Gapful Number es un número N de al menos 3 dígitos tal que es divisible por la concatenación de su primer y último dígito.
Algunos números vacíos son:
100, 105, 108, 110, 120, 121, 130, 132, 135, 140,…
Comprobar si N es un número vacío
Dado un número entero N , la tarea es verificar si N es un número vacío o no. Si N es un número vacío, escriba «Sí» , de lo contrario, escriba «No» .
Ejemplos:
Entrada: N = 108
Salida: Sí
Explicación:
108 es divisible por 18
Entrada: N = 112
Salida: No
Enfoque: La idea es crear un número (digamos num ) usando el primer y último dígito de los números dados y comprobar si N es divisible por num o no. Si N es divisible por num , entonces es un número vacío e imprime «Sí» , de lo contrario, imprime «No» .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
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
<script>
// 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
</script>
Producción:
Yes
Complejidad de tiempo: O(1)
Referencia: https://oeis.org/A108343