Dado un número, la tarea es verificar si un número es divisible por 8 o no. El número de entrada puede ser grande y puede que no sea posible almacenarlo incluso si usamos long long int.
Ejemplos:
Input : n = 1128 Output : Yes Input : n = 1124 Output : No Input : n = 363588395960667043875487 Output : No
Dado que el número de entrada puede ser muy grande, no podemos usar n % 8 para verificar si un número es divisible por 8 o no, especialmente en lenguajes como C/C++. La idea se basa en el siguiente hecho.
Un número es divisible por 8 si el número formado por sus tres últimas cifras es divisible por 8.
Ilustración:
For example, let us consider 76952 Number formed by last three digits = 952 Since 952 is divisible by 8, answer is YES.
¿Como funciona esto?
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 8 is 0 if i greater than or equal to three. Note than 10000, 1000,... etc lead to remainder 0 when divided by 8. So remainder of "7*10000 + 6*1000 + 9*100 + 5*10 + 2" divided by 8 is equivalent to remainder of following : 0 + 0 + 9*100 + 5*10 + 2 = 52 Therefore we can say that the whole number is divisible by 8 if 952 is divisible by 8.
A continuación se muestra la implementación del hecho anterior:
C++
// C++ program to find if a number is divisible by
// 8 or not
#include<bits/stdc++.h>
using namespace std;
// Function to find that number divisible by
// 8 or not
bool check(string str)
{
int n = str.length();
// Empty string
if (n == 0)
return false;
// If there is single digit
if (n == 1)
return ((str[0]-'0')%8 == 0);
// If there is double digit
if (n == 2)
return (((str[n-2]-'0')*10 + (str[n-1]-'0'))%8 == 0);
// If number formed by last three digits is
// divisible by 8.
int last = str[n-1] - '0';
int second_last = str[n-2] - '0';
int third_last = str[n-3] - '0';
return ((third_last*100 + second_last*10 + last) % 8 == 0);
}
// Driver code
int main()
{
string str = "76952";
check(str)? cout << "Yes" : cout << "No ";
return 0;
}
Java
// Java program to find if a number is
// divisible by 8 or not
class IsDivisible
{
// Function to find that number divisible by
// 8 or not
static boolean check(String str)
{
int n = str.length();
// Empty string
if (n == 0)
return false;
// If there is single digit
if (n == 1)
return ((str.charAt(0)-'0')%8 == 0);
// If there is double digit
if (n == 2)
return (((str.charAt(n-2)-'0')*10 + (str.charAt(n-1)-'0'))%8 == 0);
// If number formed by last three digits is
// divisible by 8.
int last = str.charAt(n-1) - '0';
int second_last = str.charAt(n-2) - '0';
int third_last = str.charAt(n-3) - '0';
return ((third_last*100 + second_last*10 + last) % 8 == 0);
}
// main function
public static void main (String[] args)
{
String str = "76952";
if(check(str))
System.out.println("Yes");
else
System.out.println("No");
}
}
Python3
# Python 3 program to find
# if a number is divisible
# by 8 or not
# Function to find that
# number divisible by 8
# or not
def check(st) :
n = len(st)
# Empty string
if (n == 0) :
return False
# If there is single digit
if (n == 1) :
return ((int)(st[0]) % 8 == 0)
# If there is double digit
if (n == 2) :
return ((int)(st[n - 2]) * 10 +
((int)(str[n - 1]) % 8 == 0))
# If number formed by last
# three digits is divisible
# by 8.
last = (int)(st[n - 1])
second_last = (int)(st[n - 2])
third_last = (int)(st[n - 3])
return ((third_last*100 + second_last*10 +
last) % 8 == 0)
# Driver code
st = "76952"
if(check(st)) :
print("Yes")
else :
print("No ")
# This code is contributed by Nikita tiwari
C#
// C# program to find if a number
// is divisible by 8 or not
using System;
class IsDivisible
{
// Function to find that number
// divisible by 8 or not
static bool check(String str)
{
int n = str.Length;
// Empty string
if (n == 0)
return false;
// If there is single digit
if (n == 1)
return ((str[0] - '0') %8 == 0);
// If there is double digit
if (n == 2)
return (((str[n - 2] - '0') * 10 +
(str[n - 1] - '0')) % 8 == 0);
// If number formed by last three
// digits is divisible by 8
int last = str[n - 1] - '0';
int second_last = str[n - 2] - '0';
int third_last = str[n - 3] - '0';
return ((third_last * 100 + second_last
* 10 + last) % 8 == 0);
}
// Driver Code
public static void Main ()
{
String str = "76952";
if(check(str))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This Code is contributed by Nitin Mittal.
PHP
<?php
// PHP program to find if a number
// is divisible by 8 or not
// Function to find that number
// divisible by 8 or not
function check($str)
{
$n = strlen($str);
// Empty string
if ($n == 0)
return false;
// If there is single digit
if ($n == 1)
return (($str[0] - '0') %
8 == 0);
// If there is double digit
if ($n == 2)
return ((($str[$n - 2] - '0') *
10 + ($str[$n - 1] - '0'))
% 8 == 0);
// If number formed by last three
// digits is divisible by 8.
$last = $str[$n - 1] - '0';
$second_last = $str[$n - 2] - '0';
$third_last = $str[$n - 3] - '0';
return (($third_last * 100 +
$second_last * 10 +
$last) % 8 == 0);
}
// Driver code
$str = "76952";
$x = check($str)? "Yes" : "No ";
echo($x);
// This code is contributed by Ajit.
?>
Javascript
<script>
// JavaScript program for the above approach
// Function to find that number
// divisible by 8 or not
function check(str)
{
let n = str.length;
// Empty string
if (n == 0)
return false;
// If there is single digit
if (n == 1)
return ((str[0] - '0') %8 == 0);
// If there is double digit
if (n == 2)
return (((str[n - 2] - '0') * 10 +
(str[n - 1] - '0')) % 8 == 0);
// If number formed by last three
// digits is divisible by 8
let last = str[n - 1] - '0';
let second_last = str[n - 2] - '0';
let third_last = str[n - 3] - '0';
return ((third_last * 100 + second_last
* 10 + last) % 8 == 0);
}
// Driver Code
let str = "76952";
if(check(str))
document.write("Yes");
else
document.write("No");
// This code is contributed by splevel62.
</script>
Producción:
Yes
Complejidad de tiempo : O (1), ya que no estamos usando ningún bucle para atravesar.
Espacio auxiliar: O(1), ya que no estamos utilizando ningún espacio adicional.
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA