将n表示为两个|的正好k次幂之和。套装2

给定两个整数n和k ，任务是确定是否有可能将n表示为2的k次幂之和。如果可能，则打印k个正整数，使它们为2的幂，并且它们的和等于n，否则将打印Impossible 。

例子：

Input: n = 9, k = 4
Output: 1 2 2 4
1, 2 and 4 are all powers of 2 and 1 + 2 + 2 + 4 = 9.

Input: n = 3, k = 7
Output: Impossible
It is impossible since 3 cannot be represented as sum of 7 numbers which are powers of 2.

我们在查找k个数(其幂为2且总和为N)中讨论了解决此问题的一种方法。在本文中，正在讨论另一种方法。

方法：

• 创建一个大小为k的数组arr [] ，将所有元素初始化为1，并创建一个变量sum = k 。
• 现在从arr []的最后一个元素开始
  • 如果sum + arr [i]≤n，则更新sum = sum + arr [i]，而arr [i] = arr [i] * 2 。
  • 否则跳过当前元素。
• 如果sum = n，则arr []的内容是必需的元素。
• 否则，不可能将n精确地表示为2的k次幂。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// Function to print k numbers which are powers of two
// and whose sum is equal to n
void FindAllElements(int n, int k)
{
    // Initialising the sum with k
    int sum = k;
  
    // Initialising an array A with k elements
    // and filling all elements with 1
    int A[k];
    fill(A, A + k, 1);
  
    for (int i = k - 1; i >= 0; --i) {
  
        // Iterating A[] from k-1 to 0
        while (sum + A[i] <= n) {
  
            // Update sum and A[i]
            // till sum + A[i] is less than equal to n
            sum += A[i];
            A[i] *= 2;
        }
    }
  
    // Impossible to find the combination
    if (sum != n) {
        cout << "Impossible";
    }
  
    // Possible solution is stored in A[]
    else {
        for (int i = 0; i < k; ++i)
            cout << A[i] << ' ';
    }
}
  
// Driver code
int main()
{
    int n = 12;
    int k = 6;
  
    FindAllElements(n, k);
  
    return 0;
}

Java

// Java implementation of the above approach 
import java.util.Arrays;
  
public class GfG {
      
    // Function to print k numbers which are powers of two 
    // and whose sum is equal to n 
    public static void FindAllElements(int n, int k) 
    { 
        // Initialising the sum with k 
        int sum = k; 
        
        // Initialising an array A with k elements 
        // and filling all elements with 1 
        int[] A = new int[k]; 
        Arrays.fill(A, 0, k, 1); 
          
        for (int i = k - 1; i >= 0; --i) { 
        
            // Iterating A[] from k-1 to 0 
            while (sum + A[i] <= n) { 
        
                // Update sum and A[i] 
                // till sum + A[i] is less than equal to n 
                sum += A[i]; 
                A[i] *= 2; 
            } 
        } 
        
        // Impossible to find the combination 
        if (sum != n) { 
            System.out.print("Impossible"); 
        } 
        
        // Possible solution is stored in A[] 
        else { 
            for (int i = 0; i < k; ++i) 
                System.out.print(A[i] + " "); 
        } 
    } 
      
    public static void main(String []args){
          
        int n = 12; 
        int k = 6; 
        
        FindAllElements(n, k); 
    }
}
    
// This code is contributed by Rituraj Jain

Python3

# Python 3 implementation of the above approach
  
# Function to print k numbers which are 
# powers of two and whose sum is equal to n
def FindAllElements(n, k):
      
    # Initialising the sum with k
    sum = k
  
    # Initialising an array A with k elements
    # and filling all elements with 1
    A = [1 for i in range(k)]
    i = k - 1
    while(i >= 0):
          
        # Iterating A[] from k-1 to 0
        while (sum + A[i] <= n):
              
            # Update sum and A[i] till
            # sum + A[i] is less than equal to n
            sum += A[i]
            A[i] *= 2
        i -= 1
      
    # Impossible to find the combination
    if (sum != n):
        print("Impossible")
  
    # Possible solution is stored in A[]
    else:
        for i in range(0, k, 1):
            print(A[i], end = ' ')
  
# Driver code
if __name__ == '__main__':
    n = 12
    k = 6
  
    FindAllElements(n, k)
  
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the above approach 
using System;
  
class GfG 
{
      
    // Function to print k numbers
    // which are powers of two 
    // and whose sum is equal to n 
    public static void FindAllElements(int n, int k) 
    { 
        // Initialising the sum with k 
        int sum = k; 
          
        // Initialising an array A with k elements 
        // and filling all elements with 1 
        int[] A = new int[k]; 
        for(int i = 0; i < k; i++)
            A[i] = 1;
          
        for (int i = k - 1; i >= 0; --i) 
        { 
          
            // Iterating A[] from k-1 to 0 
            while (sum + A[i] <= n) 
            { 
          
                // Update sum and A[i] 
                // till sum + A[i] is less than equal to n 
                sum += A[i]; 
                A[i] *= 2; 
            } 
        } 
          
        // Impossible to find the combination 
        if (sum != n)
        { 
            Console.Write("Impossible"); 
        } 
          
        // Possible solution is stored in A[] 
        else
        { 
            for (int i = 0; i < k; ++i) 
                Console.Write(A[i] + " "); 
        } 
    } 
      
    // Driver code
    public static void Main(String []args)
    {
          
        int n = 12; 
        int k = 6; 
          
        FindAllElements(n, k); 
    }
}
  
// This code contributed by Rajput-Ji

PHP

= 0; --$i)
    { 
  
        // Iterating A[] from k-1 to 0 
        while ($sum + $A[$i] <= $n) 
        { 
  
            // Update sum and A[i] till  
            // sum + A[i] is less than equal to n 
            $sum += $A[$i]; 
            $A[$i] *= 2; 
        } 
    } 
  
    // Impossible to find the combination 
    if ($sum != $n) 
    { 
        echo"Impossible"; 
    } 
  
    // Possible solution is stored in A[] 
    else
    { 
        for ($i = 0; $i < $k; ++$i) 
            echo $A[$i], ' '; 
    } 
} 
  
// Driver code 
$n = 12; 
$k = 6; 
  
FindAllElements($n, $k); 
  
// This code is contributed by Ryuga
?>

输出：

1 1 1 1 4 4

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