通过生成一个大小为 N 且总和为 X 的数组可能的最大中位数

给定两个正整数N和X 。任务是通过生成一个大小为N且总和为X的数组来打印可能的最大中位数

例子：

Input: N = 1, X = 7
Output: 7
Explanation: Array can be: [7], median is the 1st element, i.e., 7.

Input: N = 7, X = 18
Output: 4
Explanation: One of the possible arrays can be: [0, 1, 2, 3, 4, 4, 4]. The median = ceil(n/2)th element  = ceil(7/2) = 5th element, i.e., 4.

方法：考虑中值需要最大化，因此贪心方法可以是使中值元素位置之前的所有元素为零，并将总和X等分到其余元素中。
请按照以下步骤解决问题：

• 如果n = 1 ，则打印X 。
• 对于n >= 2。
• 创建一个变量median_pos = ceil((double)(n)/2.0)。
• 递减median_pos ，以表示索引值。
• 创建一个变量median = X/(n-median_pos) 。
• 打印中位数。

以下是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the maximum median possible
int maximizeMedian(int n, int X)
{
    // If only 1 element present
    if (n == 1) {
        return X;
    }
    else {
        // Position of median
        int median_pos = ceil((double)(n) / (2.0));
        median_pos--;
        int median = X / (n - median_pos);
        return median;
    }
    return 0;
}
 
// Driver Code
int main()
{
    int n = 1, X = 7;
    cout << maximizeMedian(n, X);
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
   
    // Function to find the maximum median possible
    static int maximizeMedian(int n, int X)
    {
       
        // If only 1 element present
        if (n == 1) {
            return X;
        }
        else {
            // Position of median
            int median_pos
                = (int)Math.ceil((double)(n) / (2.0));
            median_pos--;
            int median = X / (n - median_pos);
            return median;
        }
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int n = 1, X = 7;
        System.out.println(maximizeMedian(n, X));
    }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach
 
# Function to find the maximum median possible
def maximizeMedian(n, X):
 
    # If only 1 element present
    if (n == 1):
        return X
    else:
        # Position of median
        median_pos = (n) // (2.0)
        median_pos -= 1
        median = X // (n - median_pos)
        return median
    return 0
 
# Driver Code
 
n = 1
X = 7
print(maximizeMedian(n, X))
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
class GFG
{
   
    // Function to find the maximum median possible
    static int maximizeMedian(int n, int X)
    {
       
        // If only 1 element present
        if (n == 1) {
            return X;
        }
        else {
            // Position of median
            int median_pos
                = (int)Math.Ceiling((double)(n) / (2.0));
            median_pos--;
            int median = X / (n - median_pos);
            return median;
        }
    }
 
    // Driver Code
    public static void Main()
    {
        int n = 1, X = 7;
        Console.WriteLine(maximizeMedian(n, X));
    }
}
 
// This code is contributed by ukasp.

Javascript

       // JavaScript code for the above approach
 
       // Function to find the maximum median possible
       function maximizeMedian(n, X)
       {
        
           // If only 1 element present
           if (n == 1) {
               return X;
           }
           else
           {
            
               // Position of median
               let median_pos = Math.ceil((n) / (2.0));
               median_pos--;
               let median = X / (n - median_pos);
               return median;
           }
           return 0;
       }
 
       // Driver Code
 
       let n = 1, X = 7;
       document.write(maximizeMedian(n, X));
 
 // This code is contributed by Potta Lokesh
   

输出

7

时间复杂度： O(1)
辅助空间： O(1)