在[0，n]范围内只有1个置位的数字的计数

给定一个整数n ，任务是对范围为[0，n]的仅具有1个设置位的数字进行计数。
例子：

Input: n = 7 
Output: 3 
000, 001, 010, 011, 100, 101, 110 and 111 are the binary representation of all the numbers upto 7. 
And there are only 3 numbers ( 001, 010 and 100 ) having only 1 set bit.
Input: n = 3 
Output: 2 
 

方法：如果需要k个位来表示n，则可能有k个数，因为每次1可以位于k个不同的位置。
下面是上述方法的实现

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the required count
int count(int n)
{
 
    // To store the count of numbers
    int cnt = 0;
    int p = 1;
    while (p <= n) {
        cnt++;
 
        // Every power of 2 contains
        // only 1 set bit
        p *= 2;
    }
    return cnt;
}
 
// Driver code
int main()
{
    int n = 7;
    cout << count(n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG {
 
    // Function to return the required count
    static int count(int n)
    {
 
        // To store the count of numbers
        int cnt = 0;
        int p = 1;
        while (p <= n) {
            cnt++;
 
            // Every power of 2 contains
            // only 1 set bit
            p *= 2;
        }
        return cnt;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 7;
        System.out.print(count(n));
    }
}

C#

// C# implementation of the approach
using System;
class GFG {
 
    // Function to return the required count
    static int count(int n)
    {
 
        // To store the count of numbers
        int cnt = 0;
        int p = 1;
        while (p <= n) {
            cnt++;
 
            // Every power of 2 contains
            // only 1 set bit
            p *= 2;
        }
        return cnt;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 7;
        Console.Write(count(n));
    }
}

Python3

# Python3 implementation of the approach
 
# Function to return the required count
def count(n):
     
    # To store the count of numbers
    cnt = 0
    p = 1
    while (p <= n):
        cnt = cnt + 1
         
        # Every power of 2 contains
        # only 1 set bit
        p *= 2
    return cnt
 
# Driver code
n = 7
print(count(n));

Javascript

// Javascript implementation of the approach
 
// Function to return the required count
function count(n)
{
 
    // To store the count of numbers
    var cnt = 0;
    var p = 1;
    while (p <= n) {
        cnt++;
 
        // Every power of 2 contains
        // only 1 set bit
        p *= 2;
    }
    return cnt;
}
 
// Driver code
var n = 7;
document.write(count(n));
 
// This code is contributed by noob2000.

输出：

3

时间复杂度： O(log n)