计算将N表示为2的幂的和的方法

给定整数N ，任务是计算将N表示为2的幂的和的方式数。

例子：

Input: N = 4
Output: 4
Explanation: All possible ways to obtains sum N using powers of 2 are  {4, 2+2, 1+1+1+1, 2+1+1}.

Input: N = 5
Output: 4
Explanation: All possible ways to obtains sum N using powers of 2 are  {4 + 1, 2+2 + 1, 1+1+1+1 + 1, 2+1+1 + 1}

天真的方法：解决该问题的最简单方法是使值小于N的2的所有次幂并打印所有组合以表示总和N。

高效方法：为了优化上述方法，我们的想法是使用递归。定义一个函数f(N，K) ，该函数表示将N表示为2的幂的和的方式的数量，其中所有数量的幂均小于或等于k，其中K (= log ₂ (N) )最大的2个电源，其满足2 ^k≤ N．

If (power(2, K) ≤ N) :
     f(N, K) = f(N – power(2, K), K) + f(N, K – 1) //to check if power(2, k) can be one of the number.    
Otherwise:
      f(N, K)=f(N, K – 1)
Base cases :

• If (N = 0) f(N, K)=1 (Only 1 possible way exists to represent N)
• If (k==0) f(N, K)=1 (Only 1 possible way exists to represent N by taking 2⁰)

下面是上述方法的实现：

C++

// C++ program for above implementation
#include 
using namespace std;
int numberOfWays(int n, int k)
{
 
    // Base Cases
    if (n == 0)
        return 1;
 
    if (k == 0)
        return 1;
 
    // Check if 2^k can be used as
    // one of the numbers or not
    if (n >= pow(2, k)) {
        int curr_val = pow(2, k);
        return numberOfWays(n - curr_val, k)
               + numberOfWays(n, k - 1);
    }
    // Otherwise
    else
 
        // Count number of  ways to
        // N using 2 ^ k - 1
        return numberOfWays(n, k - 1);
}
 
// Driver Code
int main()
{
    int n = 4;
    int k = log2(n);
 
    cout << numberOfWays(n, k) << endl;
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG
{
static int numberOfWays(int n, int k)
{
 
    // Base Cases
    if (n == 0)
        return 1;
    if (k == 0)
        return 1;
 
    // Check if 2^k can be used as
    // one of the numbers or not
    if (n >= (int)Math.pow(2, k))
    {
        int curr_val = (int)Math.pow(2, k);
        return numberOfWays(n - curr_val, k)
               + numberOfWays(n, k - 1);
    }
   
    // Otherwise
    else
 
        // Count number of  ways to
        // N using 2 ^ k - 1
        return numberOfWays(n, k - 1);
}
 
// Driver code
public static void main(String[] args)
{
    int n = 4;
    int k = (int)(Math.log(n) / Math.log(2));
     System.out.println(numberOfWays(n, k));
}
}
 
// This code is contributed by susmitakundugoaldanga.

Python3

# Python3 program for above implementation
from math import log2
def numberOfWays(n, k):
 
    # Base Cases
    if (n == 0):
        return 1
    if (k == 0):
        return 1
 
    # Check if 2^k can be used as
    # one of the numbers or not
    if (n >= pow(2, k)):
        curr_val = pow(2, k)
        return numberOfWays(n - curr_val, k) + numberOfWays(n, k - 1)
     
    # Otherwise
    else:
 
        # Count number of  ways to
        # N using 2 ^ k - 1
        return numberOfWays(n, k - 1)
 
# Driver Code
if __name__ == '__main__':
    n = 4
    k = log2(n)
 
    print(numberOfWays(n, k))
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
class GFG
{
static int numberOfWays(int n, int k)
{
 
    // Base Cases
    if (n == 0)
        return 1;
    if (k == 0)
        return 1;
 
    // Check if 2^k can be used as
    // one of the numbers or not
    if (n >= (int)Math.Pow(2, k))
    {
        int curr_val = (int)Math.Pow(2, k);
        return numberOfWays(n - curr_val, k)
               + numberOfWays(n, k - 1);
    }
   
    // Otherwise
    else
 
        // Count number of  ways to
        // N using 2 ^ k - 1
        return numberOfWays(n, k - 1);
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 4;
    int k = (int)(Math.Log(n) / Math.Log(2));
     Console.WriteLine(numberOfWays(n, k));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// JavaScript program for above implementation
 
function numberOfWays(n, k)
{
 
    // Base Cases
    if (n == 0)
        return 1;
 
    if (k == 0)
        return 1;
 
    // Check if 2^k can be used as
    // one of the numbers or not
    if (n >= Math.pow(2, k)) {
        let curr_val = Math.pow(2, k);
        return numberOfWays(n - curr_val, k)
            + numberOfWays(n, k - 1);
    }
    // Otherwise
    else
 
        // Count number of ways to
        // N using 2 ^ k - 1
        return numberOfWays(n, k - 1);
}
 
// Driver Code
 
    let n = 4;
    let k = Math.log2(n);
 
    document.write(numberOfWays(n, k) + "<br>");
 
// This code is contributed by Mayank Tyagi
 

输出：

4

时间复杂度： O((logN + K) ^K )，其中K为log ₂ (N)
辅助空间： O(1)