Given a number N. The task is to find first N Iccanobif Numbers.
Iccanobif Numbers are similar to Fibonacci Numbers. The K-th Iccanobif number can be obtained by addition of previous two numbers after reversing their digits.
The first few Iccanobif Numbers are:
0, 1, 1, 2, 3, 5, 8, 13, 39, 124, 514, 836, …..
Examples:
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.
Approach: The idea is to take the first two Iccanobif Numbers as first = 0 and second = 1. Now iterate using a demoPointer N-2 times and every time find reverse of the previous two numbers using the approach discussed in: Reversing digits of a number. Find the sum of the two reversed numbers and update the variables first and second accordingly.
Below is the implementation of above approach:
C++
// C++ program to find first// N Icanobif numbers#include <bits/stdc++.h>using namespace std;// Iterative function to// reverse digits of numint reverseDigits(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 Numbersvoid 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 = reverseDigits(first); int y = reverseDigits(second); cout << x + y << " "; int temp = second; second = x + y; first = temp; } }}// Driver Codeint main(){ int N = 12; icanobifNumbers(N); return 0;} |
Java
// Java program to find first// N Icanobif numberspublic class GFG{ // Iterative function to // reverse digits of num static int reverseDigits(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 = reverseDigits(first); int y = reverseDigits(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 numdef 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 Numbersdef 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 codeN = 12icanobifNumbers(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 reverseDigits(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 = reverseDigits(first); int y = reverseDigits(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 reverseDigits($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 $x = reverseDigits($first); $y = reverseDigits($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 reverseDigits(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 = reverseDigits(first); let y = reverseDigits(second); document.write(x + y + " "); let temp = second; second = x + y; first = temp; } } } let N = 12; icanobifNumbers(N); </script> |
Output:
0 1 1 2 3 5 8 13 39 124 514 836
Time Complexity: O(NlogN)
Auxiliary Space: O(1)
Note: For the larger value of N, use numbers as strings.
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