设置位为K的倍数的数组中的元素计数

给定一个由N个元素组成的数组arr []和一个整数K ，任务是计算置位位数为K的倍数的所有元素。

例子：

Input: arr[] = {1, 2, 3, 4, 5}, K = 2
Output: 2
Explanation:
Two numbers whose setbits count is multiple of 2 are {3, 5}.

Input: arr[] = {10, 20, 30, 40}, K = 4
Output: 1
Explanation:
There number whose setbits count is multiple of 4 is {30}.

方法：

1. 遍历数组中的数字。
2. 计算数组中每个数字的设置位。
3. 检查setbits计数是否为K的倍数。

下面是上述方法的实现：

C++

// C++ implementation of above approach
  
#include 
using namespace std;
  
// Function to find the count of numbers
int find_count(vector arr, int k)
{
  
    int ans = 0;
    for (int i : arr) {
  
        // Get the set-bits count of each element
        int x = __builtin_popcount(i);
  
        // Check if the setbits count
        // is divisible by K
        if (x % k == 0)
  
            // Increment the count
            // of required numbers by 1
            ans += 1;
    }
  
    return ans;
}
  
// Driver code
int main()
{
    vector arr = { 12, 345, 2, 68, 7896 };
    int K = 2;
  
    cout << find_count(arr, K);
  
    return 0;
}

Java

// Java implementation of above approach
  
class GFG{
  
// Function to find the count of numbers
static int find_count(int []arr, int k)
{
  
    int ans = 0;
    for (int i : arr) {
  
        // Get the set-bits count of each element
        int x = Integer.bitCount(i);
  
        // Check if the setbits count
        // is divisible by K
        if (x % k == 0)
  
            // Increment the count
            // of required numbers by 1
            ans += 1;
    }
  
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    int []arr = { 12, 345, 2, 68, 7896 };
    int K = 2;
  
    System.out.print(find_count(arr, K));
  
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of above approach 
  
# Function to count total set bits
def bitsoncount(x):
    return bin(x).count('1')
  
# Function to find the count of numbers 
def find_count(arr, k) :
  
    ans = 0
    for i in arr:
  
        # Get the set-bits count of each element 
        x = bitsoncount(i)
  
        # Check if the setbits count 
        # is divisible by K 
        if (x % k == 0) :
            # Increment the count 
            # of required numbers by 1 
            ans += 1
    return ans
  
# Driver code 
arr = [ 12, 345, 2, 68, 7896 ]
K = 2
print(find_count(arr, K))
  
# This code is contributed by Sanjit_Prasad

C#

// C# implementation of above approach
using System;
  
class GFG{
  
// Function to find the count of numbers
static int find_count(int []arr, int k)
{
    int ans = 0;
    foreach(int i in arr)
    {
  
        // Get the set-bits count of each element
        int x = countSetBits(i);
  
        // Check if the setbits count
        // is divisible by K
        if (x % k == 0) 
              
            // Increment the count
            // of required numbers by 1
            ans += 1;
    }
  
    return ans;
}
  
static int countSetBits(long x)
{
    int setBits = 0;
    while (x != 0) 
    {
        x = x & (x - 1);
        setBits++;
    }
  
    return setBits;
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 12, 345, 2, 68, 7896 };
    int K = 2;
  
    Console.Write(find_count(arr, K));
}
}
// This code is contributed by sapnasingh4991

输出：

3

时间复杂度： O(N * M) ，其中N是数组的大小，M是数组中最大数字的位数。
辅助空间复杂度： O(1)