数组中元素的二进制表示形式中存在的置位计数值的乘积

给定一个由N个整数组成的数组arr [] ，任务是在每个数组元素的二进制表示形式中找到设置位数的乘积。

例子：

Input: arr[] = {3, 2, 4, 1, 5}
Output: 4
Explanation:
Binary representation of the array elements are {3, 2, 4, 1, 5} are {“11”, “10”, “100”, “1”, “101”} respectively.
Therefore, the product of count of set bits = (2 * 1 * 1 * 1 * 2) = 4.

Input: arr[] = {10, 11, 12}
Output: 12

方法：可以通过对每个数组元素的二进制表示形式中的总位数进行计数来解决给定的问题。请按照以下步骤解决问题：

• 初始化一个变量，例如product ，以存储生成的产品。
• 遍历给定数组arr []并执行以下步骤：
  • 查找整数arr [i]的设置位数，并将其存储在变量中，例如bits 。
  • 将产品的值更新为product * bits 。
• 完成上述步骤后，将结果打印为产品值。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count the
// set bits in an integer
int countbits(int n)
{
    // Stores the count of set bits
    int count = 0;
 
    // Iterate while N is not equal to 0
    while (n != 0) {
 
        // Increment count by 1
        if (n & 1)
            count++;
 
        // Divide N by 2
        n = n / 2;
    }
 
    // Return the total count obtained
    return count;
}
 
// Function to find the product
// of count of set bits present
// in each element of an array
int BitProduct(int arr[], int N)
{
    // Stores the resultant product
    int product = 1;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // Stores the count
        // of set bits of arr[i]
        int bits = countbits(arr[i]);
 
        // Update the product
        product *= bits;
    }
 
    // Return the resultant product
    return product;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 2, 4, 1, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << BitProduct(arr, N);
 
    return 0;
}

Java

// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG {
 
    // Function to count the
    // set bits in an integer
    static int countbits(int n)
    {
        // Stores the count of set bits
        int count = 0;
 
        // Iterate while N is not equal to 0
        while (n != 0) {
 
            // Increment count by 1
            if ((n & 1) != 0)
                count++;
 
            // Divide N by 2
            n = n / 2;
        }
 
        // Return the total count obtained
        return count;
    }
 
    // Function to find the product
    // of count of set bits present
    // in each element of an array
    static int BitProduct(int arr[], int N)
    {
       
        // Stores the resultant product
        int product = 1;
 
        // Traverse the array arr[]
        for (int i = 0; i < N; i++) {
 
            // Stores the count
            // of set bits of arr[i]
            int bits = countbits(arr[i]);
 
            // Update the product
            product *= bits;
        }
 
        // Return the resultant product
        return product;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 3, 2, 4, 1, 5 };
        int N = arr.length;
        System.out.print(BitProduct(arr, N));
    }
}
 
// This code is contributed by Kingash.

Python3

# Python3 program for the above approach
 
# Function to count the
# set bits in an integer
def countbits(n):
   
    # Stores the count of set bits
    count = 0
 
    # Iterate while N is not equal to 0
    while (n != 0):
 
        # Increment count by 1
        if (n & 1):
            count += 1
 
        # Divide N by 2
        n = n // 2
 
    # Return the total count obtained
    return count
 
# Function to find the product
# of count of set bits present
# in each element of an array
def BitProduct(arr, N):
   
    # Stores the resultant product
    product = 1
 
    # Traverse the array arr[]
    for i in range(N):
       
        # Stores the count
        # of set bits of arr[i]
        bits = countbits(arr[i])
 
        # Update the product
        product *= bits
 
    # Return the resultant product
    return product
 
# Driver Code
if __name__ == '__main__':
    arr = [3, 2, 4, 1, 5]
    N = len(arr)
    print(BitProduct(arr, N))
 
    # This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
 
public class GFG {
 
    // Function to count the
    // set bits in an integer
    static int countbits(int n)
    {
        // Stores the count of set bits
        int count = 0;
 
        // Iterate while N is not equal to 0
        while (n != 0) {
 
            // Increment count by 1
            if ((n & 1) != 0)
                count++;
 
            // Divide N by 2
            n = n / 2;
        }
 
        // Return the total count obtained
        return count;
    }
 
    // Function to find the product
    // of count of set bits present
    // in each element of an array
    static int BitProduct(int[] arr, int N)
    {
 
        // Stores the resultant product
        int product = 1;
 
        // Traverse the array arr[]
        for (int i = 0; i < N; i++) {
 
            // Stores the count
            // of set bits of arr[i]
            int bits = countbits(arr[i]);
 
            // Update the product
            product *= bits;
        }
 
        // Return the resultant product
        return product;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int[] arr = { 3, 2, 4, 1, 5 };
        int N = arr.Length;
        Console.Write(BitProduct(arr, N));
    }
}
 
// This code is contributed by ukasp.

输出：

4

时间复杂度： O(N * log M)，M是数组的最大元素。
辅助空间： O(1)