用2 ^ k检查二进制字符串的可除性

给定一个二进制字符串和一个数字k ，任务是检查二进制字符串是否可以被2 ^k整除。

例子 ：

Input : 11000  k = 2
Output : Yes
Explanation :
(11000)2 = (24)10
24 is evenly divisible by 2k i.e. 4.
 
Input : 10101  k = 3
Output : No
Explanation : 
(10101)2 = (21)10
21 is not evenly divisible by 2k i.e. 8.

天真的方法：将二进制字符串计算为十进制，然后简单地检查它是否可以被2 ^k整除。但是，这种方法对于长二进制字符串不可行，因为从二进制字符串计算十进制数然后将其除以2 ^k的时间复杂度将很高。

高效方法：在二进制字符串，检查最后k位。如果最后k个位全部为0，则二进制数可以被2 ^k整除，否则就不能被整除。使用此方法的时间复杂度为O(k)。

下面是该方法的实现。

C++

// C++ implementation to check whether
// given binary number is evenly
// divisible by 2^k or not
#include 
using namespace std;
  
// function to check whether
// given binary number is 
// evenly divisible by 2^k or not
bool isDivisible(char str[], int k)
{
    int n = strlen(str);
    int c = 0;
      
    // count of number of 0 from last
    for (int i = 0; i < k; i++)    
        if (str[n - i - 1] == '0')         
            c++;
      
    // if count = k, number is evenly 
    // divisible, so returns true else 
    // false
    return (c == k);
}
  
// Driver program to test above
int main()
{
    // first example
    char str1[] = "10101100";
    int k = 2;
    if (isDivisible(str1, k))
        cout << "Yes" << endl;
    else
        cout << "No"
             << "\n";
  
    // Second example
    char str2[] = "111010100";
    k = 2;
    if (isDivisible(str2, k))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
  
    return 0;
}

Java

// Java implementation to check whether
// given binary number is evenly
// divisible by 2^k or not
class GFG {
      
    // function to check whether
    // given binary number is 
    // evenly divisible by 2^k or not
    static boolean isDivisible(String str, int k)
    {
        int n = str.length();
        int c = 0;
      
        // count of number of 0 from last
        for (int i = 0; i < k; i++) 
            if (str.charAt(n - i - 1) == '0')         
                c++;
      
        // if count = k, number is evenly 
        // divisible, so returns true else 
        // false
        return (c == k);
    }
  
    // Driver program to test above
    public static void main(String args[])
    { 
          
        // first example
        String str1 = "10101100";
        int k = 2;
        if (isDivisible(str1, k) == true)
            System.out.println("Yes");
        else
            System.out.println("No");
  
    // Second example
        String str2 = "111010100";
        k = 2;
        if (isDivisible(str2, k) == true)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
  
// This code is contributed by JaideepPyne.

Python 3

# Python 3 implementation to check
# whether given binary number is
# evenly divisible by 2^k or not
  
# function to check whether
# given binary number is 
# evenly divisible by 2^k or not
def isDivisible(str, k):
    n = len(str)
    c = 0
      
    # count of number of 0 from last
    for i in range(0, k):
        if (str[n - i - 1] == '0'):     
            c += 1
      
    # if count = k, number is evenly 
    # divisible, so returns true else 
    # false
    return (c == k)
  
# Driver program to test above
# first example
str1 = "10101100"
k = 2
if (isDivisible(str1, k)):
    print("Yes")
else:
    print("No")
  
# Second example
str2 = "111010100"
k = 2
if (isDivisible(str2, k)):
    print("Yes")
else:
    print("No")
  
# This code is contributed by Smitha

C#

// C# implementation to check whether
// given binary number is evenly
// divisible by 2^k or not
using System;
  
class GFG {
      
    // function to check whether
    // given binary number is 
    // evenly divisible by 2^k or not
    static bool isDivisible(String str, int k)
    {
        int n = str.Length;
        int c = 0;
      
        // count of number of 0 from last
        for (int i = 0; i < k; i++) 
            if (str[n - i - 1] == '0')     
                c++;
      
        // if count = k, number is evenly 
        // divisible, so returns true else 
        // false
        return (c == k);
    }
  
    // Driver program to test above
    public static void Main()
    { 
          
        // first example
        String str1 = "10101100";
        int k = 2;
          
        if (isDivisible(str1, k) == true)
            Console.Write("Yes\n");
        else
            Console.Write("No");
  
        // Second example
        String str2 = "111010100";
        k = 2;
          
        if (isDivisible(str2, k) == true)
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
  
// This code is contributed by Smitha.

输出：

Yes
Yes

时间复杂度：O(k)