Dado un número n, encuentre el n-ésimo número positivo formado solo por dígitos pares (0, 2, 4, 6, 8). Los primeros números hechos de dígitos pares son 0, 2, 4, 6, 8, 20, 22, 24…….
Ejemplos:
Input : 2 Output : 2 Second number made of 0, 2, 4, 6, 8 is 2 Input : 10 Output : 28
Enfoque
ingenuo Un enfoque ingenuo es comenzar desde 0 y verificar si está compuesto solo de {0, 2, 4, 6, 8} y detenerse cuando encuentre el n-ésimo número.
C++
// Simple C++ program to find
// n-th number made of even
// digits only
#include<bits/stdc++.h>
using namespace std;
// function to calculate nth
// number made of even digits only
int findNthEvenDigitNumber(int n )
{
// variable to note how
// many such numbers have
// been found till now
int count = 0;
for (int i = 0 ; ; i++)
{
int curr = i;
// bool variable to check if
// 1, 3, 5, 7, 9 is there or not
bool isCurrEvenDigit = true ;
// checking each digit
// of the number
while (curr != 0)
{
// If 1, 3, 5, 7, 9 is found
// temp is changed to false
if (curr % 10 == 1 || curr % 10 == 3 ||
curr % 10 == 5 || curr % 10 == 7 ||
curr % 10 == 9)
isCurrEvenDigit = false;
curr = curr / 10;
}
// temp is true it means that it
// does not have 1, 3, 5, 7, 9
if (isCurrEvenDigit == true)
count++;
// If nth such number is
// found return it
if (count == n)
return i;
}
}
// Driver Code
int main()
{
cout << findNthEvenDigitNumber(2)
<< endl;
cout << findNthEvenDigitNumber(10)
<< endl;
return 0;
}
Java
// Simple Java program to
// find the n-th number made
// of even digits only
class GFG
{
// function to calculate nth
// number made of even digits only
static int findNthEvenDigitNumber(int n )
{
// variable to note how
// many such numbers have
// been found till now
int count = 0;
for (int i = 0 ; ; i++)
{
int curr = i;
// bool variable to check if
// 1, 3, 5, 7, 9 is there or not
boolean isCurrEvenDigit = true ;
// checking each digit
// of the number
while (curr != 0)
{
// If 1, 3, 5, 7, 9 is found
// temp is changed to false
if (curr % 10 == 1 || curr % 10 == 3 ||
curr % 10 == 5 || curr % 10 == 7 ||
curr % 10 == 9)
isCurrEvenDigit = false;
curr = curr / 10;
}
// temp is true it means that it
// does not have 1, 3, 5, 7, 9
if (isCurrEvenDigit == true)
count++;
// If nth such number
// is found return it
if (count == n)
return i;
}
}
// Driver Code
public static void main (String[] args)
{
System.out.println(findNthEvenDigitNumber(2));
System.out.println(findNthEvenDigitNumber(10));
}
}
Python3
# Simple Python3 program to find nth # number made of even digits only # function to calculate nth number # made of even digits only def findNthEvenDigitNumber(n): # variable to note how many such # numbers have been found till now count = 0; i = 0; while (True): curr = i; # bool variable to check if # 1, 3, 5, 7, 9 is there or not isCurrEvenDigit = True; # checking each digit of the number while (curr != 0): # If 1, 3, 5, 7, 9 is found # temp is changed to false if (curr % 10 == 1 or curr % 10 == 3 or curr % 10 == 5 or curr % 10 == 7 or curr % 10 == 9): isCurrEvenDigit = False; curr = curr // 10; # temp is true it means that it # does not have 1, 3, 5, 7, 9 if (isCurrEvenDigit == True): count += 1; # If nth such number is found, # return it if (count == n): return i; i += 1; # Driver Code print(findNthEvenDigitNumber(2)); print(findNthEvenDigitNumber(10)); # This code is contributed by mits
C#
// Simple C# program to
// find the n-th number
// made of even digits only
using System;
class GFG
{
// function to calculate nth
// number made of even digits only
static int findNthEvenDigitNumber(int n )
{
// variable to note how
// many such numbers have
// been found till now
int count = 0;
for (int i = 0 ; ; i++)
{
int curr = i;
// bool variable to check if
// 1, 3, 5, 7, 9 is there or not
bool isCurrEvenDigit = true ;
// checking each digit
// of the number
while (curr != 0)
{
// If 1, 3, 5, 7, 9 is found
// temp is changed to false
if (curr % 10 == 1 || curr % 10 == 3 ||
curr % 10 == 5 || curr % 10 == 7 ||
curr % 10 == 9 )
isCurrEvenDigit = false;
curr = curr / 10;
}
// temp is true it means that it
// does not have 1, 3, 5, 7, 9
if (isCurrEvenDigit == true)
count++;
// If nth such number
// is found return it
if (count == n)
return i;
}
}
// Driver code
public static void Main ()
{
Console.WriteLine(findNthEvenDigitNumber(2));
Console.WriteLine(findNthEvenDigitNumber(10));
}
}
// This article is contributed by vt_m.
