El problema es verificar si la representación decimal del número binario dado es divisible por 10 o no. Tenga cuidado, el número podría ser muy grande y no encajar incluso en long long int. El enfoque debe ser tal que haya cero o un número mínimo de operaciones de multiplicación y división. No hay 0 a la izquierda en la entrada.
Ejemplos:
Input : 101000 Output : Yes (101000)2 = (40)10 and 40 is divisible by 10. Input : 11000111001110 Output : Yes
Enfoque: En primer lugar, necesitamos saber que el último dígito de pow(2, i) = 2, 4, 8, 6 si i % 4 es igual a 1, 2, 3, 0 respectivamente, donde i es mayor que igual a 1. Entonces, en la representación binaria necesitamos saber la posición del dígito ‘1’ de la derecha, para saber la potencia perfecta de 2 con la que se va a multiplicar. Esto nos ayudará a obtener el último dígito de la potencia perfecta requerida de 2. Podemos sumar estos dígitos y luego verificar si el último dígito de la suma es 0 o no, lo que implica que el número es divisible por 10 o no. Tenga en cuenta que si el último dígito en la representación binaria es ‘1’, entonces representa un número impar y, por lo tanto, no es divisible por 10.
C++
// C++ implementation to check whether decimal
// representation of given binary number is
// divisible by 10 or not
#include <bits/stdc++.h>
using namespace std;
// function to check whether decimal representation
// of given binary number is divisible by 10 or not
bool isDivisibleBy10(string bin)
{
int n = bin.size();
// if last digit is '1', then
// number is not divisible by 10
if (bin[n-1] == '1')
return false;
// to accumulate the sum of last digits
// in perfect powers of 2
int sum = 0;
// traverse from the 2nd last up to 1st digit
// in 'bin'
for (int i=n-2; i>=0; i--)
{
// if digit in '1'
if (bin[i] == '1')
{
// calculate digit's position from
// the right
int posFromRight = n - i - 1;
// according to the digit's position,
// obtain the last digit of the applicable
// perfect power of 2
if (posFromRight % 4 == 1)
sum = sum + 2;
else if (posFromRight % 4 == 2)
sum = sum + 4;
else if (posFromRight % 4 == 3)
sum = sum + 8;
else if (posFromRight % 4 == 0)
sum = sum + 6;
}
}
// if last digit is 0, then
// divisible by 10
if (sum % 10 == 0)
return true;
// not divisible by 10
return false;
}
// Driver program to test above
int main()
{
string bin = "11000111001110";
if (isDivisibleBy10(bin))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java implementation to check whether decimal
// representation of given binary number is
// divisible by 10 or not
import java.util.*;
class GFG {
// function to check whether decimal
// representation of given binary number
// is divisible by 10 or not
static boolean isDivisibleBy10(String bin)
{
int n = bin.length();
// if last digit is '1', then
// number is not divisible by 10
if (bin.charAt(n - 1) == '1')
return false;
// to accumulate the sum of last
// digits in perfect powers of 2
int sum = 0;
// traverse from the 2nd last up to
// 1st digit in 'bin'
for (int i = n - 2; i >= 0; i--)
{
// if digit in '1'
if (bin.charAt(i) == '1')
{
// calculate digit's position
// from the right
int posFromRight = n - i - 1;
// according to the digit's
// position, obtain the last
// digit of the applicable
// perfect power of 2
if (posFromRight % 4 == 1)
sum = sum + 2;
else if (posFromRight % 4 == 2)
sum = sum + 4;
else if (posFromRight % 4 == 3)
sum = sum + 8;
else if (posFromRight % 4 == 0)
sum = sum + 6;
}
}
// if last digit is 0, then
// divisible by 10
if (sum % 10 == 0)
return true;
// not divisible by 10
return false;
}
/* Driver program to test above function */
public static void main(String[] args)
{
String bin = "11000111001110";
if (isDivisibleBy10(bin))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by Arnav Kr. Mandal.
Python
# Python implementation to check whether
# decimal representation of given binary
# number is divisible by 10 or not
# function to check whether decimal
# representation of given binary number
# is divisible by 10 or not
def isDivisibleBy10(bin) :
n = len(bin)
#if last digit is '1', then
# number is not divisible by 10
if (bin[n - 1] == '1') :
return False
# to accumulate the sum of last
# digits in perfect powers of 2
sum = 0
#traverse from the 2nd last up to
# 1st digit in 'bin'
i = n - 2
while i >= 0 :
# if digit in '1'
if (bin[i] == '1') :
# calculate digit's position
# from the right
posFromRight = n - i - 1
#according to the digit's
# position, obtain the last
# digit of the applicable
# perfect power of 2
if (posFromRight % 4 == 1) :
sum = sum + 2
else if (posFromRight % 4 == 2) :
sum = sum + 4
else if (posFromRight % 4 == 3) :
sum = sum + 8
else if (posFromRight % 4 == 0) :
sum = sum + 6
i = i - 1
# if last digit is 0, then
# divisible by 10
if (sum % 10 == 0) :
return True
# not divisible by 10
return False
# Driver program to test above function
bin = "11000111001110"
if (isDivisibleBy10(bin)== True) :
print("Yes")
else :
print("No")
