空缺数字

Gapful Number是至少3位数字的N ，因此可以通过将其第一位和最后一位数字串联而将其整除。
很少有缺口的数字是：

100, 105, 108, 110, 120, 121, 130, 132, 135, 140,… 
 

检查N是否为空位编号

给定一个整数N ，任务是检查N是否为空缺数。如果N是一个空位数字，则打印“是”，否则打印“否” 。
例子：

Input: N = 108 
Output: Yes 
Explanation: 
108 is divisible by 18
Input: N = 112 
Output: No 
 

方法：想法是使用给定数字的第一位和最后一位创建一个数字(例如num )，并检查N是否可被num整除。如果N可被num整除，则它是一个有空数并打印“是” ，否则打印“否” 。
下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Find the first digit
int firstDigit(int n)
{
    // Find total number of digits - 1
    int digits = (int)log10(n);
 
    // Find first digit
    n = (int)(n / pow(10, digits));
 
    // Return first digit
    return n;
}
 
// Find the last digit
int lastDigit(int n)
{
    // return the last digit
    return (n % 10);
}
 
// A function to check Gapful numbers
bool isGapful(int n)
{
    int first_dig = firstDigit(n);
    int last_dig = lastDigit(n);
 
    int concatenation = first_dig * 10
                        + last_dig;
 
    // Return true if n is gapful number
    return (n % concatenation == 0);
}
 
// Driver Code
int main()
{
    // Given Number
    int n = 108;
 
    // Function Call
    if (isGapful(n))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Find the first digit
static int firstDigit(int n)
{
 
    // Find total number of digits - 1
    int digits = (int)(Math.log(n) /
                       Math.log(10));
 
    // Find first digit
    n = (int)(n / Math.pow(10, digits));
 
    // Return first digit
    return n;
}
 
// Find the last digit
static int lastDigit(int n)
{
     
    // Return the last digit
    return (n % 10);
}
 
// A function to check Gapful numbers
static boolean isGapful(int n)
{
    int first_dig = firstDigit(n);
    int last_dig = lastDigit(n);
 
    int concatenation = first_dig * 10 +
                        last_dig;
 
    // Return true if n is gapful number
    return (n % concatenation == 0);
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given number
    int n = 108;
 
    // Function call
    if (isGapful(n))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by Pratima Pandey

Python3

# Python3 program for the above approach
import math
 
# Find the first digit
def firstDigit(n):
 
    # Find total number of digits - 1
    digits = math.log10(n)
 
    # Find first digit
    n = (n / math.pow(10, digits))
 
    # Return first digit
    return n
 
# Find the last digit
def lastDigit(n):
 
    # return the last digit
    return (n % 10)
 
# A function to check Gapful numbers
def isGapful(n):
 
    concatenation = (firstDigit(n) * 10) +\
                     lastDigit(n)
 
    # Return true if n is gapful number
    return (n % concatenation)
 
# Driver Code
if __name__=='__main__':
 
    # Given Number
    n = 108
 
    # Function Call
    if (isGapful(n)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by Ritik Bansal

C#

// C# program for the above approach
using System;
class GFG{
 
// Find the first digit
static int firstDigit(int n)
{
 
    // Find total number of digits - 1
    int digits = (int)(Math.Log(n) /
                       Math.Log(10));
 
    // Find first digit
    n = (int)(n / Math.Pow(10, digits));
 
    // Return first digit
    return n;
}
 
// Find the last digit
static int lastDigit(int n)
{
     
    // Return the last digit
    return (n % 10);
}
 
// A function to check Gapful numbers
static bool isGapful(int n)
{
    int first_dig = firstDigit(n);
    int last_dig = lastDigit(n);
 
    int concatenation = first_dig * 10 +
                        last_dig;
 
    // Return true if n is gapful number
    return (n % concatenation == 0);
}
 
// Driver code
public static void Main()
{
     
    // Given number
    int n = 108;
 
    // Function call
    if (isGapful(n))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by Code_Mech

Javascript

// Javascript program for the above approach
 
    // Find the first digit
    function firstDigit( n)
    {
 
        // Find total number of digits - 1
        let digits = parseInt( (Math.log(n) / Math.log(10)));
 
        // Find first digit
        n = parseInt( (n / Math.pow(10, digits)));
 
        // Return first digit
        return n;
    }
 
    // Find the last digit
    function lastDigit( n)
    {
 
        // Return the last digit
        return (n % 10);
    }
 
    // A function to check Gapful numbers
    function isGapful( n)
    {
        let first_dig = firstDigit(n);
        let last_dig = lastDigit(n);
 
        let concatenation = first_dig * 10 + last_dig;
 
        // Return true if n is gapful number
        return (n % concatenation == 0);
    }
 
    // Driver code
      
    // Given number
    let n = 108;
 
    // Function call
    if (isGapful(n))
        document.write("Yes");
    else
        document.write("No");
 
// This code is contributed by aashish1995

输出：

Yes

时间复杂度： O(1)
参考： https : //oeis.org/A108343