最小化 2 的幂的减法以将 N 转换为 0

给定一个正整数N ，任务是找到将N转换为 0 所需的 2 次幂减法的最小次数。

例子：

Input: 10
Output: 2
Explanation: When we subtract 8 from 10 ( 10 – (2^3) = 2) then 2 will remain. 
After that subtract 2 from 2^0 i.e., 2 – 2^0 = 0. 
Hence we are doing two operation to make the N = 0.

Input: 5
Output: 2

方法：问题的方法基于以下思想：

As we need to minimize the number of subtractions then subtract as big a a power of 2 as possible. This will be same as the number of set bits in the binary representation of N. 

请按照下图进行更好的理解。

插图：

Take N = 10

1st step: The maximum value that can be subtracted is 8
               So N = 10 – 8 = 2.

2nd step: The maximum value that can be subtracted is 2
                So N = 2 – 2 = 0.

Now see the binary representation of 10 = “1010”.
It has 2 set bits. So minimum steps required = 2

请按照以下步骤实施该方法：

• 从i = 1 迭代到 63 ：
  • 检查是否设置了 N 的二进制表示的那个位。
  • 如果设置，则增加设置位的计数。
• 返回设置位的最终计数。

以下是上述方法的实现。

C++

// C++ code to implement the approach
  
#include 
using namespace std;
  
// Function to find the number of
// operation to make the N zero
int find_no_of_set_bits(long long n)
{
    // Take variable to count the number
    // of operation
    int set_bit_count = 0;
    for (int i = 0; i < 63; i++) {
        if (n & (1LL << i)) {
            set_bit_count++;
        }
    }
    return set_bit_count;
}
  
// Driver code
int main()
{
    long long N = 10;
  
    // Function call
    cout << find_no_of_set_bits(N) << endl;
    return 0;
}

输出

2

时间复杂度： O(1)
辅助空间： O(1)