In number theory, a Leyland number is a number of the form xy + yx, where x and y are integers greater than 1 and 1 <y <= x.
Given a positive integer N. The task is to print first N Leyland number in ascending order. The first few Leyland numbers are 8, 17, 32, 54, 57, 100, …
Examples:
Input : N = 1 Output : 8 22 + 22 = 4 + 4 = 8. Input : N = 6 Output : 100
The idea to run two loop, one for x and other for y. The outer loop start with 2 to n and for each iteration of outer loop, run inner loop start from 2 t x. And store xy + yx in an array. After calculating all the value sort them and print first n numbers.
Below is the implementation of this approach:
C++
// CPP program to print first N Leyland Numbers.#include <bits/stdc++.h>#define MAX 100using namespace std;// Print first n Leyland Number.void leyland(int n){ vector<int> ans; // Outer loop for x from 2 to n. for (int x = 2; x <= n; x++) { // Inner loop for y from 2 to x. for (int y = 2; y <= x; y++) { // Calculating x^y + y^x int temp = pow(x, y) + pow(y, x); ans.push_back(temp); } } // Sorting the all Leyland Number. sort(ans.begin(), ans.end()); // Printing first n Leyland number. for (int i = 0; i < n; i++) cout << ans[i] << " ";}// Driven Programint main(){ int n = 6; leyland(n); return 0;} |
Java
// Java program to print first N // Leyland Numbers.import java.util.*;import java.lang.*;public class GFG{ private static final int MAX = 0; // Print first n Leyland Number. public static void leyland(int n) { List<Integer> ans = new ArrayList<Integer>(); // Outer loop for x from 2 to n. for (int x = 2; x <= n; x++) { // Inner loop for y from 2 to x. for (int y = 2; y <= x; y++) { // Calculating x^y + y^x int temp = (int)Math.pow(x, y) + (int)Math.pow(y, x); ans.add(temp); } } // Sorting the all Leyland Number. Collections.sort(ans); // Printing first n Leyland number. for (int i = 0; i < n; i++) System.out.print(ans.get(i) + " "); } // Driven Program public static void main(String args[]) { int n = 6; leyland(n); }}// This code is contributed by Sachin Bisht |
Python3
# Python3 program to print first N # Leyland Numbers.import math# Print first n Leyland Number.def leyland(n): ans = [] x = 2 y = 2 # Outer loop for x from 2 to n. while x <= n : # Inner loop for y from 2 to x. y = 2 while y <= x : # Calculating x^y + y^x temp = pow(x, y) + pow(y, x) ans.append(temp); y = y + 1 x = x + 1 # Sorting the all Leyland Number. ans.sort(); i = 0 # Printing first n Leyland number. while i < n : print(ans[i], end = " ") i = i + 1# Driver Coden = 6leyland(n)# This code is contributed by rishabh_jain |
C#
// C# program to print // first N Leyland Numbers.using System;using System.Collections;class GFG{ // Print first n // Leyland Number. public static void leyland(int n) { ArrayList ans = new ArrayList(); // Outer loop for x // from 2 to n. for (int x = 2; x <= n; x++) { // Inner loop for // y from 2 to x. for (int y = 2; y <= x; y++) { // Calculating x^y + y^x int temp = (int)Math.Pow(x, y) + (int)Math.Pow(y, x); ans.Add(temp); } } // Sorting the all // Leyland Number. ans.Sort(); // Printing first // n Leyland number. for (int i = 0 ; i < n; i++) { Console.Write(ans[i] + " "); } } // Driver Code public static void Main() { int n = 6; leyland(n); }}// This code is contributed by Sam007 |
PHP
<?php// PHP program to print // first N Leyland Numbers.$MAX = 100;// Print first n // Leyland Number.function leyland($n){ $ans; $index = 0; // Outer loop for // x from 2 to n. for ($x = 2; $x <= $n; $x++) { // Inner loop for // y from 2 to x. for ($y = 2; $y <= $x; $y++) { // Calculating x^y + y^x $temp = pow($x, $y) + pow($y, $x); $ans[$index] = $temp; $index++; } } // Sorting the all // Leyland Number. sort($ans); // Printing first // n Leyland number. for ($i = 0; $i < $n; $i++) echo $ans[$i]. " ";}// Driver Code$n = 6;leyland($n); // This code is contributed// by mits?> |
Javascript
<script>// Javascript program to print// first N Leyland Numbers.let MAX = 100;// Print first n// Leyland Number.function leyland(n){ let ans = []; let index = 0; // Outer loop for // x from 2 to n. for (let x = 2; x <= n; x++) { // Inner loop for // y from 2 to x. for (let y = 2; y <= x; y++) { // Calculating x^y + y^x let temp = Math.pow(x, y) + Math.pow(y, x); ans[index] = temp; index++; } } // Sorting the all // Leyland Number. console.log(ans) ans = ans.sort((a, b)=>a-b); console.log(ans) // Printing first // n Leyland number. for (let i = 0; i < n; i++) document.write(ans[i] + " ");}// Driver Codelet n = 6;leyland(n); // This code is contributed// by _saurabh_jaiswal</script> |
Output:
8 17 32 54 57 100
Time complexity: O(n*nlogn+nlogn)
Auxiliary space: O(n)
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