通过加 1 或乘以 2 以最小步长将 0 转换为 N

给定一个正整数N ，任务是找到将数字0转换为N所需的最小加法运算次数，以便在每个运算中任何数字都可以乘以 2或将值加 1 。

例子：

Input: N = 6
Output: 1
Explanation:
Following are the operations performed to convert 0 to 6:
Add 1          –> 0 + 1 = 1.
Multiply 2  –> 1 * 2 = 2.
Add 1          –> 2 + 1 = 3. 
Multiply 2  –> 3 * 2 = 6.
Therefore number of addition operations = 2.  

Input: N = 3
Output: 2

方法：这个问题可以通过使用位操作技术来解决。在N的二进制数表示中，当N变为奇数时（这意味着设置了N的最低有效位）对每个位进行运算，然后执行加法运算。否则，乘以 2 。给定问题的最终逻辑是找到 N 中设置的位数。

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to count number of
// set bits in N
int minimumAdditionOperation(
    unsigned long long int N)
{
 
    // Stores the count of set bits
    int count = 0;
 
    while (N) {
 
        // If N is odd, then it
        // a set bit
        if (N & 1 == 1) {
            count++;
        }
        N = N >> 1;
    }
 
    // Return the result
    return count;
}
 
// Driver Code
int main()
{
    int N = 6;
    cout << minimumAdditionOperation(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG {
 
    // Function to count number of
    // set bits in N
    static int minimumAdditionOperation(int N)
    {
 
        // Stores the count of set bits
        int count = 0;
 
        while (N > 0) {
 
            // If N is odd, then it
            // a set bit
            if (N % 2 == 1) {
                count++;
            }
            N = N >> 1;
        }
 
        // Return the result
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 6;
        System.out.println(minimumAdditionOperation(N));
    }
}
 
// This code is contributed by dwivediyash

Python3

# python program for above approach
 
# Function to count number of
# set bits in N
def minimumAdditionOperation(N):
 
    # Stores the count of set bits
    count = 0
 
    while (N):
 
        # If N is odd, then it
        # a set bit
        if (N & 1 == 1):
            count += 1
 
        N = N >> 1
 
    # Return the result
    return count
 
# Driver Code
if __name__ == "__main__":
 
    N = 6
    print(minimumAdditionOperation(N))
 
    # This code is contributed by rakeshsahni.

C#

// C# program for above approach
using System;
public class GFG{
     
    // Function to count number of
    // set bits in N
    static int minimumAdditionOperation(int N)
    {
     
        // Stores the count of set bits
        int count = 0;
     
        while (N != 0) {
     
            // If N is odd, then it
            // a set bit
            if ((N & 1) == 1) {
                count++;
            }
            N = N >> 1;
        }
     
        // Return the result
        return count;
    }
     
    // Driver Code
    static public void Main (){
        int N = 6;
        Console.Write(minimumAdditionOperation(N));
     
    }
}
 
// This code is contributed by AnkThon

Javascript

// Javascript program for above approach
 
// Function to count number of
// set bits in N
function minimumAdditionOperation(N)
{
 
  // Stores the count of set bits
  let count = 0;
 
  while (N)
  {
   
    // If N is odd, then it
    // a set bit
    if (N & (1 == 1)) {
      count++;
    }
    N = N >> 1;
  }
 
  // Return the result
  return count;
}
 
// Driver Code
let N = 6;
document.write(minimumAdditionOperation(N));
 
// This code is contributed by saurabh_jaiswal.

输出：

2

时间复杂度： O(log N)
辅助空间： O(1)