分别找到奇数和偶数数组元素的最接近的奇数和偶数完美平方

给定大小为N的数组arr [] ，每个数组元素的任务是打印具有相同奇偶校验的最接近的完美正方形。

例子：

Input: arr[ ] = {6, 3, 2, 15}
Output: 4 1 4 9
Explanation:
The nearest even perfect square of arr[0] (= 6) is 4.
The nearest odd perfect square of arr[1] (= 3) is 1.
The nearest even perfect square of arr[2] (= 2) is 4
The nearest odd perfect square of arr[3] (= 15) is 9.

Input: arr[ ] = {31, 18, 64}
Output: 25 16 64

方法：请按照以下步骤解决问题：

• 遍历数组并执行以下操作：
  • 找到当前数组元素的平方根，并将其存储在变量中，例如sr 。
  • 如果sr与arr [i]具有相同的奇偶性，则sr * sr是最接近的理想平方。
  • 否则， (sr +1) ²是最接近的理想平方。
  • 打印在上述步骤中获得的最接近的理想平方。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to find the nearest even and odd
// perfect squares for even and odd array elements
void nearestPerfectSquare(int arr[], int N)
{
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Calculate square root of
        // current array element
        int sr = sqrt(arr[i]);
 
        // If both are of same parity
        if ((sr & 1) == (arr[i] & 1))
            cout << sr * sr << " ";
 
      // Otherwise
        else {
            sr++;
            cout << sr * sr << " ";
        }
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 6, 3, 2, 15 };
    int N = sizeof(arr) / sizeof(arr[0]);
    nearestPerfectSquare(arr, N);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG
{
 
// Function to find the nearest even and odd
// perfect squares for even and odd array elements
static void nearestPerfectSquare(int arr[], int N)
{
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Calculate square root of
        // current array element
        int sr = (int)Math.sqrt(arr[i]);
 
        // If both are of same parity
        if ((sr & 1) == (arr[i] & 1))
            System.out.print((sr * sr) + " ");
 
      // Otherwise
        else {
            sr++;
            System.out.print((sr * sr) + " ");
        }
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 6, 3, 2, 15 };
    int N = arr.length;
    nearestPerfectSquare(arr, N);
}
}
 
// This code is contributed by souravghosh0416.

Python3

# Python3 program for the above approach
import math
 
# Function to find the nearest even and odd
# perfect squares for even and odd array elements
def nearestPerfectSquare(arr, N) :
 
    # Traverse the array
    for i in range(N):
 
        # Calculate square root of
        # current array element
        sr = int(math.sqrt(arr[i]))
 
        # If both are of same parity
        if ((sr & 1) == (arr[i] & 1)) :
            print(sr * sr, end = " ")
 
      # Otherwise
        else :
            sr += 1
            print(sr * sr, end = " ")
         
# Driver Code
arr = [ 6, 3, 2, 15 ]
N = len(arr)
nearestPerfectSquare(arr, N)
 
# This code is contributed by sanjoy_62.

C#

// C# program for the above approach
using System;
 
class GFG{
 
  // Function to find the nearest even and odd
  // perfect squares for even and odd array elements
  static void nearestPerfectSquare(int[] arr, int N)
  {
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
      // Calculate square root of
      // current array element
      int sr = (int)Math.Sqrt(arr[i]);
 
      // If both are of same parity
      if ((sr & 1) == (arr[i] & 1))
        Console.Write((sr * sr) + " ");
 
      // Otherwise
      else {
        sr++;
        Console.Write((sr * sr) + " ");
      }
    }
  }
   
  // Driver Code
  static public void Main()
  {
    int[] arr = { 6, 3, 2, 15 };
    int N = arr.Length;
    nearestPerfectSquare(arr, N);
  }
}
 
// This code is contributed by splevel62.

输出：

4 1 4 9

时间复杂度： O(N * M)
辅助空间： O(1)