Dado un número N, la tarea es encontrar el N-ésimo término de la serie:
1, 2, 11, 12, 21…
Ejemplos:
Input : N = 2 Output : 2 Input : N = 5 Output : 21
Planteamiento:
La idea se basa en que el valor del último dígito se alterna en la serie. Por ejemplo, si el último dígito del número i es 1, entonces el último dígito de los números (i-1) y (i+1) debe ser 2.
Por lo tanto, cree una array de tamaño (n+1) y empuje 1 y 2 (Estos dos son siempre los dos primeros elementos de la serie) a él.
Por tanto, el i-ésimo término del arreglo es:
1) Si i es impar,
arr[i] = arr[i/2]*10 + 1;
2) Si i es par,
arr[i] = arr[(i/2)-1]*10 + 2;
Por fin regresa arr[n].
A continuación se muestra la implementación de la idea anterior:
C++
// C++ program to find
// the N-th term in the series
// 1, 2, 11, 12, 21...
#include <bits/stdc++.h>
using namespace std;
// Function to find N-th number in series
int printNthElement(int N)
{
// create an array of size (N+1)
int arr[N + 1];
arr[1] = 1;
arr[2] = 2;
for (int i = 3; i <= N; i++) {
// If i is odd
if (i % 2 != 0) {
arr[i] = arr[i / 2] * 10 + 1;
}
else {
arr[i] = arr[(i / 2) - 1] * 10 + 2;
}
}
return arr[N];
}
// Driver code
int main()
{
// Get N
int N = 5;
// Get Nth term
cout << printNthElement(N);
return 0;
}
C
// C program to find
// the N-th term in the series
// 1, 2, 11, 12, 21...
#include <stdio.h>
// Function to find N-th number in series
int printNthElement(int N)
{
// create an array of size (N+1)
int arr[N + 1];
arr[1] = 1;
arr[2] = 2;
for (int i = 3; i <= N; i++) {
// If i is odd
if (i % 2 != 0) {
arr[i] = arr[i / 2] * 10 + 1;
}
else {
arr[i] = arr[(i / 2) - 1] * 10 + 2;
}
}
return arr[N];
}
// Driver code
int main()
{
// Get N
int N = 5;
// Get Nth term
printf("%d",printNthElement(N));
return 0;
}
// This code is contributed by kothavvsaakash.
Java
// Java program to find
// the N-th term in the series
// 1, 2, 11, 12, 21...
class FindNth {
// Function to find n-th number in series
static int printNthElement(int n)
{
// create an array of size (n+1)
int arr[] = new int[n + 1];
arr[1] = 1;
arr[2] = 2;
for (int i = 3; i <= n; i++) {
// If i is odd
if (i % 2 != 0)
arr[i] = arr[i / 2] * 10 + 1;
else
arr[i] = arr[(i / 2) - 1] * 10 + 2;
}
return arr[n];
}
// main function
public static void main(String[] args)
{
int n = 5;
System.out.println(printNthElement(n));
}
}
Python3
# Python3 program to find # the N-th term in the series # 1, 2, 11, 12, 21... # Return n-th number in series def printNthElement(n) : # create an array of size (n + 1) arr =[0] * (n + 1); arr[1] = 1 arr[2] = 2 for i in range(3, n + 1) : # If i is odd if (i % 2 != 0) : arr[i] = arr[i // 2] * 10 + 1 else : arr[i] = arr[(i // 2) - 1] * 10 + 2 return arr[n] # Driver code n = 5 print(printNthElement(n))
C#
// C# program to find
// the N-th term in the series
// 1, 2, 11, 12, 21...
using System;
class GFG
{
// Function to find n-th
// number in series
static int printNthElement(int n)
{
// create an array of size (n+1)
int []arr = new int[n + 1];
arr[1] = 1;
arr[2] = 2;
for (int i = 3; i <= n; i++)
{
// If i is odd
if (i % 2 != 0)
arr[i] = arr[i / 2] * 10 + 1;
else
arr[i] = arr[(i / 2) - 1] * 10 + 2;
}
return arr[n];
}
// Driver Code
public static void Main()
{
int n = 5;
Console.WriteLine(printNthElement(n));
}
}
// This code is contributed
// by inder_verma
PHP
<?php
// PHP program to find
// the N-th term in the series
// 1, 2, 11, 12, 21...
// Function to find N-th
// number in series
function printNthElement($N)
{
// create an array of size (N+1)
$arr = array($N + 1);
$arr[1] = 1;
$arr[2] = 2;
for ( $i = 3; $i <= $N; $i++)
{
// If i is odd
if ($i % 2 != 0)
{
$arr[$i] = $arr[$i / 2] *
10 + 1;
}
else
{
$arr[$i] = $arr[($i / 2) - 1] *
10 + 2;
}
}
return $arr[$N];
}
// Driver code
$N = 5;
// Get Nth term
echo printNthElement($N);
// This code is contributed
// by Mahadev99
?>
Javascript
<script>
// Javascript program to find
// the N-th term in the series
// 1, 2, 11, 12, 21...
// Function to find n-th number in series
function printNthElement(n)
{
// create an array of size (n+1)
let arr = new Array(n + 1);
arr[1] = 1;
arr[2] = 2;
for (let i = 3; i <= n; i++) {
// If i is odd
if (i % 2 != 0)
arr[i] = arr[parseInt(i / 2, 10)] * 10 + 1;
else
arr[i] = arr[parseInt(i / 2, 10) - 1] * 10 + 2;
}
return arr[n];
}
let n = 5;
document.write(printNthElement(n));
// This code is contributed by vaibhavrabadiya117.
</script>
Producción:
21
Publicación traducida automáticamente
Artículo escrito por pawan_asipu y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA