将奇数定位的元素替换为其立方体，偶数定位的元素替换其正方形

给定n个元素的数组arr [] ，任务是用其立方体替换所有奇数位置的元素，并用其正方形替换偶数位置的元素，即结果数组必须为{arr [0] ³ ，arr [1] ² ，arr [2] ³ ，arr [3] ² ，…} 。
例子：

Input: arr[]= {2, 3, 4, 5} 
Output: 8 9 64 25 
Updated array will be {2³, 3², 4³, 5²} -> {8, 9, 64, 25}
Input: arr[] = {3, 4, 5, 2} 
Output: 27 16 125 4  

方法：对于数组arr [i]的任何元素，仅当(i + 1)为奇数时才是奇数位置，因为索引从0开始。现在，遍历数组，将所有奇数定位的元素替换为它们的立方体，甚至将定位的元素替换为其正方形。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define ll long long int
 
// Utility function to print
// the contents of an array
void printArr(ll arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
 
// Function to update the array
void updateArr(ll arr[], int n)
{
    for (int i = 0; i < n; i++) {
 
        // In case of even positioned element
        if ((i + 1) % 2 == 0)
            arr[i] = (ll)pow(arr[i], 2);
 
        // Odd positioned element
        else
            arr[i] = (ll)pow(arr[i], 3);
    }
 
    // Print the updated array
    printArr(arr, n);
}
 
// Driver code
int main()
{
    ll arr[] = { 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    updateArr(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.lang.Math;
 
class GFG
{
     
// Utility function to print
// the contents of an array
static void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        System.out.print(arr[i] + " ");
}
 
// Function to update the array
static void updateArr(int arr[], int n)
{
    for (int i = 0; i < n; i++)
    {
 
        // In case of even positioned element
        if ((i + 1) % 2 == 0)
            arr[i] = (int)Math.pow(arr[i], 2);
 
        // Odd positioned element
        else
            arr[i] = (int)Math.pow(arr[i], 3);
    }
 
    // Print the updated array
    printArr(arr, n);
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 3, 4, 5, 6 };
    int n = arr.length;
 
    updateArr(arr, n);
}
}
 
// This code is contributed
// by Code_Mech.

Python3

# Python3 implementation of the approach
 
# Utility function to print
# the contents of an array
def printArr(arr,n):
    for i in range(n):
        print(arr[i], end = " ")
 
# Function to update the array
def updateArr(arr, n):
    for i in range(n):
 
        # In case of even positioned element
        if ((i + 1) % 2 == 0):
            arr[i] = pow(arr[i], 2)
 
        # Odd positioned element
        else:
            arr[i] = pow(arr[i], 3)
     
    # Print the updated array
    printArr(arr, n)
 
# Driver code
arr = [ 2, 3, 4, 5, 6 ]
n = len(arr)
 
updateArr(arr, n)
 
# This code is contributed
# by mohit kumar

C#

// C# implementation of the approach
using System;
     
class GFG
{
     
// Utility function to print
// the contents of an array
static void printArr(int []arr, int n)
{
    for (int i = 0; i < n; i++)
        Console.Write(arr[i] + " ");
}
 
// Function to update the array
static void updateArr(int []arr, int n)
{
    for (int i = 0; i < n; i++)
    {
 
        // In case of even positioned element
        if ((i + 1) % 2 == 0)
            arr[i] = (int)Math.Pow(arr[i], 2);
 
        // Odd positioned element
        else
            arr[i] = (int)Math.Pow(arr[i], 3);
    }
 
    // Print the updated array
    printArr(arr, n);
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 2, 3, 4, 5, 6 };
    int n = arr.Length;
 
    updateArr(arr, n);
}
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

// javascript implementation of the approach
 
    // Utility function to print
    // the contents of an array
    function printArr(arr , n) {
        for (i = 0; i < n; i++)
            document.write(arr[i] + " ");
    }
 
    // Function to update the array
    function updateArr(arr , n) {
        for (i = 0; i < n; i++) {
 
            // In case of even positioned element
            if ((i + 1) % 2 == 0)
                arr[i] = parseInt( Math.pow(arr[i], 2));
 
            // Odd positioned element
            else
                arr[i] = parseInt( Math.pow(arr[i], 3));
        }
 
        // Prvar the updated array
        printArr(arr, n);
    }
 
    // Driver code
     
        var arr = [ 2, 3, 4, 5, 6 ];
        var n = arr.length;
 
        updateArr(arr, n);
 
// This code contributed by gauravrajput1

输出：

8 9 64 25 216

时间复杂度： O(n)

辅助空间： O(1)

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