从偶数和对中删除 Max 和从奇数对中删除 Min 后 2^N 个整数的最后剩余值
给定一个非负整数N ，表示从 1 开始的连续整数相互配对的前2 N个正整数，任务是在执行以下操作后找到最后剩余的数：
- 考虑尚未从2 N个整数中删除的两个连续整数对 (x, y)。
- 如果两个整数 x + y 之和为偶数，则删除 x（最小值）。
- 如果两个整数 x + y 之和为奇数，则删除 y（最大值）。
例子：
Input: N = 2
Output: 3
Explanation:
For N = 2 we have 2^N i.e; 4 integers and 4/2 pairs which are as follows-:
(1, 2), (3, 4)
Now after performing the operations:
1 + 2 = 3 which is odd, delete y. So remains 1.
3 + 4 = 7 which is odd, delete y. So remains 3
Now the remaining numbers are 1 and 3 and the pair is (1, 3).
Sum = 1 + 3 = 4 which is even, so delete 1.
So the last remaining element is 3.
Input: N = 3
Output: 7
Explanation:
For N = 2 we have 2^N i.e; 8 integers and 8/2 pairs which are as follows-:
(1, 2), (3, 4), (5, 6) and (7, 8)
Now after performing the operations:
1 + 2 = 3 which is odd, delete y. So remains 1.
3 + 4 = 7 which is odd, delete y. So remains 3.
5 + 6 = 11 which is odd, delete y. So remains 5.
7 + 8 = 15 which is odd, delete y. So remains 7.
Now the remaining numbers are 1, 3, 5 and 7 and the pairs are (1, 3), (5, 7).
1 + 3 = 4 which is even, delete x. So remains 3.
5 + 7 = 12 which is even, delete x. So remains 7.
Now the remaining numbers are 3 and 7 and the pair is (3, 7)
3 + 7 = 10 which is even, so delete 3.
So the last remaining element is 7.
方法：给定的问题可以使用以下数学观察来解决：
- The idea is to consider when the sum is odd or even while adding two numbers.
- If we add one even and one odd number then sum will always be an odd number
- If we add two odd numbers or two even numbers then sum will always be an even number.
基于以上观察，可以得出问题的解决方案如下：
- 第一次，每对都有一个奇数和一个偶数，其中偶数较大。所以总和总是奇数，偶数被删除。
- 在接下来的步骤中，剩余的数字都是奇数，因此对的总和将始终是偶数。因此，从第一个（较小的）开始的替代奇数在每个步骤中都会被删除。
- 从上面的观察可以清楚地看出，所有奇数中的最大值将是最后一个，即2 N -1
下面是上述方法的实现：
C++
// C++ program to implement the approach
#include
using namespace std;
// Function to calculate
// the last remaining number
long long int lastdigit(int N)
{
// 1< Java
// Java program to implement the approach
import java.io.*;
class GFG {
// Function to calculate
// the last remaining number
static long lastdigit(int N)
{
// 1<C#
// C# program to implement the approach
using System;
public class GFG{
// Function to calculate
// the last remaining number
static long lastdigit(int N)
{
// 1<7时间复杂度： O(1)
辅助空间： O(1)