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📜  N中的置位和未置位计数之间的绝对差

📅  最后修改于: 2021-04-26 05:57:57             🧑  作者: Mango

先决条件: STL库中的位集函数
给定数字N ,任务是找到该给定数字的置位和未置位位数的绝对差。

例子:

方法:

  1. 计算给定数字的二进制表示形式中的总位数。
  2. 使用在STL库中定义位集合函数,有效设置位计数的数量。
  3. 然后,我们将从总位数中减去设置的位数,以获得未设置的位数。

下面是上述方法的实现:

C++
// C++ program for the above approach
#include 
using namespace std;
 
// Max size of bitset
const int sz = 64;
 
// Function to return the total bits
// in the binary representation
// of a number
int totalbits(int N)
{
    return (int)(1 + log2(N));
}
 
// Function to calculate the
// absolute difference
int absoluteDifference(int N)
{
    bitset arr(N);
 
    int total_bits = totalbits(N);
 
    // Calculate the number of
    // set bits
    int set_bits = arr.count();
 
    // Calculate the number of
    // unset bits
    int unset_bits = total_bits
                     - set_bits;
 
    int ans = abs(set_bits
                  - unset_bits);
 
    // Return the absolute difference
    return ans;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 14;
 
    // Function Call
    cout << absoluteDifference(N);
    return 0;
}


Java
// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Max size of bitset
static final int sz = 64;
 
// Function to return the total bits
// in the binary representation
// of a number
static int totalbits(int N)
{
    return (1 + (int)(Math.log(N) /
                      Math.log(2)));
}
 
// Function to calculate the
// absolute difference
static int absoluteDifference(int N)
{
    int arr = N;
 
    int total_bits = totalbits(N);
 
    // Calculate the number of
    // set bits
    int set_bits = countSetBits(arr);
 
    // Calculate the number of
    // unset bits
    int unset_bits = total_bits - set_bits;
 
    int ans = Math.abs(set_bits - unset_bits);
 
    // Return the absolute difference
    return ans;
}
 
static int countSetBits(int n)
{
    int count = 0;
    while (n > 0)
    {
        n &= (n - 1);
        count++;
    }
    return count;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given Number
    int N = 14;
 
    // Function Call
    System.out.println(absoluteDifference(N));
}
}
 
// This code is contributed by offbeat


Python3
# Python3 program for the above approach
import math
 
# Max size of bitset
sz = 64
 
# Function to return the total bits
# in the binary representation
# of a number
def totalbits(N) :
 
    return (1 + (int)(math.log(N) / math.log(2)))
 
# Function to calculate the
# absolute difference
def absoluteDifference(N) :
 
    arr = N
 
    total_bits = totalbits(N)
 
    # Calculate the number of
    # set bits
    set_bits = countSetBits(arr)
 
    # Calculate the number of
    # unset bits
    unset_bits = total_bits - set_bits
 
    ans = abs(set_bits - unset_bits)
 
    # Return the absolute difference
    return ans
 
def countSetBits(n) :
 
    count = 0
    while (n > 0) :
     
        n = n & (n - 1)
        count += 1
     
    return count
 
# Given Number
N = 14
 
# Function Call
print(absoluteDifference(N))
 
# This code is contributed by divyesh072019


C#
// C# program for the above approach
using System;
class GFG{
      
    // Function to return the total bits
    // in the binary representation
    // of a number
    static int totalbits(int N)
    {
        return (1 + (int)(Math.Log(N) /
                          Math.Log(2)));
    }
      
    // Function to calculate the
    // absolute difference
    static int absoluteDifference(int N)
    {
        int arr = N;
      
        int total_bits = totalbits(N);
      
        // Calculate the number of
        // set bits
        int set_bits = countSetBits(arr);
      
        // Calculate the number of
        // unset bits
        int unset_bits = total_bits - set_bits;
      
        int ans = Math.Abs(set_bits - unset_bits);
      
        // Return the absolute difference
        return ans;
    }
      
    static int countSetBits(int n)
    {
        int count = 0;
        while (n > 0)
        {
            n &= (n - 1);
            count++;
        }
        return count;
    }
 
  // Driver code
  static void Main() {
       
        // Given Number
        int N = 14;
      
        // Function Call
        Console.WriteLine(absoluteDifference(N));
  }
}
 
// This code is contributed by divyeshrabadiya07


Javascript


输出:
2

时间复杂度: O(log N)