每次成功搜索后通过将元素加倍来重复搜索

给定一个数组“ a []”和整数“ b”。查找b是否存在于a []中。如果存在，则将b的值加倍并再次搜索。我们重复这些步骤，直到找不到b。最后，我们返回b的值。

例子：

Input : a[] = {1, 2, 3}
          b = 1 
Output :4
Initially we start with b = 1. Since 
it is present in array, it becomes 2.
Now 2 is also present in array b becomes 4 .
Since 4 is not present, we return 4.

Input : a[] = {1 3 5 2 12}
          b = 3 
Output :6

问题来源：在Yatra.com在线测试中询问

1)对输入数组进行排序。
2)继续进行二进制搜索并加倍，直到不存在该元素。

下面的代码在STL中使用binary_search()

C++

// C++ program to repeatedly search an element by
// doubling it after every successful search
#include 
using namespace std;
  
int findElement(int a[], int n, int b)
{
    // Sort the given array so that binary search
    // can be applied on it
    sort(a, a + n);
  
    int max = a[n - 1]; // Maximum array element
  
    while (b < max) {
  
        // search for the element b present or
        // not in array
        if (binary_search(a, a + n, b))
            b *= 2;
        else
            return b;
    }
  
    return b;
}
  
// Driver code
int main()
{
    int a[] = { 1, 2, 3 };
    int n = sizeof(a) / sizeof(a[0]);
    int b = 1;
    cout << findElement(a, n, b);
    return 0;
}

Java

// Java program to repeatedly search an element by
// doubling it after every successful search
import java.util.Arrays;
public class Test4 {
  
    static int findElement(int a[], int n, int b)
    {
        // Sort the given array so that binary search
        // can be applied on it
        Arrays.sort(a);
  
        int max = a[n - 1]; // Maximum array element
  
        while (b < max) {
  
            // search for the element b present or
            // not in array
            if (Arrays.binarySearch(a, b) > -1)
                b *= 2;
            else
                return b;
        }
  
        return b;
    }
  
    // Driver code
    public static void main(String args[])
    {
        int a[] = { 1, 2, 3 };
        int n = a.length;
        int b = 1;
        System.out.println(findElement(a, n, b));
    }
}
// This article is contributed by Sumit Ghosh

Python

# Python program to repeatedly search an element by
# doubling it after every successful search
  
def binary_search(a, x, lo = 0, hi = None):
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo + hi)//2
        midval = a[mid]
        if midval < x:
            lo = mid + 1
        elif midval > x: 
            hi = mid
        else:
            return mid
    return -1
  
def findElement(a, n, b):
     
    # Sort the given array so that binary search
    # can be applied on it
    a.sort()
   
    mx = a[n - 1] # Maximum array element
   
    while (b < max):
          
        # search for the element b present or
        # not in array
        if (binary_search(a, b, 0, n) != -1):
            b *= 2
        else:
            return b
    return b
   
# Driver code
a = [ 1, 2, 3 ]
n = len(a)
b = 1
print findElement(a, n, b)
  
# This code is contributed by Sachin Bisht

C#

// C# program to repeatedly search an
// element by doubling it after every
// successful search
using System;
  
public class GFG {
  
    static int findElement(int[] a,
                           int n, int b)
    {
  
        // Sort the given array so that
        // binary search can be applied
        // on it
        Array.Sort(a);
  
        // Maximum array element
        int max = a[n - 1];
  
        while (b < max) {
  
            // search for the element b
            // present or not in array
            if (Array.BinarySearch(a, b) > -1)
                b *= 2;
            else
                return b;
        }
  
        return b;
    }
  
    // Driver code
    public static void Main()
    {
        int[] a = { 1, 2, 3 };
        int n = a.Length;
        int b = 1;
        Console.WriteLine(findElement(a, n, b));
    }
}
  
// This code is contributed by vt_m.

PHP

 $x)
            $hi = $mid;
        else
            return $mid;
    }
    return -1;
}
  
function findElement($a, $n, $b)
{
// Sort the given array so that binary search
// can be applied on it
    sort($a);
  
    $mx = $a[$n - 1]; // Maximum array element
  
while ($b < max($a)) {
  
// search for the element b present or
// not in array
    if (binary_search($a, $b, 0, $n) != -1)
        $b *= 2;
    else
        return $b;
}
return $b;
}
  
// Driver code
$a = array(1, 2, 3 );
$n = count($a);
$b = 1;
echo findElement($a, $n, $b);
  
// This code is contributed by Srathore
?>

输出：

4