Dado un número N, la tarea es encontrar los números requeridos que constan de solo 0 y 1 dígito cuya suma sea igual a N.
Ejemplo:
Input: 9 Output: 1 1 1 1 1 1 1 1 1 Only numbers smaller than or equal to 9 with digits 0 and 1 only are 0 and 1 itself. So to get 9, we have to add 1 - 9 times. Input: 31 Output: 11 10 10
Acercarse:
- Inicialice el producto p a 1 y m a cero.
- Cree el vector que almacena los recuentos enteros resultantes de 0 y 1.
- Haga un bucle para N y verifique si N es múltiplo de 10, si es así, obtenga el decimal y actualice p multiplicando 10 y almacene este valor en un vector y disminuya N por m, haga esto para cada decimal e imprima el tamaño total del vector.
- Finalmente recorre el vector e imprime los elementos.
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the numbers
int findNumbers(int N)
{
// Initialize vector array that store
// result.
vector<int> v;
// Get the each decimal and find its
// count store in vector.
while (N) {
int n = N, m = 0, p = 1;
while (n) {
// find decimal
if (n % 10)
m += p;
n /= 10;
p *= 10;
}
v.push_back(m);
// Decrement N by m for each decimal
N -= m;
}
// Loop for each element of vector
// And print its content.
for (int i = 0; i < v.size(); i++)
cout << " " << v[i];
return 0;
}
// Driver code
int main()
{
int N = 31;
findNumbers(N);
return 0;
}
Java
// Java implementation of the above approach
import java.util.*;
public class GfG{
// Function to count the numbers
public static int findNumbers(int N)
{
// Initialize vector array that store
// result.
ArrayList<Integer> v = new ArrayList<Integer>();
// Get the each decimal and find its
// count store in vector.
while (N > 0) {
int n = N, m = 0, p = 1;
while (n > 0) {
// find decimal
if (n % 10 != 0)
m += p;
n /= 10;
p *= 10;
}
v.add(m);
// Decrement N by m for each decimal
N -= m;
}
// Loop for each element of vector
// And print its content.
for (int i = 0; i < v.size(); i++)
System.out.print(" " + v.get(i));
return 0;
}
public static void main(String []args){
int N = 31;
findNumbers(N);
}
}
// This code is contributed by Rituraj Jain
Python3
# Python 3 implementation of # the above approach # Function to count the numbers def findNumbers(N) : # Initialize vector array that # store result. v = []; # Get the each decimal and find # its count store in vector. while (N) : n, m, p = N, 0, 1 while (n) : # find decimal if (n % 10) : m += p n //= 10 p *= 10 v.append(m); # Decrement N by m for # each decimal N -= m # Loop for each element of vector # And print its content. for i in range(len(v)) : print(v[i], end = " ") # Driver Code if __name__ == "__main__" : N = 31 findNumbers(N) # This code is contributed by Ryuga
C#
// C# implementation of the above approach
using System;
using System.Collections;
class GfG
{
// Function to count the numbers
public static int findNumbers(int N)
{
// Initialize vector array that store
// result.
ArrayList v = new ArrayList();
// Get the each decimal and find its
// count store in vector.
while (N > 0)
{
int n = N, m = 0, p = 1;
while (n > 0)
{
// find decimal
if (n % 10 != 0)
m += p;
n /= 10;
p *= 10;
}
v.Add(m);
// Decrement N by m for each decimal
N -= m;
}
// Loop for each element of vector
// And print its content.
for (int i = 0; i < v.Count; i++)
Console.Write(" " + v[i]);
return 0;
}
// Driver code
public static void Main()
{
int N = 31;
findNumbers(N);
}
}
// This code is contributed by PrinciRaj1992
PHP
<?php
// PHP implementation of the
// above approach
// Function to count the numbers
function findNumbers($N)
{
// Initialize vector array
// that store result.
$v = array();
// Get the each decimal and find
// its count store in vector.
while ($N)
{
$n = $N;
$m = 0;
$p = 1;
while ($n)
{
// find decimal
if ($n % 10)
$m += $p;
$n /= 10;
$p *= 10;
}
array_push($v,$m);
// Decrement N by m for
// each decimal
$N -= $m;
}
// Loop for each element of vector
// And print its content.
for ($i = 0; $i < sizeof($v); $i++)
echo " " , $v[$i];
return 0;
}
// Driver code
$N = 31;
findNumbers($N);
// This code is contributed
// by ChitraNayal
?>
Javascript
<script>
// Javascript implementation of the above approach
// Function to count the numbers
function findNumbers(N)
{
// Initialize vector array that store
// result.
let v = [];
// Get the each decimal and find its
// count store in vector.
while (N) {
let n = N, m = 0, p = 1;
while (n) {
// find decimal
if (n % 10)
m += p;
n = parseInt(n/10);
p *= 10;
}
v.push(m);
// Decrement N by m for each decimal
N -= m;
}
// Loop for each element of vector
// And print its content.
for (let i = 0; i < v.length; i++)
document.write(" " + v[i]);
return 0;
}
// Driver code
let N = 31;
findNumbers(N);
// This code is contributed by souravmahato34.
</script>
Producción:
11 10 10
Complejidad de tiempo : O(log(N))
Espacio auxiliar : O(log(N))