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

Javascript

// javascript program to find the minimum
// difference possible by splitting
// all powers of 2 up to N into two 
// sets of equal size 
 
    function MinDiff(n)
    {
 
        // Store the largest
        var val = parseInt( Math.pow(2, n));
 
        // Form two separate groups
        var sep = n / 2;
 
        // Initialize the sum
        // for two groups as 0
        var grp1 = 0;
        var 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 = 1; i < sep; i++)
            grp1 = grp1 + parseInt( Math.pow(2, i));
 
        // Sum of remaining
        // n / 2 - 1 elements
        for (i = sep; i < n; i++)
            grp2 = grp2 + parseInt( Math.pow(2, i));
 
        // Min difference between
        // the two groups
        document.write(Math.abs(grp1 - grp2));
    }
 
    // Driver Code
        var n = 4;
        MinDiff(n);
 
// This code is contributed by gauravrajput1

输出：

6

时间复杂度： O(N)
辅助空间： O(1)