PHP
<?php
// Simple C++ program to find
// nth number made of even
// digits only
// function to calculate nth
// number made of even digits only
function findNthEvenDigitNumber($n )
{
// variable to note how
// many such numbers have
// been found till now
$count = 0;
for ($i = 0 ; ; $i++)
{
$curr = $i;
// bool variable to check if
// 1, 3, 5, 7, 9 is there or not
$isCurrEvenDigit = true ;
// checking each digit
// of the number
while ($curr != 0)
{
// If 1, 3, 5, 7, 9 is found
// temp is changed to false
if ($curr % 10 == 1 || $curr % 10 == 3 ||
$curr % 10 == 5 || $curr % 10 == 7 ||
$curr % 10 == 9)
$isCurrEvenDigit = false;
$curr = $curr / 10;
}
// temp is true it means that it
// does not have 1, 3, 5, 7, 9
if ($isCurrEvenDigit == true)
$count++;
// If nth such number
// is found return it
if ($count == $n)
return $i;
}
}
// Driver Code
echo findNthEvenDigitNumber(2),"\n" ;
echo findNthEvenDigitNumber(10) ;
// This code is contributed by nitin mittal
?>
Javascript
<script>
// Simple JavaScript program to find
// n-th number made of even
// digits only
// Function to calculate nth
// number made of even digits only
function findNthEvenDigitNumber(n)
{
// Variable to note how
// many such numbers have
// been found till now
let count = 0;
for(let i = 0;; i++)
{
let curr = i;
// Bool variable to check if
// 1, 3, 5, 7, 9 is there or not
let isCurrEvenDigit = true;
// Checking each digit
// of the number
while (curr != 0)
{
// If 1, 3, 5, 7, 9 is found
// temp is changed to false
if (curr % 10 == 1 || curr % 10 == 3 ||
curr % 10 == 5 || curr % 10 == 7 ||
curr % 10 == 9)
isCurrEvenDigit = false;
curr = Math.floor(curr / 10);
}
// temp is true it means that it
// does not have 1, 3, 5, 7, 9
if (isCurrEvenDigit === true)
count++;
// If nth such number is
// found return it
if (count === n)
return i;
}
}
// Driver Code
document.write(findNthEvenDigitNumber(2) + "<br>");
document.write(findNthEvenDigitNumber(10) + "<br>");
// This code is contributed by Manoj.
</script>
Producción :
2 28
Complejidad de tiempo: O(n * log 10 n), donde n representa el entero dado.
Espacio auxiliar: O(1), no se requiere espacio adicional, por lo que es una constante.
Enfoque eficiente
Necesitamos encontrar números hechos de 5 dígitos, 0, 2, 4, 6 y 8. Cuando convertimos un número en un número de base 5, solo estará hecho de números {0, 1, 2, 3, 4} . Se puede ver claramente que cada dígito en el conjunto de dígitos requerido {0, 2, 4, 6, 8} es el doble del dígito en el índice correspondiente del conjunto de dígitos de base 5. Entonces, para encontrar el número n-ésimo compuesto solo por dígitos pares, siga los pasos mencionados a continuación
. Paso 1: Convierta n en n-1, esto se hace para excluir el cero.
Paso 2: Convierta n en un número decimal de base 5.
Paso 3: Multiplique el número encontrado arriba por 2. Este es el número requerido
C++
// Efficient C++ program to
// find n-th number made of
// even digits only
#include<bits/stdc++.h>
using namespace std;
// function to find nth number
// made of even digits only
int findNthEvenDigitNumber(int n)
{
// If n=1 return 0
if (n == 1)
return 0;
// vector to store the digits
// when converted into base 5
vector< int> v;
// Reduce n to n-1 to exclude 0
n = n - 1;
// Reduce n to base 5
// number and store digits
while (n > 0)
{
// pushing the digits
// into vector
v.push_back(n % 5);
n = n / 5;
}
// variable to represent the
// number after converting it
// to base 5. Since the digits
// are be in reverse order,
// we traverse vector from back
int result = 0;
for (int i = v.size() - 1; i >= 0; i--)
{
result = result * 10;
result = result + v[i];
}
// return 2*result (to convert
// digits 0, 1, 2, 3, 4 to
// 0, 2, 4, 6, 8.
return 2*result;
}
// Driver Code
int main()
{
cout << findNthEvenDigitNumber(2)
<< endl;
cout << findNthEvenDigitNumber(10)
<< endl;
return 0;
}
Java
import java.util.*;
// Efficient Java program to
// find n-th number made of
// even digits only
class GFG {
// function to find nth number
// made of even digits only
static int findNthEvenDigitNumber(int n) {
// If n=1 return 0
if (n == 1) {
return 0;
}
// vector to store the digits
// when converted into base 5
Vector< Integer> v = new Vector<>();
// Reduce n to n-1 to exclude 0
n = n - 1;
// Reduce n to base 5
// number and store digits
while (n > 0) {
// pushing the digits
// into vector
v.add(n % 5);
n = n / 5;
}
// variable to represent the
// number after converting it
// to base 5. Since the digits
// are be in reverse order,
// we traverse vector from back
int result = 0;
for (int i = v.size() - 1; i >= 0; i--) {
result = result * 10;
result = result + v.get(i);
}
// return 2*result (to convert
// digits 0, 1, 2, 3, 4 to
// 0, 2, 4, 6, 8.
