Dado un número N , la tarea es verificar si el número es divisible por 43 o no.
Ejemplos:
Entrada: N = 2795
Salida: sí
Explicación:
43 * 65 = 2795
Entrada: N = 11094
Salida: sí
Explicación:
43 * 258 = 11094
Enfoque: La prueba de divisibilidad de 43 es:
- Extraiga el último dígito.
- Agregue 13 * último dígito del número restante obtenido después de eliminar el último dígito.
- Repita los pasos anteriores hasta obtener un número de dos dígitos, o cero.
- Si el número de dos dígitos es divisible por 43, o es 0, entonces el número original también es divisible por 43.
Por ejemplo:
If N = 11739 Step 1: N = 11739 Last digit = 9 Remaining number = 1173 Adding 13 times last digit Resultant number = 1173 + 13*9 = 1290 Step 2: N = 1290 Since 129 is divisible by 43 as 43 * 3 = 129 Therefore N = 11739 is also divisible by 43
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to check whether a number // is divisible by 43 or not #include<bits/stdc++.h> #include<stdlib.h> using namespace std; // Function to check if the number is divisible by 43 or not bool isDivisible(int n) { int d; // While there are at least two digits while (n / 100) { // Extracting the last d = n % 10; // Truncating the number n /= 10; // adding thirteen times the last // digit to the remaining number n = abs(n+(d * 13)); } // Finally return if the two-digit // number is divisible by 43 or not return (n % 43 == 0) ; } // Driver Code int main() { int N = 2795; if (isDivisible(N)) cout<<"Yes"<<endl ; else cout<<"No"<<endl ; return 0; } // This code is contributed by ANKITKUMAR34
Java
// Java program to check whether a number // is divisible by 43 or not class GFG { // Function to check if the number is divisible by 43 or not static boolean isDivisible(int n) { int d; // While there are at least two digits while ((n / 100) > 0) { // Extracting the last d = n % 10; // Truncating the number n /= 10; // adding thirteen times the last // digit to the remaining number n = Math.abs(n+(d * 13)); } // Finally return if the two-digit // number is divisible by 43 or not return (n % 43 == 0) ; } // Driver Code public static void main(String[] args) { int N = 2795; if (isDivisible(N)) System.out.print("Yes"); else System.out.print("No"); } } // This code is contributed by PrinciRaj1992
Python 3
# Python program to check whether a number # is divisible by 43 or not # Function to check if the number is # divisible by 43 or not def isDivisible(n) : # While there are at least two digits while n // 100 : # Extracting the last d = n % 10 # Truncating the number n //= 10 # Adding thirteen times the last # digit to the remaining number n = abs(n+(d * 13)) # Finally return if the two-digit # number is divisible by 43 or not return (n % 43 == 0) # Driver Code if __name__ == "__main__" : N = 2795 if (isDivisible(N)): print("Yes") else : print("No")
C#
// C# program to check whether a number // is divisible by 43 or not using System; class GFG { // Function to check if the number is divisible by 43 or not static bool isDivisible(int n) { int d; // While there are at least two digits while (n / 100 > 0) { // Extracting the last d = n % 10; // Truncating the number n /= 10; // adding thirteen times the last // digit to the remaining number n = Math.Abs(n + (d * 13)); } // Finally return if the two-digit // number is divisible by 43 or not return (n % 43 == 0) ; } // Driver Code public static void Main() { int N = 2795; if (isDivisible(N)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } } // This code is contributed by AbhiThakur
Javascript
<script> //javascript program to check whether a number // is divisible by 43 or not // Function to check if the number is divisible by 43 or not function isDivisible(n) { let d; // While there are at least two digits while(parseInt(n/100) > 0) { // Extracting the last d = n % 10; // Truncating the number n = parseInt(n / 10) // adding thirteen times the last // digit to the remaining number n = Math.abs(n+(d * 13)); } // Finally return if the two-digit // number is divisible by 43 or not return (n % 43 == 0) ; } // Driver Code let N = 2795; if (isDivisible(N)) document.write("Yes"); else document.write("No"); // This code is contributed by vaibhavrabadiya117. </script>
Producción:
Yes
Complejidad del tiempo: O(log 10 N )
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por virusbuddha y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA