在 N 的表示中计数包含一个设置位作为最高有效位的基数

给定一个正整数N ，任务是计算不同基数的数量，在这些基数中，当表示N时，发现N的最高有效位是一个设置位。

例子：

Input: N = 6
Output: 4
Explanation: The number 6 can be represented in 5 different bases, i.e. base 2, base 3, base 4, base 5, base 6.

1. (6)₁₀  in base 2: (110)₂
2. (6)₁₀  in base 3: (20)₃
3. (6)₁₀  in base 4: (12)₄
4. (6)₁₀  in base 5: (11)₅
5. (6)₁₀  in base 6: (10)₆

The base representation for (6)₁₀ in base 2, base 4, base 5, base 6 starts with 1. Hence, the required count of bases is 4.

Input: N = 5
Output: 4

方法：给定问题可以通过为每个可能的碱基找到给定数N的 MSB 并计算那些MSB为1的碱基来解决。请按照以下步骤解决问题：

• 初始化一个变量，比如count为0 ，以存储所需的结果。
• 使用变量B遍历范围[2, N]并执行以下步骤：
  • 存储base B的最高功率 需要在变量P中表示数字N 。这可以通过找到（以B为底的log N ）的值来轻松实现，即log _B (N)被截断为最接近的整数。
  • 要查找log _B (N)的值，请使用 log 属性： log _B (N) = log(N)/log(B)
  • 存储N的最高位 通过将N除以(B) ^P 。如果它等于1 ，则将计数值增加 1。
• 完成上述步骤后，打印count的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count bases
// having MSB of N as a set bit
int countOfBase(int N)
{
    // Store the required count
    int count = 0;
 
    // Iterate over the range [2, N]
    for (int i = 2; i <= N; ++i) {
 
        int highestPower
            = (int)(log(N) / log(i));
 
        // Store the MSB of N
        int firstDigit = N / (int)pow(
                                 i, highestPower);
 
        // If MSB is 1, then increment
        // the count by 1
        if (firstDigit == 1) {
            ++count;
        }
    }
 
    // Return the count
    return count;
}
 
// Driver Code
int main()
{
    int N = 6;
    cout << countOfBase(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG {
     
 public static int countOfBase(int N)
{
    
    // Store the required count
    int count = 0;
 
    // Iterate over the range [2, N]
    for (int i = 2; i <= N; ++i) {
 
        int highestPower
            = (int)(Math.log(N) /Math.log(i));
 
        // Store the MSB of N
        int firstDigit = N / (int)Math.pow(
                                 i, highestPower);
 
        // If MSB is 1, then increment
        // the count by 1
        if (firstDigit == 1) {
            ++count;
        }
    }
 
    // Return the count
    return count;
}
 
  // DRIVER CODE
    public static void main (String[] args)
    {
       int N = 6;
 
        System.out.println(countOfBase(N));
    }
}
 
 // This code is contributed by Potta Lokesh

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count bases
// having MSB of N as a set bit
static int countOfBase(int N)
{
     
    // Store the required count
    int count = 0;
     
    // Iterate over the range [2, N]
    for(int i = 2; i <= N; ++i)
    {
     
        int highestPower = (int)(Math.Log(N) /
                                 Math.Log(i));
         
        // Store the MSB of N
        int firstDigit = N / (int)Math.Pow(
                                 i, highestPower);
         
        // If MSB is 1, then increment
        // the count by 1
        if (firstDigit == 1)
        {
            ++count;
        }
    }
     
    // Return the count
    return count;
}
 
// Driver Code
public static void Main()
{
    int N = 6;
     
    Console.Write(countOfBase(N));
}
}
 
// This code is contributed by ipg2016107

Javascript

        // JavaScript implementation of the approach
        function countOfBase(N)
        {
         
            // Store the required count
            let count = 0;
 
            // Iterate over the range [2, N]
            for (let i = 2; i <= N; ++i) {
 
                let highestPower
                    = parseInt(Math.log(N) / Math.log(i));
 
                // Store the MSB of N
                let firstDigit = parseInt(N / Math.pow(
                    i, highestPower));
 
                // If MSB is 1, then increment
                // the count by 1
                if (firstDigit == 1) {
                    ++count;
                }
            }
 
            // Return the count
            return count;
        }
         
        // Driver code
        let N = 6;
        document.write(countOfBase(N))
 
// This code is contributed by Potta Lokesh
 
 
    

Python3

# Python 3 program for the above approach
import math
 
# Function to count bases
# having MSB of N as a set bit
def countOfBase(N) :
     
    # Store the required count
    count = 0
 
    # Iterate over the range [2, N]
    for i in range(2, N+1):
 
        highestPower = int(math.log(N) / math.log(i))
 
        # Store the MSB of N
        firstDigit = int(N / int(math.pow(i, highestPower)))
 
        # If MSB is 1, then increment
        # the count by 1
        if (firstDigit == 1) :
            count += 1
         
     
 
    # Return the count
    return count
 
 
# Driver Code
 
N = 6
print(countOfBase(N))

输出：

4

时间复杂度： O(N)
辅助空间： O(1)