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📜  计算从1到N的所有数字中的总未设置位

📅  最后修改于: 2021-04-21 22:41:07             🧑  作者: Mango

给定一个正整数N ，任务是计算从1到N的所有数字的二进制表示形式中未设置位的总数。请注意，前导零不会被算作未设置的位。
例子：

方法：

  1. 将循环从1迭代到N。
  2. 当数字大于0时，将其除以2，然后检查余数。
  3. 如果余数等于0，则将count的值增加1 。

下面是上述方法的实现：

C++
// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the count of unset
// bits in the binary representation of
// all the numbers from 1 to n
int countUnsetBits(int n)
{
 
    // To store the count of unset bits
    int cnt = 0;
 
    // For every integer from the range [1, n]
    for (int i = 1; i <= n; i++) {
 
        // A copy of the current integer
        int temp = i;
 
        // Count of unset bits in
        // the current integer
        while (temp) {
 
            // If current bit is unset
            if (temp % 2 == 0)
                cnt++;
 
            temp = temp / 2;
        }
    }
    return cnt;
}
 
// Driver code
int main()
{
    int n = 5;
 
    cout << countUnsetBits(n);
 
    return 0;
}

Java
// Java implementation of the approach
class GFG
{
 
    // Function to return the count of unset
    // bits in the binary representation of
    // all the numbers from 1 to n
    static int countUnsetBits(int n)
    {
 
        // To store the count of unset bits
        int cnt = 0;
 
        // For every integer from the range [1, n]
        for (int i = 1; i <= n; i++)
        {
 
            // A copy of the current integer
            int temp = i;
 
            // Count of unset bits in
            // the current integer
            while (temp > 0)
            {
 
                // If current bit is unset
                if (temp % 2 == 0)
                {
                    cnt = cnt + 1;
                }
 
                temp = temp / 2;
            }
        }
        return cnt;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 5;
        System.out.println(countUnsetBits(n));
    }
}
 
// This code is contributed by 29AjayKumar

Python3
# Python3 implementation of the approach
 
# Function to return the count of unset
# bits in the binary representation of
# all the numbers from 1 to n
def countUnsetBits(n) :
 
    # To store the count of unset bits
    cnt = 0;
 
    # For every integer from the range [1, n]
    for i in range(1, n + 1) :
         
        # A copy of the current integer
        temp = i;
 
        # Count of unset bits in
        # the current integer
        while (temp) :
 
            # If current bit is unset
            if (temp % 2 == 0) :
                cnt += 1;
 
            temp = temp // 2;
 
    return cnt;
 
# Driver code
if __name__ == "__main__" :
 
    n = 5;
 
    print(countUnsetBits(n));
     
# This code is contributed by AnkitRai01

C#
// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the count of unset
    // bits in the binary representation of
    // all the numbers from 1 to n
    static int countUnsetBits(int n)
    {
     
        // To store the count of unset bits
        int cnt = 0;
     
        // For every integer from the range [1, n]
        for (int i = 1; i <= n; i++)
        {
     
            // A copy of the current integer
            int temp = i;
     
            // Count of unset bits in
            // the current integer
            while (temp > 0)
            {
     
                // If current bit is unset
                if (temp % 2 == 0)
                    cnt = cnt + 1;
     
                temp = temp / 2;
            }
        }
        return cnt;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 5;
        Console.Write(countUnsetBits(n));
    }
}
 
// This code is contributed by Sanjit_Prasad

Javascript

输出： 
4