Dado un número, debe representar este número como la suma del número mínimo de números pseudobinarios posibles . Se dice que un número es pseudobinario si su número decimal consta de solo dos dígitos (0 y 1). Ejemplo: 11,10,101 son todos números pseudobinarios.
Ejemplos:-
Input : 44 Output : 11 11 11 11 Explanation : 44 can be represented as sum of minimum 4 psuedobinary numbers as 11+11+11+11 Input : 31 Output : 11 10 10 Explanation : 31 can be represented as sum of minimum 3 psuedobinary numbers as 11+10+10
La idea para hacer esto es primero observar cuidadosamente que necesitamos calcular el número mínimo de números pseudobinarios posibles. Para hacer esto, encontramos un nuevo número m tal que si para un lugar en el número n dado, el dígito no es cero, entonces el dígito en ese lugar en m es 1, de lo contrario, cero. Por ejemplo, si n = 5102, entonces m será 1101. Luego imprimiremos este número m y restaremos m de n. Seguiremos repitiendo estos pasos hasta que n sea mayor que cero.
C++
// C++ program to represent a given
// number as sum of minimum possible
// psuedobinary numbers
#include<iostream>
using namespace std;
// function to represent a given
// number as sum of minimum possible
// psuedobinary numbers
void psuedoBinary(int n)
{
// Repeat below steps until n > 0
while (n > 0)
{
// calculate m (A number that has same
// number of digits as n, but has 1 in
// place of non-zero digits 0 in place
// of 0 digits)
int temp = n, m = 0, p = 1;
while (temp)
{
int rem = temp % 10;
temp = temp / 10;
if (rem != 0)
m += p;
p *= 10;
}
cout << m << " ";
// subtract m from n
n = n - m;
}
}
// Driver code
int main()
{
int n = 31;
psuedoBinary(n);
return 0;
}
Java
// Java program to represent a given
// number as sum of minimum possible
// psuedobinary numbers
import java.util.*;
import java.lang.*;
class GFG
{
public static void psuedoBinary(int n)
{
// Repeat below steps until n > 0
while (n != 0)
{
// calculate m (A number that has same
// number of digits as n, but has 1 in
// place of non-zero digits 0 in place
// of 0 digits)
int temp = n, m = 0, p = 1;
while(temp != 0)
{
int rem = temp % 10;
temp = temp / 10;
if (rem != 0)
m += p;
p *= 10;
}
System.out.print(m + " ");
// subtract m from n
n = n - m;
}
System.out.println(" ");
}
// Driver code
public static void main(String[] args)
{
int n = 31;
psuedoBinary(n);
}
}
// This code is contributed by Mohit Gupta_OMG
Python3
# Python3 program to represent # a given number as sum of # minimum possible psuedobinary # numbers # function to represent a # given number as sum of # minimum possible # psuedobinary numbers def psuedoBinary(n): # Repeat below steps # until n > 0 while (n > 0): # calculate m (A number # that has same number # of digits as n, but # has 1 in place of non-zero # digits 0 in place of 0 digits) temp = n; m = 0; p = 1; while (temp): rem = temp % 10; temp = int(temp / 10); if (rem != 0): m += p; p *= 10; print(m,end=" "); # subtract m from n n = n - m; # Driver code n = 31; psuedoBinary(n); # This code is contributed # by mits.
C#
// C# program to represent a given
// number as sum of minimum possible
// psuedobinary numbers
using System;
class GFG
{
public static void psuedoBinary(int n)
{
// Repeat below steps until n > 0
while (n != 0)
{
// calculate m (A number that has same
// number of digits as n, but has 1 in
// place of non-zero digits 0 in place
// of 0 digits)
int temp = n, m = 0, p = 1;
while(temp != 0)
{
int rem = temp % 10;
temp = temp / 10;
if (rem != 0)
m += p;
p *= 10;
}
Console.Write(m + " ");
// subtract m from n
n = n - m;
}
Console.Write(" ");
}
// Driver code
public static void Main()
{
int n = 31;
psuedoBinary(n);
}
}
// This code is contributed by nitin mittal
PHP
<?php
// PHP program to represent a
// given number as sum of minimum
// possible psuedobinary numbers
// Function to represent a
// given number as sum of minimum
// possible psuedobinary numbers
function psuedoBinary($n)
{
// Repeat below steps until n > 0
while ($n > 0)
{
// calculate m (A number
// that has same number of
// digits as n, but has 1
// in place of non-zero
// digits 0 in place of 0
// digits)
$temp = $n; $m = 0; $p = 1;
while ($temp)
{
$rem = $temp % 10;
$temp = $temp / 10;
if ($rem != 0)
$m += $p;
$p *= 10;
}
echo $m , " ";
// subtract m from n
$n = $n - $m;
}
}
// Driver code
$n = 31;
psuedoBinary($n);
// This code is contributed
// by nitin mittal.
?>
Javascript
<script>
// JavaScript program to represent a given
// number as sum of minimum possible
// psuedobinary numbers
function psuedoBinary( n)
{
// Repeat below steps until n > 0
while (n != 0)
{
// calculate m (A number that has same
// number of digits as n, but has 1 in
// place of non-zero digits 0 in place
// of 0 digits)
var temp = n, m = 0, p = 1;
while (temp != 0) {
var rem = temp % 10;
temp = parseInt(temp / 10);
if (rem != 0)
m += p;
p *= 10;
}
document.write(m + " ");
// subtract m from n
n = n - m;
}
document.write(" ");
}
// Driver code
var n = 31;
psuedoBinary(n);
// This code is contributed by Amit Katiyar
</script>
Producción:
11 10 10
Complejidad de tiempo : O (log n)
Espacio auxiliar : O (1)
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA