可以通过给定的运算使2的幂数的计数

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📜 可以通过给定的运算使2的幂数的计数

📅 最后修改于: 2021-04-22 04:14:50 🧑 作者: Mango

给定一个数组arr [] ，任务是对可以通过以下操作将其乘以2的幂进行计数：
如果1的乘方尚未为2的幂，则可以最多将其添加一次。

例子：

Input: arr[] = {2, 3, 7, 9, 15}
Output: 4
3, 7 and 15 can be made a power of 2 by adding 1, and 2 is already a power of 2

Input: arr[] = {5, 6, 9, 3, 1}
Output: 2

为什么编程需要懂一点英语

方法：遍历数组并检查当前元素是否为2的幂，如果是，则更新count = count + 1 。如果不是2的幂，则检查一个大的元素，即arr [i] + 1 。要检查元素是否为2的幂：

天真的方法是将元素重复除以2，直到得到0或1作为余数。如果余数是1，则其2的幂，否则不是2的幂。
高效的方法：如果X＆(X – 1)= 0，则X为2的幂。
假设X = 16 = 10000，并且X – 1 = 15 = 01111，那么X＆(X – 1)= 10000＆01111 = 0，即X = 16的2的幂。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function that returns true if x is a power of 2
bool isPowerOfTwo(int x)
{
    if (x == 0)
        return false;
 
    // If x & (x-1) = 0 then x is a power of 2
    if (!(x & (x - 1)))
        return true;
    else
        return false;
}
 
// Function to return the required count
int countNum(int a[], int n)
{
    int count = 0;
 
    for (int i = 0; i < n; i++) {
 
        // If a[i] or (a[i]+1) is a power of 2
        if (isPowerOfTwo(a[i]) || isPowerOfTwo(a[i] + 1))
            count++;
    }
 
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 5, 6, 9, 3, 1 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << countNum(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function that returns true if x is a power of 2
static boolean isPowerOfTwo(int x)
{
    if (x == 0)
        return false;
 
    // If x & (x-1) = 0 then x is a power of 2
    if ((x & (x - 1)) == 0)
        return true;
    else
        return false;
}
 
// Function to return the required count
static int countNum(int a[], int n)
{
    int count = 0;
 
    for (int i = 0; i < n; i++) {
 
        // If a[i] or (a[i]+1) is a power of 2
        if (isPowerOfTwo(a[i]) || isPowerOfTwo(a[i] + 1))
            count++;
    }
 
    return count;
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = { 5, 6, 9, 3, 1 };
 
    int n = arr.length;
 
    System.out.println(countNum(arr, n));
 
}
}
 
// This code is contributed by
// Sahil_Shelangia

Python3

# Python 3 implementation of the approach
 
# Function that returns true if x
# is a power of 2
def isPowerOfTwo(x):
    if (x == 0):
        return False
 
    # If x & (x-1) = 0 then x is a power of 2
    if ((x & (x - 1)) == 0):
        return True
    else:
        return False
 
# Function to return the required count
def countNum(a, n):
    count = 0
 
    for i in range(0, n, 1):
         
        # If a[i] or (a[i]+1) is a power of 2
        if (isPowerOfTwo(a[i]) or
            isPowerOfTwo(a[i] + 1)):
            count += 1
 
    return count
 
# Driver code
if __name__ == '__main__':
    arr = [5, 6, 9, 3, 1]
 
    n = len(arr)
 
    print(countNum(arr, n))
 
# This code is contributed by
# Sanjit_Prasad

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function that returns true if x is a power of 2
static bool isPowerOfTwo(int x)
{
    if (x == 0)
        return false;
 
    // If x & (x-1) = 0 then x is a power of 2
    if ((x & (x - 1)) == 0)
        return true;
    else
        return false;
}
 
// Function to return the required count
static int countNum(int[] a, int n)
{
    int count = 0;
 
    for (int i = 0; i < n; i++)
    {
        // If a[i] or (a[i]+1) is a power of 2
        if (isPowerOfTwo(a[i]) || isPowerOfTwo(a[i] + 1))
            count++;
    }
 
    return count;
}
 
// Driver code
public static void Main()
{
    int[] arr = { 5, 6, 9, 3, 1 };
    int n = arr.Length;
    Console.WriteLine(countNum(arr, n));
 
}
}
 
// This code is contributed by
// Mukul Singh

PHP

Javascript

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