return 2 * result;
}
// Driver Code
public static void main(String[] args) {
System.out.println(findNthEvenDigitNumber(2));
System.out.println(findNthEvenDigitNumber(10));
}
}
// This code is contributed by PrinciRaj1992
Python3
# Efficient Python 3 program to find n-th # number made of even digits only # function to find nth number made of # even digits only def findNthEvenDigitNumber( n): # If n = 1 return 0 if (n == 1): return 0 # vector to store the digits # when converted into base 5 v = [] # Reduce n to n-1 to exclude 0 n = n - 1 # Reduce n to base 5 number and # store digits while (n > 0): # pushing the digits into vector v.append(n % 5) n = n // 5 # variable to represent the number # after converting it to base 5. # Since the digits are be in reverse # order, we traverse vector from back result = 0 for i in range(len(v) - 1, -1, -1): result = result * 10 result = result + v[i] # return 2*result (to convert # digits 0, 1, 2, 3, 4 to # 0, 2, 4, 6, 8. return 2 * result # Driver Code if __name__ == "__main__": print(findNthEvenDigitNumber(2)) print(findNthEvenDigitNumber(10)) # This code is contributed by ita_c
C#
// Efficient C# program to
// find n-th number made of
// even digits only
using System;
using System.Collections;
class GFG {
// function to find nth number
// made of even digits only
static int findNthEvenDigitNumber(int n)
{
// If n=1 return 0
if (n == 1)
{
return 0;
}
// vector to store the digits
// when converted into base 5
ArrayList v = new ArrayList();
// Reduce n to n-1 to exclude 0
n = n - 1;
// Reduce n to base 5
// number and store digits
while (n > 0)
{
// pushing the digits
// into vector
v.Add(n % 5);
n = n / 5;
}
// variable to represent the
// number after converting it
// to base 5. Since the digits
// are be in reverse order,
// we traverse vector from back
int result = 0;
for (int i = v.Count - 1; i >= 0; i--)
{
result = result * 10;
result = result + (int)v[i];
}
// return 2*result (to convert
// digits 0, 1, 2, 3, 4 to
// 0, 2, 4, 6, 8.
return 2 * result;
}
// Driver Code
public static void Main()
{
Console.WriteLine(findNthEvenDigitNumber(2));
Console.WriteLine(findNthEvenDigitNumber(10));
}
}
// This code is contributed by 29AjayKumar
PHP
<?php
// Efficient PHP program to find n-th
// number made of even digits only
// function to find nth number
// made of even digits only
function findNthEvenDigitNumber($n)
{
// If n=1 return 0
if ($n == 1)
return 0;
// vector to store the digits
// when converted into base 5
$v = array();
// Reduce n to n-1 to exclude 0
$n = $n - 1;
// Reduce n to base 5
// number and store digits
while ($n > 0)
{
// pushing the digits
// into vector
array_push($v, $n % 5);
$n = (int)($n / 5);
}
// variable to represent the number
// after converting it to base 5.
// Since the digits are be in
// reverse order, we traverse vector
// from back
$result = 0;
for ($i = count($v) - 1; $i >= 0; $i--)
{
$result = $result * 10;
$result = $result + $v[$i];
}
// return 2*result (to convert
// digits 0, 1, 2, 3, 4 to
// 0, 2, 4, 6, 8.
return 2 * $result;
}
// Driver Code
echo findNthEvenDigitNumber(2) . "\n";
echo findNthEvenDigitNumber(10) . "\n"
// This code is contributed by mits
?>
Javascript
<script>
// Efficient Javascript program to
// find n-th number made of
// even digits only
// function to find nth number
// made of even digits only
function findNthEvenDigitNumber(n)
{
// If n=1 return 0
if (n == 1) {
return 0;
}
// vector to store the digits
// when converted into base 5
let v = [];
// Reduce n to n-1 to exclude 0
n = n - 1;
// Reduce n to base 5
// number and store digits
while (n > 0) {
// pushing the digits
// into vector
v.push(n % 5);
n = Math.floor(n / 5);
}
// variable to represent the
// number after converting it
// to base 5. Since the digits
// are be in reverse order,
// we traverse vector from back
let result = 0;
for (let i = v.length - 1; i >= 0; i--) {
result = result * 10;
result = result + v[i];
}
// return 2*result (to convert
// digits 0, 1, 2, 3, 4 to
// 0, 2, 4, 6, 8.
return 2 * result;
}
// Driver Code
document.write(findNthEvenDigitNumber(2)+"<br>");
document.write(findNthEvenDigitNumber(10));
// This code is contributed by rag2127
</script>
Producción :
2 28
Complejidad de tiempo: O (log 5 (n)), donde n es el número entero dado.
Espacio auxiliar: O(log 5 (n)), donde n es el entero dado.
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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