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。任务是为所有有效的 i 和b[0]+b[1]+b[2]+…+b[n–1]= 0生成一个数组 B[] ，使得 B[i] 要么是⌊a[i/2⌋要么是⌈a[i/2⌉。
示例:
输入: A[] = {1，2，-5，3，-1} 输出:0 1-2 1 0 T6】输入: A[] = {3，-5，-7，9，2，-2} 输出: 1 -2 -4 5 1 -1
方法:对于偶数整数，可以安全地假设 B[i] 将是 A[i] / 2 但是对于奇数整数，为了保持和等于零，取正好一半奇数整数的上限和正好另一半奇数整数的下限。既然奇–奇=偶且偶–偶=偶且 0 也是偶，可以说 A[] 将始终包含偶数个奇整数，这样和就可以是 0 。所以对于有效的输入，总会有答案。 以下是上述方法的实施:
C++
// 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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<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>
Output:
1 -2 -4 5 1 -1
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