Dado un número N. La tarea es encontrar primero N Números Iccanobif .
Los números de Iccanobif son similares a los números de Fibonacci. El número K-th Iccanobif se puede obtener sumando los dos números anteriores después de invertir sus dígitos.
Los primeros números de Iccanobif son:
0, 1, 1, 2, 3, 5, 8, 13, 39, 124, 514, 836, …..
Ejemplos :
Input : N = 5 Output : 0 1 1 2 3 Input : N = 9 Output : 0 1 1 2 3 5 8 13 39 Explanation: Upto 8th term, adding previous two terms is required, as there is an only single digit. For 9th term, adding 31(reversing 8th term) and 8 will give 39.
Enfoque: La idea es tomar los primeros dos números de Iccanobif como primero = 0 y segundo = 1 . Ahora itere usando un demoPointer N-2 veces y cada vez encuentre el reverso de los dos números anteriores usando el enfoque discutido en: Inversión de dígitos de un número . Encuentre la suma de los dos números invertidos y actualice las variables primero y segundo según corresponda.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find first
// N Icanobif numbers
#include <bits/stdc++.h>
using namespace std;
// Iterative function to
// reverse digits of num
int reversDigits(int num)
{
int rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
// Function to print first
// N Icanobif Numbers
void icanobifNumbers(int N)
{
// Initialize first, second numbers
int first = 0, second = 1;
if (N == 1)
cout << first;
else if (N == 2)
cout << first << " " << second;
else {
// Print first two numbers
cout << first << " " << second << " ";
for (int i = 3; i <= N; i++) {
// Reversing digit of previous
// two terms and adding them
int x = reversDigits(first);
int y = reversDigits(second);
cout << x + y << " ";
int temp = second;
second = x + y;
first = temp;
}
}
}
// Driver Code
int main()
{
int N = 12;
icanobifNumbers(N);
return 0;
}
Java
// Java program to find first
// N Icanobif numbers
public class GFG{
// Iterative function to
// reverse digits of num
static int reversDigits(int num)
{
int rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
// Function to print first
// N Icanobif Numbers
static void icanobifNumbers(int N)
{
// Initialize first, second numbers
int first = 0, second = 1;
if (N == 1)
System.out.print(first);
else if (N == 2)
System.out.print(first + " " + second);
else {
// Print first two numbers
System.out.print(first + " " + second + " ");
for (int i = 3; i <= N; i++) {
// Reversing digit of previous
// two terms and adding them
int x = reversDigits(first);
int y = reversDigits(second);
System.out.print(x + y + " ");
int temp = second;
second = x + y;
first = temp;
}
}
}
// Driver Code
public static void main(String []args){
int N = 12;
icanobifNumbers(N);
}
// This code is contributed by ANKITRAI1
}
Python3
# Python 3 program to find first # N Icanobif numbers # Iterative function to # reverse digits of num def reversedigit(num): rev_num = 0 while num > 0: rev_num = rev_num * 10 + num % 10 num = num // 10 return rev_num # Function to print first # N Icanobif Numbers def icanobifNumbers(N): # Initialize first, second numbers first = 0 second = 1 if N == 1: print(first) elif N == 2: print(first, second) else: # Print first two numbers print(first, second, end = " ") for i in range(3, N + 1): # Reversing digit of previous # two terms and adding them x = reversedigit(first) y = reversedigit(second) print(x + y, end = " ") temp = second second = x + y first = temp # Driver code N = 12 icanobifNumbers(N) # This code is contributed by Shrikant13
C#
// C# program to find first
// N Icanobif numbers
using System;
public class GFG{
// Iterative function to
// reverse digits of num
static int reversDigits(int num)
{
int rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
// Function to print first
// N Icanobif Numbers
static void icanobifNumbers(int N)
{
// Initialize first, second numbers
int first = 0, second = 1;
if (N == 1)
Console.Write(first);
else if (N == 2)
Console.Write(first + " " + second);
else {
// Print first two numbers
Console.Write(first + " " + second + " ");
for (int i = 3; i <= N; i++) {
// Reversing digit of previous
// two terms and adding them
int x = reversDigits(first);
int y = reversDigits(second);
Console.Write(x + y + " ");
int temp = second;
second = x + y;
first = temp;
}
}
}
// Driver Code
public static void Main(){
int N = 12;
icanobifNumbers(N);
}
}
PHP
<?php
// PHP program to find first N
// Icanobif numbers
// Iterative function to reverse
// digits of num
function reversDigits($num)
{
$rev_num = 0;
while ($num > 0)
{
$rev_num = ($rev_num * 10) +
($num % 10);
$num = (int)( $num / 10);
}
return $rev_num;
}
// Function to print first
// N Icanobif Numbers
function icanobifNumbers($N)
{
// Initialize first, second numbers
$first = 0;
$second = 1;
if ($N == 1)
echo $first;
else if ($N == 2)
echo $first , " ", $second;
else
{
// Print first two numbers
echo $first, " " , $second, " ";
for ($i = 3; $i <= $N; $i++)
{
// Reversing digit of previous
// two terms and adding them
$x = reversDigits($first);
$y = reversDigits($second);
echo ($x + $y), " ";
$temp = $second;
$second = $x + $y;
$first = $temp;
}
}
}
// Driver Code
$N = 12;
icanobifNumbers($N);
// This code is contributed by Tushil.
?>
Javascript
<script>
// Javascript program to find first
// N Icanobif numbers
// Iterative function to
// reverse digits of num
function reversDigits(num)
{
let rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10 + num % 10;
num = parseInt(num / 10, 10);
}
return rev_num;
}
// Function to print first
// N Icanobif Numbers
function icanobifNumbers(N)
{
// Initialize first, second numbers
let first = 0, second = 1;
if (N == 1)
document.write(first);
else if (N == 2)
document.write(first + " " + second);
else {
// Print first two numbers
document.write(first + " " + second + " ");
for (let i = 3; i <= N; i++) {
// Reversing digit of previous
// two terms and adding them
let x = reversDigits(first);
let y = reversDigits(second);
document.write(x + y + " ");
let temp = second;
second = x + y;
first = temp;
}
}
}
let N = 12;
icanobifNumbers(N);
</script>
Producción:
0 1 1 2 3 5 8 13 39 124 514 836
Complejidad de tiempo: O (NlogN)
Espacio Auxiliar: O(1)
Nota : para el valor mayor de N, use números como strings.