将2的N次幂分成两个子集，以使它们的和差最小

给定偶数N ，任务是将2的所有N次幂均分成两组，以使它们的总和之差最小。

例子：

Input: n = 4
Output: 6
Explanation:
Here n = 4 which means we have 2¹, 2², 2³, 2⁴. The most optimal way to divide it into two groups with equal element is 2⁴ + 2¹ in one group and 2² + 2³ in another group giving a minimum possible difference of 6.

Input: n = 8
Output: 30
Explanation:
Here n = 8 which means we have 2¹, 2², 2³, 2⁴, 2⁵, 2⁶, 2⁷, 2⁸. The most optimal way to divide it into two groups with equal element is 2⁸ + 2¹ + 2² + 2³ in one group and 2⁴ + 2⁵ + 2⁶ + 2⁷ in another group giving a minimum possible difference of 30.

方法：要解决上述问题，我们必须遵循以下步骤：

• 在第一组中，添加最大元素2 ^N。
• 将一组中的第一个数字相加后，再向该组中添加N / 2 – 1个元素，其中元素必须从2的最小幂开始或从序列的开头开始。

  例如：如果N = 4，则将2 ⁴添加到第一组中，并添加N / 2 – 1 ，即，该组中再增加1个元素，这是最小的，即2 ¹ 。

• 序列的其余元素构成第二组的元素。
• 计算两个组的总和，然后找到两个组之间的绝对差，最终将是最小的差。

下面是上述方法的实现：

C++

// C++ program to find the minimum
// difference possible by splitting
// all powers of 2 up to N into two 
// sets of equal size 
#include
using namespace std;
  
void MinDiff(int n)
{
  
    // Store the largest
    int val = pow(2, n);
      
    // Form two separate groups
    int sep = n / 2;
      
    // Initialize the sum 
    // for two groups as 0
    int grp1 = 0;
    int grp2 = 0;
      
    // Insert 2 ^ n in the
    // first group
    grp1 = grp1 + val;
      
    // Calculate the sum of
    // least n / 2 -1 elements
    // added to the first set 
    for(int i = 1; i < sep; i++)
       grp1 = grp1 + pow(2, i);
          
    // Sum of remaining
    // n / 2 - 1 elements
    for(int i = sep; i < n; i++)
       grp2 = grp2 + pow(2, i);
          
    // Min Difference between
    // the two groups
    cout << (abs(grp1 - grp2));
} 
  
// Driver code
int main()
{
    int n = 4;
    MinDiff(n); 
}
  
// This code is contributed by Bhupendra_Singh

Java

// Java program to find the minimum 
// difference possible by splitting 
// all powers of 2 up to N into two  
// sets of equal size  
import java.lang.Math; 
  
class GFG{ 
  
public static void MinDiff(int n) 
{ 
  
    // Store the largest 
    int val = (int)Math.pow(2, n); 
      
    // Form two separate groups 
    int sep = n / 2; 
      
    // Initialize the sum 
    // for two groups as 0 
    int grp1 = 0; 
    int grp2 = 0; 
      
    // Insert 2 ^ n in the 
    // first group 
    grp1 = grp1 + val; 
      
    // Calculate the sum of 
    // least n / 2 -1 elements 
    // added to the first set 
    for(int i = 1; i < sep; i++) 
       grp1 = grp1 + (int)Math.pow(2, i); 
          
    // Sum of remaining 
    // n / 2 - 1 elements 
    for(int i = sep; i < n; i++) 
       grp2 = grp2 + (int)Math.pow(2, i); 
          
    // Min difference between 
    // the two groups 
    System.out.println(Math.abs(grp1 - grp2)); 
} 
  
// Driver Code 
public static void main(String args[]) 
{ 
    int n = 4; 
      
    MinDiff(n); 
} 
} 
  
// This code is contributed by grand_master

Python3

# Python3 program to find the minimum
# difference possible by splitting
# all powers of 2 up to N into two 
# sets of equal size  
  
def MinDiff(n):
      
    # Store the largest
    val = 2 ** n
      
    # Form two separate groups
    sep = n // 2
      
    # Initialize the sum 
    # for two groups as 0
    grp1 = 0
    grp2 = 0
      
    # Insert 2 ^ n in the
    # first group
    grp1 = grp1 + val
      
    # Calculate the sum of
    # least n / 2 -1 elements
    # added to the first set 
    for i in range(1, sep):
        grp1 = grp1 + 2 ** i
          
      
    # sum of remaining
    # n / 2 - 1 elements
    for i in range(sep, n):
        grp2 = grp2 + 2 ** i
          
    # Min Difference between
    # the two groups
    print(abs(grp1 - grp2))     
      
# Driver code
if __name__=='__main__':
    n = 4
    MinDiff(n)

C#

// C# program to find the minimum 
// difference possible by splitting 
// all powers of 2 up to N into two 
// sets of equal size 
using System;
class GFG{ 
  
public static void MinDiff(int n) 
{ 
  
    // Store the largest 
    int val = (int)Math.Pow(2, n); 
      
    // Form two separate groups 
    int sep = n / 2; 
      
    // Initialize the sum 
    // for two groups as 0 
    int grp1 = 0; 
    int grp2 = 0; 
      
    // Insert 2 ^ n in the 
    // first group 
    grp1 = grp1 + val; 
      
    // Calculate the sum of 
    // least n / 2 -1 elements 
    // added to the first set 
    for(int i = 1; i < sep; i++) 
    grp1 = grp1 + (int)Math.Pow(2, i); 
          
    // Sum of remaining 
    // n / 2 - 1 elements 
    for(int i = sep; i < n; i++) 
    grp2 = grp2 + (int)Math.Pow(2, i); 
          
    // Min difference between 
    // the two groups 
    Console.Write(Math.Abs(grp1 - grp2)); 
} 
  
// Driver Code 
public static void Main() 
{ 
    int n = 4; 
      
    MinDiff(n); 
} 
} 
  
// This code is contributed by Akanksha_Rai

输出：

6

时间复杂度： O(N)
辅助空间： O(1)