找到系列 2、2、4、4、6、8、8 中缺失的第 N 项。

给定一个数N，找出序列2、2、4、4、6、8、8……中的第 N 项

例子：

Input: N = 1
Output: 2

Input: N = 2
Output: 2

Input: N = 5
Output: 6

方法：仔细观察给定的系列，可以看出它由两个不同的系列组成，在不同的位置梳理在一起。

Original Series: 2, 2, 4, 4, 6, 8, 8, …
—————————————————————–
Considering only even places: 
=> 2, 2, 4, 4, 6, 8, 8, … 
=> 2, _, 4, _, 6, _, 8, … (considering 0-based indexing) 
=> 2, 4, 6, 8, … (an AP series)
As it can be seen, the numbers at even places forms an Arithmetic Progression in itself.
—————————————————————–
Similarly, Considering only odd places: 
=> 2, 2, 4, 4, 6, 8, 8, … 
=> _, 2, _, 4, _, 8, … (considering 0-based indexing) 
=> 2, 4, 8, … (a GP series)
As it can be seen, the numbers at odd places forms a Geometric Progression in itself.
 

因此，可以说给定序列是 2 个不同序列的混合，其中 AP 的所有项都位于偶数位置，而 GP 的所有项都位于奇数位置。
因此，在上述观察的基础上，按照以下步骤找到给定序列的第 N 个缺失项：

1. 如果 n 是奇数：
   1. 所需的项是 AP [2, 4, 6, 8 …] 的 ( ⁿ +1)/2 项。
   2. AP的^第n项可以计算为a + (n – 1)d。
2. 如果 n 是偶数：
   1. 所需项是 GP [2, 4, 8 ...] 的^{第 n} /2 项。
   2. GP的^第n项可以计算为a*r ^n-1

下面是上述方法的实现：

C++

// C++ code to find the n-th term
// of series 2, 2, 4, 4, 6, 8, 8...
#include 
#include 
using namespace std;
 
// Driver code
int main()
{
    int n = 15;
    int term;
 
    // If n is even
    if (n % 2 == 0) {
        term = n / 2;
 
        // a(first term) = 2
        // r(common ratio) = 2
        cout << 2 * pow(2, term - 1);
    }
 
    // If n is odd
    else {
        term = (n + 1) / 2;
 
        // a(first term) = 2
        // d(common ratio) = 2
        cout << 2 + (term - 1) * 2;
    }
}

Java

// Java code for the above approach
import java.io.*;
 
class GFG {
  public static void main(String[] args)
  {
    int n = 15;
    int term;
 
    // If n is even
    if (n % 2 == 0) {
      term = n / 2;
 
      // a(first term) = 2
      // r(common ratio) = 2
      System.out.println(2 * Math.pow(2, term - 1));
    }
 
    // If n is odd
    else {
      term = (n + 1) / 2;
 
      // a(first term) = 2
      // d(common ratio) = 2
      System.out.println(2 + (term - 1) * 2);
    }
  }
}
 
// This code is contributed by Potta Lokesh

Python

# Python code to find the n-th term
# of series 2, 2, 4, 4, 6, 8, 8...
 
# Driver code
n = 15;
term = 0;
 
# If n is even
if (n % 2 == 0):
    term = n / 2;
 
    # a(first term) = 2
    # r(common ratio) = 2
    print(2 * int(pow(2, term - 1)));
 
# If n is odd
else:
    term = (n + 1) / 2;
 
    # a(first term) = 2
    # d(common ratio) = 2
    print(2 + int((term - 1) * 2));
     
# This code is contributed by Samim Hossain Mondal.

C#

// C# code for the above approach
using System;
 
class GFG {
    public static void Main(string[] args)
    {
        int n = 15;
        int term;
 
        // If n is even
        if (n % 2 == 0) {
            term = n / 2;
 
            // a(first term) = 2
            // r(common ratio) = 2
            Console.WriteLine(2 * Math.Pow(2, term - 1));
        }
 
        // If n is odd
        else {
            term = (n + 1) / 2;
 
            // a(first term) = 2
            // d(common ratio) = 2
            Console.WriteLine(2 + (term - 1) * 2);
        }
    }
}
 
// This code is contributed by ukasp.

Javascript

// Javascript code to find the n-th term
// of series 2, 2, 4, 4, 6, 8, 8...
 
// Driver code
let n = 15;
let term;
 
// If n is even
if (n % 2 == 0) {
    term = n / 2;
 
    // a(first term) = 2
    // r(common ratio) = 2
    document.write(2 * Math.pow(2, term - 1));
}
 
// If n is odd
else {
    term = (n + 1) / 2;
 
    // a(first term) = 2
    // d(common ratio) = 2
    document.write(2 + (term - 1) * 2);
}
 
// This code is contributed by gfgking.

输出：

16

时间复杂度： O(1)
辅助空间： O(1)