使数组的所有元素与给定操作相等

给定一个由n个整数和一个整数k组成的数组arr [] 。任务是使arr []的所有元素与给定的操作相等。在一次操作中，可以将x≤k的任何非负数(可以是浮点值)添加到数组的任何元素，并且k将更新为k = k – x 。打印有可能是其他打印无。

例子：

Input: k = 8, arr[] = {1, 2, 3, 4}
Output: Yes
1 + 3.5 = 4.5
2 + 2.5 = 4.5
3 + 1.5 = 4.5
4 + 0.5 = 4.5
3.5 + 2.5 + 1.5 + 0.5 = 8 = k

Input: k = 2, arr[] = {1, 2, 3, 4}
Output: -1

方法：由于任务是使数组的所有元素相等，并且加法的总和必须恰好是k 。只有一个单一的值可以使所有这些元素相等，即(sum(arr)+ k)/ n 。如果数组中的某个元素已经大于该值，则答案不存在，否则打印“是” 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function that returns true if all the elements
// of the array can be made equal
// with the given operation
bool isPossible(int n, int k, int arr[])
{
  
    // To store the sum of the array elements
    // and the maximum element from the array
    int sum = arr[0], maxVal = arr[0];
  
    for (int i = 1; i < n; i++) {
        sum += arr[i];
        maxVal = max(maxVal, arr[i]);
    }
  
    if ((float)maxVal > (float)(sum + k) / n)
        return false;
  
    return true;
}
  
// Driver code
int main()
{
    int k = 8;
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    if (isPossible(n, k, arr))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

Java

//Java implementation of the approach
import java.io.*;
  
class GFG 
{
      
// Function that returns true if all 
// the elements of the array can be 
// made equal with the given operation
static boolean isPossible(int n, int k, int arr[])
{
  
    // To store the sum of the array elements
    // and the maximum element from the array
    int sum = arr[0];
    int maxVal = arr[0];
  
    for (int i = 1; i < n; i++) 
    {
        sum += arr[i];
        maxVal = Math.max(maxVal, arr[i]);
    }
  
    if ((float)maxVal > (float)(sum + k) / n)
        return false;
  
    return true;
}
  
    // Driver code
    public static void main (String[] args) 
    {
      
        int k = 8;
        int arr[] = { 1, 2, 3, 4 };
        int n = arr.length;
  
        if (isPossible(n, k, arr))
            System.out.println ("Yes");
        else
            System.out.println( "No");
    }
}
  
// This code is contributed by @Tushil.

Python3

# Python 3 implementation of the approach
  
# Function that returns true if all 
# the elements of the array can be 
# made equal with the given operation
def isPossible(n, k, arr):
      
    # To store the sum of the array elements
    # and the maximum element from the array
    sum = arr[0]
    maxVal = arr[0];
  
    for i in range(1, n):
        sum += arr[i]
        maxVal = max(maxVal, arr[i])
  
  
    if (int(maxVal)> int((sum + k) / n)):
        return False
  
    return True
  
# Driver code
if __name__ == '__main__':
    k = 8
    arr = [1, 2, 3, 4]
    n = len(arr)
  
    if (isPossible(n, k, arr)):
        print("Yes")
    else:
        print("No")
  
  
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach 
using System;
  
class GFG 
{ 
      
    // Function that returns true if all 
    // the elements of the array can be 
    // made equal with the given operation 
    static bool isPossible(int n, 
                        int k, int []arr) 
    { 
      
        // To store the sum of the array elements 
        // and the maximum element from the array 
        int sum = arr[0]; 
        int maxVal = arr[0]; 
          
        for (int i = 1; i < n; i++) 
        { 
            sum += arr[i]; 
            maxVal = Math.Max(maxVal, arr[i]); 
        } 
      
        if ((float)maxVal > (float)(sum + k) / n) 
            return false; 
      
        return true; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
          
        int k = 8; 
        int []arr = { 1, 2, 3, 4 }; 
        int n = arr.Length; 
      
        if (isPossible(n, k, arr)) 
            Console.WriteLine("Yes"); 
        else
            Console.WriteLine( "No"); 
    } 
} 
  
// This code is contributed by Ryuga

PHP

 (float)($sum + $k) / $n)
        return false;
  
    return true;
}
  
    // Driver code
    $k = 8;
    $arr = array( 1, 2, 3, 4 );
    $n = sizeof($arr) / sizeof($arr[0]);
  
    if (isPossible($n, $k, $arr))
        echo "Yes";
    else
        echo "No";
  
# This code is contributed by akt_miit.
?>

输出：

Yes

时间复杂度： O(n)