检查排序数组中是否存在数字除以 2 的幂的 ceil

给定一个已排序的数组arr[]和一个整数K ，任务是检查数组中是否存在数字K除以某个 2 的幂的单元格。
注意：如果没有这样的元素打印-1。
例子：

Input: arr[] = {3, 5, 7, 8, 10}, K = 4 
Output: -1 
Explanation: 
There is no such element.
Input: arr[] = {1, 2, 3, 5, 7, 8}, K = 4 
Output: 2 
Explanation: 
When 4 is divided by 2 to the power 1, there exist an element in the array.

方法：这个想法是尝试 2 的每个幂，并检查是否存在从幂为 0 开始的那个数的 ceil。检查元素是否存在于数组中可以使用二分搜索来完成。
下面是上述方法的实现：

C++14

// C++14 implementation to check
// if a number divided by power
// of two exist in the sorted array
#include 
using namespace std;
 
// Function to find there exist a
// number or not in the array
int findNumberDivByPowerofTwo(int ar[],
                              int k, int n)
{
    int found = -1, m = k;
 
    // Loop to check if there exist
    // a number by divided by power of 2
    while (m > 0)
    {
        int l = 0;
        int r = n - 1;
 
        // Binary Search
        while (l <= r)
        {
            int mid = (l + r) / 2;
 
            if (ar[mid] == m)
            {
                found = m;
                break;
            }
            else if (ar[mid] > m)
            {
                r = mid - 1;
            }
            else if (ar[mid] < m)
            {
                l = mid + 1;
            }
        }
 
        // Condition to check the number
        // is found in the array or not
        if (found != -1)
        {
            break;
        }
 
        // Otherwise divide the number
        // by increasing the one more
        // power of 2
        m = m / 2;
    }
    return found;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 5, 7, 8, 10 };
    int k = 4, n = 5;
     
    cout << findNumberDivByPowerofTwo(arr, k, n);
}
 
// This code is contributed by code_hunt

Java

// Java implementation to check
// if a number divided by power
// of two exist in the sorted array
 
import java.util.Scanner;
 
public class GreeksForGreeksQuestions {
 
    // Function to find there exist a
    // number or not in the array
    static int findNumberDivByPowerofTwo(
        int[] ar, int k, int n)
    {
        int found = -1, m = k;
 
        // Loop to check if there exist
        // a number by divided by power of 2
        while (m > 0) {
            int l = 0;
            int r = n - 1;
 
            // Binary Search
            while (l <= r) {
                int mid = (l + r) / 2;
 
                if (ar[mid] == m) {
                    found = m;
                    break;
                }
                else if (ar[mid] > m) {
                    r = mid - 1;
                }
                else if (ar[mid] < m) {
                    l = mid + 1;
                }
            }
 
            // Condition to check the number
            // is found in the array or not
            if (found != -1) {
                break;
            }
 
            // Otherwise divide the number
            // by increasing the one more
            // power of 2
            m = m / 2;
        }
 
        return found;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 3, 5, 7, 8, 10 };
        int k = 4, n = 5;
 
        System.out.println(
            findNumberDivByPowerofTwo(
                arr, k, n));
    }
}

Python3

# Python3 implementation to check
# if a number divided by power
# of two exist in the sorted array
 
# Function to find there exist a
# number or not in the array
def findNumberDivByPowerofTwo(ar, k, n):
     
    found = -1
    m = k
 
    # Loop to check if there exist
    # a number by divided by power of 2
    while (m > 0):
        l = 0
        r = n - 1
 
        # Binary Search
        while (l <= r):
            mid = (l + r) // 2
 
            if (ar[mid] == m):
                found = m
                break
             
            elif (ar[mid] > m):
                r = mid - 1
             
            elif (ar[mid] < m):
                l = mid + 1
             
        # Condition to check the number
        # is found in the array or not
        if (found != -1):
            break
         
        # Otherwise divide the number
        # by increasing the one more
        # power of 2
        m = m // 2
     
    return found
 
# Driver Code
arr = [ 3, 5, 7, 8, 10 ]
k = 4
n = 5
 
print(findNumberDivByPowerofTwo(arr, k, n))
 
# This code is contributed by code_hunt

C#

// C# implementation to check
// if a number divided by power
// of two exist in the sorted array
using System;
 
class GFG{
     
// Function to find there exist a
// number or not in the array
static int findNumberDivByPowerofTwo(int[] ar, int k,
                                     int n)
{
    int found = -1, m = k;
 
    // Loop to check if there exist
    // a number by divided by power of 2
    while (m > 0)
    {
        int l = 0;
        int r = n - 1;
 
        // Binary Search
        while (l <= r)
        {
            int mid = (l + r) / 2;
 
            if (ar[mid] == m)
            {
                found = m;
                break;
            }
            else if (ar[mid] > m)
            {
                r = mid - 1;
            }
            else if (ar[mid] < m)
            {
                l = mid + 1;
            }
        }
 
        // Condition to check the number
        // is found in the array or not
        if (found != -1)
        {
            break;
        }
 
        // Otherwise divide the number
        // by increasing the one more
        // power of 2
        m = m / 2;
    }
    return found;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 3, 5, 7, 8, 10 };
    int k = 4, n = 5;
 
    Console.WriteLine(findNumberDivByPowerofTwo(
                      arr, k, n));
}
}
 
// This code is contributed by princi singh

Javascript

// javascript implementation to check
// if a number divided by power
// of two exist in the sorted array
 
 
    // Function to find there exist a
    // number or not in the array
    function findNumberDivByPowerofTwo(ar , k , n) {
        var found = -1, m = k;
 
        // Loop to check if there exist
        // a number by divided by power of 2
        while (m > 0) {
            var l = 0;
            var r = n - 1;
 
            // Binary Search
            while (l <= r) {
                var mid = parseInt((l + r) / 2);
 
                if (ar[mid] == m) {
                    found = m;
                    break;
                } else if (ar[mid] > m) {
                    r = mid - 1;
                } else if (ar[mid] < m) {
                    l = mid + 1;
                }
            }
 
            // Condition to check the number
            // is found in the array or not
            if (found != -1) {
                break;
            }
 
            // Otherwise divide the number
            // by increasing the one more
            // power of 2
            m = parseInt(m / 2);
        }
 
        return found;
    }
 
    // Driver Code
     
        var arr = [ 3, 5, 7, 8, 10 ];
        var k = 4, n = 5;
 
        document.write(findNumberDivByPowerofTwo(arr, k, n));
 
// This code contributed by gauravrajput1

输出：

-1

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