# This code is contributed by Nikita Tiwari.
C#
// C# implementation to check whether decimal
// representation of given binary number is
// divisible by 10 or not
using System;
class GFG {
// function to check whether decimal
// representation of given binary number
// is divisible by 10 or not
static bool isDivisibleBy10(String bin)
{
int n = bin.Length;
// if last digit is '1', then
// number is not divisible by 10
if (bin[n - 1] == '1')
return false;
// to accumulate the sum of last
// digits in perfect powers of 2
int sum = 0;
// traverse from the 2nd last up to
// 1st digit in 'bin'
for (int i = n - 2; i >= 0; i--) {
// if digit in '1'
if (bin[i] == '1') {
// calculate digit's position
// from the right
int posFromRight = n - i - 1;
// according to the digit's
// position, obtain the last
// digit of the applicable
// perfect power of 2
if (posFromRight % 4 == 1)
sum = sum + 2;
else if (posFromRight % 4 == 2)
sum = sum + 4;
else if (posFromRight % 4 == 3)
sum = sum + 8;
else if (posFromRight % 4 == 0)
sum = sum + 6;
}
}
// if last digit is 0, then
// divisible by 10
if (sum % 10 == 0)
return true;
// not divisible by 10
return false;
}
/* Driver program to test above function */
public static void Main()
{
String bin = "11000111001110";
if (isDivisibleBy10(bin))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Sam007
PHP
<?php
// PHP implementation to
// check whether decimal
// representation of given
// binary number is divisible
// by 10 or not
// function to check whether
// decimal representation of
// given binary number is
// divisible by 10 or not
function isDivisibleBy10($bin)
{
$n = strlen($bin);
// if last digit is '1',
// then number is not
// divisible by 10
if ($bin[$n - 1] == '1')
return false;
// to accumulate the sum
// of last digits in
// perfect powers of 2
$sum = 0;
// traverse from the 2nd
// last up to 1st digit
// in 'bin'
for ($i = $n - 2; $i >= 0; $i--)
{
// if digit in '1'
if ($bin[$i] == '1')
{
// calculate digit's
// position from the right
$posFromRight = $n - $i - 1;
// according to the digit's
// position, obtain the last
// digit of the applicable
// perfect power of 2
if ($posFromRight % 4 == 1)
$sum = $sum + 2;
else if ($posFromRight % 4 == 2)
$sum = $sum + 4;
else if ($posFromRight % 4 == 3)
$sum = $sum + 8;
else if ($posFromRight % 4 == 0)
$sum = $sum + 6;
}
}
// if last digit is 0, then
// divisible by 10
if ($sum % 10 == 0)
return true;
// not divisible by 10
return false;
}
// Driver Code
$bin = "11000111001110";
if(isDivisibleBy10($bin))
echo "Yes";
else
echo "No";
// This code is contributed by mits.
?>
Javascript
<script>
// Javascript implementation to check whether decimal
// representation of given binary number is
// divisible by 10 or not
// function to check whether decimal
// representation of given binary number
// is divisible by 10 or not
function isDivisibleBy10(bin)
{
let n = bin.length;
// if last digit is '1', then
// number is not divisible by 10
if (bin[n - 1] == '1')
return false;
// to accumulate the sum of last
// digits in perfect powers of 2
let sum = 0;
// traverse from the 2nd last up to
// 1st digit in 'bin'
for (let i = n - 2; i >= 0; i--)
{
// if digit in '1'
if (bin[i] == '1')
{
// calculate digit's position
// from the right
let posFromRight = n - i - 1;
// according to the digit's
// position, obtain the last
// digit of the applicable
// perfect power of 2
if (posFromRight % 4 == 1)
sum = sum + 2;
else if (posFromRight % 4 == 2)
sum = sum + 4;
else if (posFromRight % 4 == 3)
sum = sum + 8;
else if (posFromRight % 4 == 0)
sum = sum + 6;
}
}
// if last digit is 0, then
// divisible by 10
if (sum % 10 == 0)
return true;
// not divisible by 10
return false;
}
// driver function
let bin = "11000111001110";
if (isDivisibleBy10(bin))
document.write("Yes");
else
document.write("No");
</script>
Producción:
Yes
Método: convierta la string binaria dada en decimal usando la función int y luego verifique si es divisible por 10 o no usando la división de módulo.
Python3
# Python code to check
# decimal representation of
# a given binary string is
# divisible by 10 or not
str1 = "101000"
# converting binary string in to
# decimal number using int function
decnum = int(str1, 2)
# checking if number is divisible by 10
# or not if divisible print yes else no
if decnum % 10 == 0:
print("Yes")
else:
print("No")
# this code is contributed by gangarajula laxmi
Java
// java code to check
// decimal representation of
// a given binary string is
// divisible by 10 or not
import java.io.*;
class GFG {
public static void main (String[] args) {
String s="1010";
//converting binary string in to
//decimal number using Convert.ToInt function
int n=Integer.parseInt(s,2);
if (n%10==0)
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
Javascript
<script>
// JavaScript code for the above approach
str1 = "101000"
// converting binary string in to
// decimal number using int function
decnum = Number.parseInt(str1, 2)
// checking if number is divisible by 10
// or not if divisible print yes else no
if (decnum % 10 == 0)
document.write("Yes")
else
document.write("No")
// This code is contributed by Potta Lokesh
</script>
C#
// C# code to check
// decimal representation of
// a given binary string is
// divisible by 10 or not
using System;
public class GFG{
static public void Main (){
string s="1010";
//converting binary string in to
//decimal number using Convert.ToInt function
int n=Convert.ToInt32(s,2);
if (n%10==0)
{
Console.Write("Yes");
}
else
{
Console.Write("No");
}
}
}
Producción
Yes
Complejidad temporal: O(n), donde n es el número de dígitos del número binario.
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA