var s = maxBottles(n, e);
 
document.write(parseInt(s) + "\n");
 
// This code contributed by Princi Singh
 

输出：

13

时间复杂度： O(N)
辅助空间： O(1)