生成一个 N 长度数组，该数组具有相等的计数和两个奇偶校验的元素之和
给定一个整数N(3≤N≤105)，任务是生成一个由 N 个不同的正元素组成的数组，这样双方元素的个数和和和即偶数和奇数是相同的。如果无法构建这样的阵列，请打印 -1 。
示例:
输入: N = 8 输出: 2、4、6、8、1、3、5、11
输入: 6 输出: -1 说明:为了奇偶数组元素的个数相等，数组中必须存在 3 个奇偶元素。但是，3 个 od 元素之和总是奇数。因此，不可能生成这样的数组。
方法:按照以下步骤解决问题:
- 如果 N 为奇数或 (N / 2) 为奇数，则打印 -1 。
- 否则:
- 从 2 开始打印 N/2 偶数值，并将计数存储在某个变量中，比如sumever。
- 打印N/2–1从 1 开始的奇数，并将总和存储在另一个变量中，比如 SumOdd 。
- 最后打印sum 偶数–sum 奇数，也是奇数。
下面是上述方法的实现:
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Fun dtion to print the
// required sequence
void Print(int N)
{
if ((N / 2) % 2 == 1
|| (N % 2 == 1)) {
cout << -1 << endl;
return;
}
// Stores count of even
// and odd elements
int CurEven = 2, CurOdd = 1;
// Stores sum of even
// and odd elements
int SumOdd = 0, SumEven = 0;
// Print N / 2 even elements
for (int i = 0; i < (N / 2); i++) {
cout << CurEven << " ";
SumEven += CurEven;
CurEven += 2;
}
// Print N / 2 - 1 odd elements
for (int i = 0; i < N / 2 - 1; i++) {
cout << CurOdd << " ";
SumOdd += CurOdd;
CurOdd += 2;
}
CurOdd = SumEven - SumOdd;
// Print final odd element
cout << CurOdd << '\n';
}
// Driver Code
int main()
{
int N = 12;
Print(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program to implement
// the above approach
import java.io.*;
class GFG
{
// Fun dtion to print the
// required sequence
static void Print(int N)
{
if ((N / 2) % 2 == 1 || (N % 2 == 1))
{
System.out.print(-1);
return;
}
// Stores count of even
// and odd elements
int CurEven = 2, CurOdd = 1;
// Stores sum of even
// and odd elements
int SumOdd = 0, SumEven = 0;
// Print N / 2 even elements
for (int i = 0; i < (N / 2); i++)
{
System.out.print(CurEven + " ");
SumEven += CurEven;
CurEven += 2;
}
// Print N / 2 - 1 odd elements
for (int i = 0; i < N / 2 - 1; i++)
{
System.out.print(CurOdd + " ");
SumOdd += CurOdd;
CurOdd += 2;
}
CurOdd = SumEven - SumOdd;
// Print final odd element
System.out.println(CurOdd);
}
// Driver Code
public static void main (String[] args)
{
int N = 12;
Print(N);
}
}
// This code is contributed by Dharanendra L V.
Python 3
# Python Program to implement
# the above approach
# Fun dtion to print the
# required sequence
def Print(N):
if ((N / 2) % 2 or (N % 2)):
print(-1)
return
# Stores count of even
# and odd elements
CurEven = 2
CurOdd = 1
# Stores sum of even
# and odd elements
SumOdd = 0
SumEven = 0
# Print N / 2 even elements
for i in range(N // 2):
print(CurEven,end=" ")
SumEven += CurEven
CurEven += 2
# Print N / 2 - 1 odd elements
for i in range( (N//2) - 1):
print(CurOdd, end=" ")
SumOdd += CurOdd
CurOdd += 2
CurOdd = SumEven - SumOdd
# Print final odd element
print(CurOdd)
# Driver Code
N = 12
Print(N)
# This code is contributed by rohitsingh07052
C
// C# Program to implement
// the above approach
using System;
class GFG
{
// Fun dtion to print the
// required sequence
static void Print(int N)
{
if ((N / 2) % 2 == 1 || (N % 2 == 1))
{
Console.WriteLine(-1);
return;
}
// Stores count of even
// and odd elements
int CurEven = 2, CurOdd = 1;
// Stores sum of even
// and odd elements
int SumOdd = 0, SumEven = 0;
// Print N / 2 even elements
for (int i = 0; i < (N / 2); i++)
{
Console.Write(CurEven + " ");
SumEven += CurEven;
CurEven += 2;
}
// Print N / 2 - 1 odd elements
for (int i = 0; i < N / 2 - 1; i++)
{
Console.Write(CurOdd + " ");
SumOdd += CurOdd;
CurOdd += 2;
}
CurOdd = SumEven - SumOdd;
// Print final odd element
Console.WriteLine(CurOdd);
}
// Driver Code
public static void Main()
{
int N = 12;
Print(N);
}
}
// This code is contributed by chitranayal.
java 描述语言
<script>
// Javascript Program to implement
// the above approach
// Fun dtion to print the
// required sequence
function Print(N)
{
if ((N / 2) % 2 == 1 || (N % 2 == 1))
{
document.write(-1);
return;
}
// Stores count of even
// and odd elements
var CurEven = 2, CurOdd = 1;
// Stores sum of even
// and odd elements
var SumOdd = 0, SumEven = 0;
// Print N / 2 even elements
for(var i = 0; i < (N / 2); i++)
{
document.write(CurEven + " ");
SumEven += CurEven;
CurEven += 2;
}
// Print N / 2 - 1 odd elements
for(var i = 0; i < N / 2 - 1; i++)
{
document.write(CurOdd + " ");
SumOdd += CurOdd;
CurOdd += 2;
}
CurOdd = SumEven - SumOdd;
// Print final odd element
document.write(CurOdd);
}
// Driver Code
var N = 12;
Print(N);
// This code is contributed by Ankita saini
</script>
Output:
2 4 6 8 10 12 1 3 5 7 9 17
时间复杂度:O(N) T3】辅助空间 O(1)
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