Dado un número, la única operación permitida es multiplicar el número por 2. Calcula el número mínimo de operaciones para hacer que el número sea divisible por 10.
NOTA: Si no es posible convertir, imprime -1.
Ejemplos:
Entrada: 10
Salida: 0
Como el número dado es divisible por 10,
la respuesta es 0.
Entrada: 1
Salida: -1
Como al multiplicar con 2, dado el no. no se puede
convertir en un número que sea divisible por 10,
por lo tanto, la respuesta es -1.
Enfoque: cualquier número dado es divisible por 10 solo si el último dígito del número es 0. Para este problema, extraiga el último dígito del número de entrada y compruébelo de las siguientes maneras:
1) Si el último dígito es 0, entonces ya es divisible por 10, por lo que el número mínimo de pasos es 0.
2) Si el último dígito es 5, al multiplicarlo por 2 una vez lo hará divisible por 10, por lo que el número mínimo de pasos es 1.
3) Si el El último dígito es un número par o impar (aparte de 0 y 5), luego multiplicarlo por 2 cualquier número de veces solo producirá un número par, por lo que nunca podemos hacerlo divisible por 10. Por lo tanto, el número de pasos es -1.
C++
// C++ code for finding
// number of operations
#include <bits/stdc++.h>
using namespace std;
int multiplyBy2(int n)
{
int rem, value;
// Find the last digit or remainder
rem = n % 10;
switch (rem) {
// If the last digit is 0
case 0:
value = 0;
break;
// If the last digit is 5
case 5:
value = 1;
break;
// If last digit is other
// than 0 and 5.
default:
value = -1;
}
return value;
}
// Driver code
int main()
{
int n = 28;
cout << multiplyBy2(n) << endl;
n = 255;
cout << multiplyBy2(n) << endl;
return 0;
}
Java
// JAVA code for finding
// number of operations
import java.io.*;
class GFG
{
static int multiplyBy2(int n)
{
int rem, value;
// Find the last digit
// or remainder
rem = n % 10;
switch (rem)
{
// If the last digit is 0
case 0:
value = 0;
break;
// If the last digit is 5
case 5:
value = 1;
break;
// If last digit is other
// than 0 and 5.
default:
value = -1;
}
return value;
}
// Driver code
public static void main (String[] args)
{
int n = 28;
System.out.println(multiplyBy2(n));
n = 255;
System.out.println(multiplyBy2(n));
}
}
// This code is contributed
// by shiv_bhakt.
Python3
# Python3 code for finding number
# of operations
def dig(argu):
switcher = {
0: 0,
5: 1,
}
return switcher.get(argu, -1)
def multiplyBy2(n):
# Find the last digit or remainder
rem = n % 10;
return dig(rem);
# Driver code
n = 28;
print(multiplyBy2(n));
n = 255;
print(multiplyBy2(n));
# This code is contributed by mits
C#
// C# code for finding
// number of operations
using System;
class GFG
{
static int multiplyBy2(int n)
{
int rem, value;
// Find the last digit
// or remainder
rem = n % 10;
switch (rem)
{
// If the last
// digit is 0
case 0:
value = 0;
break;
// If the last
// digit is 5
case 5:
value = 1;
break;
// If last digit is
// other than 0 and 5.
default:
value = -1;
break;
}
return value;
}
// Driver code
public static void Main ()
{
int n = 28;
Console.WriteLine(multiplyBy2(n));
n = 255;
Console.WriteLine(multiplyBy2(n));
}
}
// This code is contributed
// by shiv_bhakt.
PHP
<?php
// PHP code for finding
// number of operations
function multiplyBy2($n)
{
$rem;
$value;
// Find the last digit or remainder
$rem = $n % 10;
switch ($rem) {
// If the last digit is 0
case 0:
$value = 0;
break;
// If the last digit is 5
case 5:
$value = 1;
break;
// If last digit is other
// than 0 and 5.
default:
$value = -1;
}
return $value;
}
// Driver code
$n = 28;
echo multiplyBy2($n),"\n";
$n = 255;
echo multiplyBy2($n),"\n";
// This code is contributed by aj_36
?>
Javascript
<script>
// Javascript code for finding
// number of operations
function multiplyBy2(n)
{
var rem, value;
// Find the last digit
// or remainder
rem = n % 10;
switch (rem) {
// If the last digit is 0
case 0:
value = 0;
break;
// If the last digit is 5
case 5:
value = 1;
break;
// If last digit is other
// than 0 and 5.
default:
value = -1;
}
return value;
}
// Driver code
var n = 28;
document.write(multiplyBy2(n)+"<br/>");
n = 255;
document.write(multiplyBy2(n));
// This code is contributed by todaysgaurav
</script>
Producción:
-1 1
Método: Usar if-else anidado
C++
// C++ code to multiply given number
// by 2 such that it is divisible by 10
// using nested if else
#include <iostream>
using namespace std;
int main() {
int n = 10;
// checking if the number is divisible
// by 10 or not if divisible print 0
if (n % 10 == 0)
cout << 0;
// if not divisible by 10 then multiply
// it by 2 and check again
else{
n = n*2;
if (n % 10 == 0)
cout<<0;
else
cout<<-1;
}
return 0;
}
// This code is contributed by akashish__
Java
// Java code to multiply given number
// by 2 such that it is divisible by 10
// using nested if else
import java.io.*;
class GFG {
public static void main (String[] args) {
int n = 10;
// checking if the number is divisible
// by 10 or not if divisible print 0
if (n % 10 == 0)
System.out.println("0");
// if not divisible by 10 then multiply
// it by 2 and check again
else{
n = n*2;
if (n % 10 == 0)
System.out.println("0");
else
System.out.println("-1");
}
}
}
// This code is contributed by laxmigangarajula03.
Python3
# Python code to multiply given number # by 2 such that it is divisible by 10 # using nested if else n = 10 # checking if the number is divisible # by 10 or not if divisible print 0 if n % 10 == 0: print(0) # if not divisible by 10 then multiply # it by 2 and check again else: n = n*2 if n % 10 == 0: print(0) else: print(-1) # this code is contributed by gangarajula laxmi
C#
// C# code to multiply given number
// by 2 such that it is divisible by 10
// using nested if else
using System;
public class GFG{
static public void Main (){
// Code
int n = 10;
// checking if the number is divisible
// by 10 or not if divisible print 0
if (n % 10 == 0)
{ Console.WriteLine("0"); }
// if not divisible by 10 then multiply
// it by 2 and check again
else{
n = n*2;
if (n % 10 == 0)
{ Console.WriteLine("0"); }
else
{ Console.WriteLine("-1"); }
}
}
}
// This code is contributed by akashish__
Javascript
<script>
// JavaScript code for the above approach
let n = 10;
// checking if the number is divisible
// by 10 or not if divisible print 0
if (n % 10 == 0)
document.write(0);
// if not divisible by 10 then multiply
// it by 2 and check again
else {
n = n * 2;
if (n % 10 == 0)
document.write(0);
else
document.write(-1);
}
// This code is contributed by Potta Lokesh
</script>
Producción
0
Publicación traducida automáticamente
Artículo escrito por RohanDeoli y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA