检查一个大数是否可以被16整除

给定一个数字，任务是检查一个数字是否可被16整除。输入数字可能很大，即使我们使用long long int，也可能无法存储。

例子：

Input  : n = 1128
Output : No

Input  : n = 11216
Output : Yes

Input  : n = 1124273542764284287
Output : No

由于输入数字可能很大，因此我们不能使用n％16来检查数字是否可以被16整除，尤其是在C / C++之类的语言中。这个想法是基于以下事实。

A number is divisible by 16 if number formed 
by last four digits of it is divisible by 16.

插图：

For example, let us consider 769616 
Number formed by last four digits = 9616
Since 9522 is divisible by 16, answer is YES.

这是如何运作的?

Let us consider 76952, we can write it as
76942 = 7*10000 + 6*1000 + 9*100 + 5*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 16 is 0 if i greater 
than or equal to four. Note that 10000, 
100000,... etc lead to remainder 0 when divided by 16.

So remainder of "7*10000 + 6*1000 + 9*100 + 
5*10 + 2" divided by 16 is equivalent to remainder 
of following : 
0 + 6*1000 + 9*100 + 5*10 + 2 = 6952
Therefore we can say that the whole number is 
divisible by 16 if 6952 is divisible by 16.

C++

// C++ program to find if a number
// is divisible by 16 or not
#include
using namespace std;
  
// Function to find that
// number divisible by 16 or not
bool check(string str)
{
    int n = str.length();
  
    // Empty string
    if (n == 0 && n == 1)
        return false;
  
    // If there is double digit
    if (n == 2)
        return (((str[n-2]-'0')*10 +
                 (str[n-1]-'0'))%16 == 0);
  
    // If there is triple digit
    if(n == 3)
         return ( ((str[n-3]-'0')*100 +
                   (str[n-2]-'0')*10 +
                   (str[n-1]-'0'))%16 == 0);
  
  
    // If number formed by last four
    // digits is divisible by 16.
    int last = str[n-1] - '0';
    int second_last = str[n-2] - '0';
    int third_last = str[n-3] - '0';
    int fourth_last = str[n-4] - '0';
    return ((fourth_last*1000 + third_last*100 +
             second_last*10 + last) % 16 == 0);
}
  
// Driver code
int main()
{
    string str = "769528";
    check(str)?  cout << "Yes" : cout << "No ";
    return 0;
}

Java

// Java program to find if a number
// is divisible by 16 or not
import java.io.*;
  
class GFG {
    // Function to find that
    // number divisible by 16 or not
    static boolean check(String str)
    {
        int n = str.length();
       
        // Empty string
        if (n == 0 && n == 1)
            return false;
       
        // If there is double digit
        if (n == 2)
            return (((str.charAt(n-2)-'0')*10 +
                     (str.charAt(n-1)-'0'))%16 == 0);
       
        // If there is triple digit
        if(n == 3)
             return ( ((str.charAt(n-3)-'0')*100 +
                       (str.charAt(n-2)-'0')*10 +
                       (str.charAt(n-1)-'0'))%16 == 0);
       
       
        // If number formed by last
        // four digits is divisible by 16.
        int last = str.charAt(n-1) - '0';
        int second_last = str.charAt(n-2) - '0';
        int third_last = str.charAt(n-3) - '0';
        int fourth_last = str.charAt(n-4) - '0';
        return ((fourth_last*1000 + third_last*100 
                + second_last*10 + last) % 16 == 0);
    }
       
    // Driver code
    public static void main(String args[])
    {
        String str = "769528";
        if(check(str))
            System.out.println("Yes");
        else
            System.out.println("No ");
    }
}
  
// This code is contributed by Nikita Tiwari.

Python3

# Python 3 program to find
# if a number is divisible
# by 16 or not
  
# Function to find that
# number divisible by
# 16 or not
def check(st) :
    n = len(st) 
      
    # Empty string
    if (n == 0 and n == 1) :
        return False
   
    # If there is double digit
    if (n == 2) :
        return ((int)(st[n-2])*10 +
                ((int)(st[n-1])%16 == 0))
   
    # If there is triple digit
    if(n == 3) :
        return ( ((int)(st[n-3])*100 +
                   (int)(st[n-2])*10 +
                   (int)(st[n-1]))%16 == 0)
   
   
    # If number formed by last
    # four digits is divisible
    # by 16.
    last = (int)(st[n-1])
    second_last = (int)(st[n-2])
    third_last = (int)(st[n-3])
    fourth_last = (int)(st[n-4])
    return ((fourth_last*1000 + third_last*100
            + second_last*10 + last) % 16 == 0)
  
  
# Driver code
st = "769528"
if(check(st)) :
    print("Yes")
else :
    print("No")
      
  
# This code is conributed by Nikita Tiwari.

C#

// C# program to find if a number
// is divisible by 16 or not
using System;
  
class GFG {
      
    // Function to find that number 
    // divisible by 16 or not
    static bool check(String str)
    {
        int n = str.Length;
      
        // Empty string
        if (n == 0 && n == 1)
            return false;
      
        // If there is double digit
        if (n == 2)
            return (((str[n - 2] - '0') * 10 +
                (str[n - 1] - '0')) % 16 == 0);
      
        // If there is triple digit
        if(n == 3)
            return (((str[n - 3] - '0') * 100 +
                     (str[n - 2] - '0') * 10 +
                     (str[n - 1] - '0')) % 16 == 0);
      
      
        // If number formed by last
        // four digits is divisible by 16.
        int last = str[n - 1] - '0';
        int second_last = str[n - 2] - '0';
        int third_last = str[n - 3] - '0';
        int fourth_last = str[n - 4] - '0';
        return ((fourth_last * 1000 + third_last * 100
            + second_last * 10 + last) % 16 == 0);
    }
      
    // Driver code
    public static void Main()
    {
        String str = "769528";
        if(check(str))
            Console.Write("Yes");
        else
            Console.Write("No ");
    }
}
  
// This code is contributed by Nitin Mittal.

输出：

No