检查一个大数是否可以被一个 2 的幂的数整除

给定一个字符串str和数字K形式的大数，任务是检查字符串str形成的数字是否能被K整除，其中K是 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 。那么根据2的完全幂的整除规则，如果给定数的最后X位能被K整除，则给定数能被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

Javascript

 
// Javascript program for the above approach
 
// Function to check divisibility
function checkIfDivisible(str, num)
{
     
    // Calculate the number of digits in num
    let powerOf2 = (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
    let i, number = 0;
    let len = str.length;
 
    for(i = len - powerOf2; i < len; i++)
    {
        number += (str[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
     
    // Given number
    let str = "213467756564";
    let num = 4;
     
    // Function call
    if (checkIfDivisible(str, num))
        document.write("Yes");
    else
        document.write("No");
   

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)