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 []总是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_GangwarPython3
# 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 KumarC#
// 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 29AjayKumarJavascript
输出：
1 -2 -4 5 1 -1如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。