分别找出最近的奇、偶数组元素的奇、偶完美平方
原文:https://www . geeksforgeeks . org/分别查找奇数和偶数数组元素的最近奇数和偶数完美平方/
给定一个大小为 N 的数组 arr[ ] ，每个数组元素的任务是打印具有相同奇偶性的最近的完美正方形。
示例:
输入 : arr[ ] = {6，3，2，15} 输出 : 4 1 4 9 解释: arr0最近的偶完美平方是 4。 arr1最近的奇完美平方是 1。 arr2的最近偶完美平方是 4 arr3的最近奇完美平方是 9。
输入 : arr[ ] = {31，18，64} 输出 : 25 16 64
方法:按照以下步骤解决问题:
- 遍历数组并执行以下操作:
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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.
Python 3
# 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.
java 描述语言
<script>
// Javascript program to implement
// the above approach
// Function to find the nearest even and odd
// perfect squares for even and odd array elements
function nearestPerfectSquare(arr, N)
{
// Traverse the array
for(let i = 0; i < N; i++)
{
// Calculate square root of
// current array element
let sr = Math.floor(Math.sqrt(arr[i]));
// If both are of same parity
if ((sr & 1) == (arr[i] & 1))
document.write((sr * sr) + " ");
// Otherwise
else
{
sr++;
document.write((sr * sr) + " ");
}
}
}
// Driver code
// Given array
let arr = [ 6, 3, 2, 15 ];
let N = arr.length;
nearestPerfectSquare(arr, N);
// This code is contributed by target_2
</script>
Output:
4 1 4 9
时间复杂度: O(N * M) 辅助空间: O(1)
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