检查大数是否可以被2的整数除

给定一个以字符串str和数字K形式存在的大数字，任务是检查由字符串str形成的数字是否可被K整除，其中K为2的幂。
例子：

Input: str = “5426987513245621541524288”, num = 64 
Output: Yes 
Explanation: 
Since log₂(64) = 6, so the number formed by the last 6 digits from the string str is divisible by 64 .
Input: str = “21346775656413259795656497974113461254”, num = 4 
Output: No 
Explanation: 
Since log₂(4)=2, the number formed by the last 2 digits from the string str is not divisible by 4. 
 

方法：
由于K是2的完美幂。令K可以表示为2 ^X。然后根据2的幂的除数规则，如果给定数的最后X个数可被K整除，则给定数可被K整除。否则，不能被K整除。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include  
using namespace std;
  
// Function to check divisibility
bool checkIfDivisible(string str,
                      long long int num)
{
  
    // Calculate the number of digits in num
    long long int powerOf2 = log2(num);
  
    // Check if the length of
    // the string is less than
    // the powerOf2 then
    // return false
    if (str.length() < powerOf2)
        return false;
  
    // Check if the powerOf2 is 0
    // that means the given number
    // is 1 and as every number
    // is divisible by 1 so return true
    if (powerOf2 == 0)
        return true;
  
    // Find the number which is
    // formed by the last n digits
    // of the string where n=powerOf2
    long long int i, number = 0;
    int len = str.length();
  
    for (i = len - powerOf2; i < len; i++) {
        number += (str[i] - '0')
                  * pow(10,
                        powerOf2 - 1);
        powerOf2--;
    }
  
    // Check if the number formed is
    // divisible by input num or not
    if (number % num)
        return false;
    else
        return true;
}
  
// Driver Code
int main()
{
    // Given number
    string str = "213467756564";
    long long int num = 4;
  
    // Function Call
    if (checkIfDivisible(str, num))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

Java

// Java program for the above approach
class GFG{ 
  
// Function to check divisibility
static boolean checkIfDivisible(String str,
                                long num)
{
      
    // Calculate the number of digits in num
    long powerOf2 = (int)(Math.log(num) /
                          Math.log(2));
  
    // Check if the length of
    // the string is less than
    // the powerOf2 then
    // return false
    if (str.length() < powerOf2)
        return false;
  
    // Check if the powerOf2 is 0
    // that means the given number
    // is 1 and as every number
    // is divisible by 1 so return true
    if (powerOf2 == 0)
        return true;
  
    // Find the number which is
    // formed by the last n digits
    // of the string where n=powerOf2
    long i, number = 0;
    int len = str.length();
  
    for(i = len - powerOf2; i < len; i++) 
    {
        number += (str.charAt((int)i) - '0') * 
                   Math.pow(10, powerOf2 - 1);
        powerOf2--;
    }
  
    // Check if the number formed is
    // divisible by input num or not
    if (number % num != 0)
        return false;
    else
        return true;
}
  
// Driver Code
public static void main(String[] args) 
{
      
    // Given number
    String str = "213467756564";
    long num = 4;
      
    // Function call
    if (checkIfDivisible(str, num))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
  
// This code is contributed by rutvik_56

Python3

# Python3 program for the above approach 
from math import log2
  
# Function to check divisibility 
def checkIfDivisible(string, num): 
  
    # Calculate the number of digits in num 
    powerOf2 = int(log2(num)); 
  
    # Check if the length of 
    # the string is less than 
    # the powerOf2 then 
    # return false 
    if (len(string) < powerOf2):
        return False; 
  
    # Check if the powerOf2 is 0 
    # that means the given number 
    # is 1 and as every number 
    # is divisible by 1 so return true 
    if (powerOf2 == 0):
        return True; 
  
    # Find the number which is 
    # formed by the last n digits 
    # of the string where n=powerOf2 
    number = 0; 
    length = len(string); 
  
    for i in range(length - powerOf2, length): 
        number += ((ord(string[i]) - ord('0')) *
                  (10 ** (powerOf2 - 1))); 
          
        powerOf2 -= 1; 
  
    # Check if the number formed is 
    # divisible by input num or not 
    if (number % num):
        return False; 
    else :
        return True; 
  
# Driver Code 
if __name__ == "__main__" : 
  
    # Given number 
    string = "213467756564"; 
    num = 4; 
  
    # Function Call 
    if (checkIfDivisible(string, num)):
        print("Yes"); 
    else :
        print("No"); 
  
# This code is contributed by AnkitRai01

C#

// C# program for the above approach
using System;
  
class GFG{ 
  
// Function to check divisibility
static bool checkIfDivisible(String str,
                             long num)
{
      
    // Calculate the number of digits in num
    long powerOf2 = (int)(Math.Log(num) /
                          Math.Log(2));
  
    // Check if the length of
    // the string is less than
    // the powerOf2 then
    // return false
    if (str.Length < powerOf2)
        return false;
  
    // Check if the powerOf2 is 0
    // that means the given number
    // is 1 and as every number
    // is divisible by 1 so return true
    if (powerOf2 == 0)
        return true;
  
    // Find the number which is
    // formed by the last n digits
    // of the string where n=powerOf2
    long i, number = 0;
    int len = str.Length;
  
    for(i = len - powerOf2; i < len; i++) 
    {
        number += (long)((str[(int)i] - '0') * 
                Math.Pow(10, powerOf2 - 1));
        powerOf2--;
    }
  
    // Check if the number formed is
    // divisible by input num or not
    if (number % num != 0)
        return false;
    else
        return true;
}
  
// Driver Code
public static void Main(String[] args) 
{
      
    // Given number
    String str = "213467756564";
    long num = 4;
      
    // Function call
    if (checkIfDivisible(str, num))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
  
// This code is contributed by amal kumar choubey

输出：

Yes

时间复杂度： O(Len)，其中Len是字符串的长度。
辅助空间： O(log ₂ K)