Genere una array B[] a partir de la array dada A[] que satisfaga las condiciones dadas

Dada una array A[] de N enteros tal que A[0] + A[1] + A[2] + … A[N – 1] = 0 . La tarea es generar una array B[] tal que B[i] sea ⌊A[i] / 2⌋ o ⌈A[i] / 2⌉ para todas las i válidas y B[0] + B[1] + B[2] + … + B[N – 1] = 0 .
 

Ejemplos: 

Entrada: A[] = {1, 2, -5, 3, -1} 
Salida: 0 1 -2 1 0
Entrada: A[] = {3, -5, -7, 9, 2, -2} 
Salida : 1 -2 -4 5 1 -1  

Enfoque: Para enteros pares, es seguro asumir que B[i] será A[i] / 2 pero para enteros impares, para mantener la suma igual a cero, tome el techo de exactamente la mitad de los enteros impares y el piso de exactamente otros medio enteros impares. Dado que Impar – Impar = Par y Par – Par = Par y 0 también es Par, se puede decir que A[] siempre contendrá un número par de enteros impares, por lo que la suma puede ser 0 . Entonces, para una entrada válida, siempre habrá una respuesta.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Utility function to print
// the array elements
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
 
// Function to generate and print
// the required array
void generateArr(int arr[], int n)
{
 
    // To switch the ceil and floor
    // function alternatively
    bool flip = true;
 
    // For every element of the array
    for (int i = 0; i < n; i++) {
 
        // If the number is odd then print the ceil
        // or floor value after division by 2
        if (arr[i] & 1) {
 
            // Use the ceil and floor alternatively
            if (flip ^= true)
                cout << ceil((float)(arr[i]) / 2.0) << " ";
            else
                cout << floor((float)(arr[i]) / 2.0) << " ";
        }
 
        // If arr[i] is even then it will
        // be completely divisible by 2
        else {
            cout << arr[i] / 2 << " ";
        }
    }
}
 
// Driver code
int main()
{
    int arr[] = { 3, -5, -7, 9, 2, -2 };
    int n = sizeof(arr) / sizeof(int);
 
    generateArr(arr, n);
 
    return 0;
}

// Java implementation of the approach
// Utility function to print
// the array elements
import java.util.*;
import java.lang.*;
 
class GFG
{
 
static void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        System.out.print(arr[i] + " ");
}
 
// Function to generate and print
// the required array
static void generateArr(int arr[], int n)
{
 
    // To switch the ceil and floor
    // function alternatively
    boolean flip = true;
 
    // For every element of the array
    for (int i = 0; i < n; i++)
    {
 
        // If the number is odd then print the ceil
        // or floor value after division by 2
        if ((arr[i] & 1) != 0)
        {
 
            // Use the ceil and floor alternatively
            if (flip ^= true)
                System.out.print((int)(Math.ceil(arr[i] /
                                            2.0)) + " ");
            else
                System.out.print((int)(Math.floor(arr[i] /
                                            2.0)) + " ");
        }
 
        // If arr[i] is even then it will
        // be completely divisible by 2
        else
        {
            System.out.print(arr[i] / 2 +" ");
        }
    }
}
 
// Driver code
public static void main(String []args)
{
    int arr[] = { 3, -5, -7, 9, 2, -2 };
    int n = arr.length;
 
    generateArr(arr, n);
}
}
 
// This code is contributed by Surendra_Gangwar

# Python3 implementation of the approach
from math import ceil, floor
 
# Utility function to print
# the array elements
def printArr(arr, n):
    for i in range(n):
        print(arr[i], end = " ")
 
# Function to generate and print
# the required array
def generateArr(arr, n):
 
    # To switch the ceil and floor
    # function alternatively
    flip = True
 
    # For every element of the array
    for i in range(n):
 
        # If the number is odd then print the ceil
        # or floor value after division by 2
        if (arr[i] & 1):
 
            # Use the ceil and floor alternatively
            flip ^= True
            if (flip):
                print(int(ceil((arr[i]) / 2)),
                                   end = " ")
            else:
                print(int(floor((arr[i]) / 2)),
                                    end = " ")
 
        # If arr[i] is even then it will
        # be completely divisible by 2
        else:
            print(int(arr[i] / 2), end = " ")
 
# Driver code
arr = [3, -5, -7, 9, 2, -2]
n = len(arr)
 
generateArr(arr, n)
 
# This code is contributed by Mohit Kumar

// C# implementation of the approach
// Utility function to print
// the array elements
using System;
using System.Collections.Generic;
 
class GFG
{
 
static void printArr(int []arr, int n)
{
    for (int i = 0; i < n; i++)
        Console.Write(arr[i] + " ");
}
 
// Function to generate and print
// the required array
static void generateArr(int []arr, int n)
{
 
    // To switch the ceil and floor
    // function alternatively
    bool flip = true;
 
    // For every element of the array
    for (int i = 0; i < n; i++)
    {
 
        // If the number is odd then print the ceil
        // or floor value after division by 2
        if ((arr[i] & 1) != 0)
        {
 
            // Use the ceil and floor alternatively
            if (flip ^= true)
                Console.Write((int)(Math.Ceiling(arr[i] /
                                            2.0)) + " ");
            else
                Console.Write((int)(Math.Floor(arr[i] /
                                            2.0)) + " ");
        }
 
        // If arr[i] is even then it will
        // be completely divisible by 2
        else
        {
            Console.Write(arr[i] / 2 +" ");
        }
    }
}
 
// Driver code
public static void Main(String []args)
{
    int []arr = { 3, -5, -7, 9, 2, -2 };
    int n = arr.Length;
 
    generateArr(arr, n);
}
}
 
// This code is contributed by 29AjayKumar

<script>
// javascript implementation of the approach
// Utility function to print
// the array elements   
function printArr(arr , n) {
        for (i = 0; i < n; i++)
            document.write(arr[i] + " ");
    }
 
    // Function to generate and print
    // the required array
    function generateArr(arr , n) {
 
        // To switch the ceil and floor
        // function alternatively
        var flip = true;
 
        // For every element of the array
        for (i = 0; i < n; i++) {
 
            // If the number is odd then print the ceil
            // or floor value after division by 2
            if ((arr[i] & 1) != 0) {
 
                // Use the ceil and floor alternatively
                if (flip ^= true)
                    document.write(parseInt( (Math.ceil(arr[i] / 2.0))) + " ");
                else
                    document.write(parseInt( (Math.floor(arr[i] / 2.0))) + " ");
            }
 
            // If arr[i] is even then it will
            // be completely divisible by 2
            else {
                document.write(arr[i] / 2 + " ");
            }
        }
    }
 
    // Driver code
     
        var arr = [ 3, -5, -7, 9, 2, -2 ];
        var n = arr.length;
 
        generateArr(arr, n);
 
// This code is contributed by todaysgaurav
</script>

1 -2 -4 5 1 -1

Complejidad de tiempo: O(n)

Espacio Auxiliar: O(1)