从满足给定条件的给定数组A []生成数组B []

给定一个由N个整数组成的数组A [] ，使得A [0] + A [1] + A [2] +…A [N – 1] = 0 。任务是为所有有效i和B [0] + B [1] +生成一个数组B [] ，使B [i]为⌊A[i] /2⌋或⌈A[i] / 2] B [2] +…+ B [N – 1] = 0 。

例子：

Input: A[] = {1, 2, -5, 3, -1}
Output: 0 1 -2 1 0

Input: A[] = {3, -5, -7, 9, 2, -2}
Output: 1 -2 -4 5 1 -1

方法：对于偶数整数，可以安全地假设B [i]将为A [i] / 2，但是对于奇数整数，要保持总和等于零，取正整数的半个ceil和正整数的下限其他半个奇数整数。由于Odd – Odd =偶数–偶数–偶数=偶数且0也是偶数，因此可以说A []将始终包含偶数个奇数整数，因此总和可以为0 。因此，对于有效的输入，总会有答案。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

// 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

# 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#

// 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

输出：

1 -2 -4 5 1 -1