Given two integers N and X. Make a number in such a way that the number contains the first and last digit occurring in 
.
Examples :
Input : N = 10, X = 5 Output : 1010101010 Explanation : 10^1 = 10 10^2 = 100 10^3 = 1000 10^4 = 10000 10^5 = 100000 Take First and Last Digit of each Power to get required number. Input : N = 19, X = 4 Output : 19316911 Explanation : 19^1 = 19 19^2 = 361 19^3 = 6859 19^4 = 130321 Take First and Last Digit of each Power to get required number.
Recommended: Please solve it on “PRACTICE“first, before moving on to the solution.
Approach :
- Calculate all Powers of N from 1 to X one by one.
- Store the Output in power[] array.
- Store power[0] i.e, last_digit and power[power_size – 1] i.e, unit_digit from power[] array to result[] array.
- Print the result[] Array.
Below is the implementation of the above approach :
C++
// C++ program to find number formed by// corner digits of powers.#include <bits/stdc++.h>using namespace std;// Find next power by multiplying N with// current powervoid nextPower(int N, vector<int> &power){ int carry = 0; for (int i=0 ; i < power.size(); i++) { int prod = (power[i] * N) + carry ; // Store digits of Power one by one. power[i] = prod % 10 ; // Calculate carry. carry = prod / 10 ; } while (carry) { // Store carry in Power array. power.push_back(carry % 10); carry = carry / 10 ; }}// Prints number formed by corner digits of// powers of N.void printPowerNumber(int X, int N){ // Storing N raised to power 0 vector<int> power; power.push_back(1); // Initializing empty result vector<int> res; // One by one compute next powers and // add their corner digits. for (int i=1; i<=X; i++) { // Call Function that store power // in Power array. nextPower(N, power) ; // Store unit and last digits of // power in res. res.push_back(power.back()); res.push_back(power.front()); } for (int i=0 ; i < res.size(); i++) cout << res[i] ;}// Driver Codeint main(){ int N = 19 , X = 4; printPowerNumber(X, N); return 0 ;} |
Java
// Java program to find number formed by// corner digits of powers.import java.io.*;import java.util.*; public class GFG { static List<Integer> power = new ArrayList<Integer>(); // Find next power by multiplying N // with current power static void nextPower(Integer N) { Integer carry = 0; for (int i = 0; i < power.size(); i++) { Integer prod = (power.get(i) * N) + carry ; // Store digits of Power one by one. power.set(i,prod % 10); // Calculate carry. carry = prod / 10 ; } while (carry >= 1) { // Store carry in Power array. power.add(carry % 10); carry = carry / 10 ; } } // Prints number formed by corner digits of // powers of N. static void printPowerNumber(int X, int N) { // Storing N raised to power 0 power.add(1); // Initializing empty result List<Integer> res = new ArrayList<Integer>(); // One by one compute next powers and // add their corner digits. for (int i = 1; i <= X; i++) { // Call Function that store power // in Power array. nextPower(N) ; // Store unit and last digits of // power in res. res.add(power.get(power.size() - 1)); res.add(power.get(0)); } for (int i = 0 ; i < res.size(); i++) System.out.print(res.get(i)) ; } // Driver Code public static void main(String args[]) { Integer N = 19 , X = 4; printPowerNumber(X, N); }} // This code is contributed by Manish Shaw// (manishshaw1) |
Python3
# Python3 program to find # number formed by # corner digits of powers. # Storing N raised to power 0power = []# Find next power by multiplying # N with current powerdef nextPower(N) : global power carry = 0 for i in range(0, len(power)) : prod = (power[i] * N) + carry # Store digits of # Power one by one. power[i] = prod % 10 # Calculate carry. carry = (int)(prod / 10) while (carry) : # Store carry in Power array. power.append(carry % 10) carry = (int)(carry / 10) # Prints number formed by corner # digits of powers of N.def printPowerNumber(X, N) : global power power.append(1) # Initializing empty result res = [] # One by one compute next powers # and add their corner digits. for i in range(1, X+1) : # Call Function that store # power in Power array. nextPower(N) # Store unit and last # digits of power in res. res.append(power[-1]) res.append(power[0]) for i in range(0, len(res)) : print (res[i], end="") # Driver CodeN = 19X = 4printPowerNumber(X, N) # This code is contributed by # Manish Shaw(manishshaw1) |
C#
// C# program to find number formed by// corner digits of powers.using System;using System.Collections.Generic;using System.Linq;using System.Collections;class GFG { // Find next power by multiplying N // with current power static void nextPower(int N, ref List<int> power) { int carry = 0; for (int i = 0; i < power.Count; i++) { int prod = (power[i] * N) + carry ; // Store digits of Power one by one. power[i] = prod % 10 ; // Calculate carry. carry = prod / 10 ; } while (carry >= 1) { // Store carry in Power array. power.Add(carry % 10); carry = carry / 10 ; } } // Prints number formed by corner digits of // powers of N. static void printPowerNumber(int X, int N) { // Storing N raised to power 0 List<int> power = new List<int>(); power.Add(1); // Initializing empty result List<int> res = new List<int>(); // One by one compute next powers and // add their corner digits. for (int i = 1; i <= X; i++) { // Call Function that store power // in Power array. nextPower(N, ref power) ; // Store unit and last digits of // power in res. res.Add(power.Last()); res.Add(power.First()); } for (int i = 0 ; i < res.Count; i++) Console.Write(res[i]) ; } // Driver Code public static void Main() { int N = 19 , X = 4; printPowerNumber(X, N); }}// This code is contributed by Manish Shaw// (manishshaw1) |
PHP
<?php// PHP program to find // number formed by // corner digits of powers.// Find next power by multiplying // N with current powerfunction nextPower($N, &$power){ $carry = 0; for ($i = 0 ; $i < count($power); $i++) { $prod = ($power[$i] * $N) + $carry ; // Store digits of // Power one by one. $power[$i] = $prod % 10 ; // Calculate carry. $carry = (int)($prod / 10) ; } while ($carry) { // Store carry in Power array. array_push($power, $carry % 10); $carry = (int)($carry / 10) ; }}// Prints number formed by corner // digits of powers of N.function printPowerNumber($X, $N){ // Storing N raised to power 0 $power = array(); array_push($power, 1); // Initializing empty result $res = array(); // One by one compute next powers // and add their corner digits. for ($i = 1; $i <= $X; $i++) { // Call Function that store // power in Power array. nextPower($N, $power) ; // Store unit and last // digits of power in res. array_push($res, $power[count($power) - 1]); array_push($res, $power[0]); } for ($i = 0 ; $i < count($res); $i++) echo ($res[$i]) ;} // Driver Code$N = 19; $X = 4;printPowerNumber($X, $N);// This code is contributed by // Manish Shaw(manishshaw1)?> |
Javascript
<script>// JavaScript program to find number formed by// corner digits of powers.// Find next power by multiplying N with// current powerfunction nextPower(N, power){ var carry = 0; for (var i=0 ; i < power.length; i++) { var prod = (power[i] * N) + carry ; // Store digits of Power one by one. power[i] = prod % 10 ; // Calculate carry. carry = parseInt(prod / 10); } while (carry>=1) { // Store carry in Power array. power.push(carry % 10); carry = parseInt(carry / 10); } return power;}// Prints number formed by corner digits of// powers of N.function printPowerNumber( X, N){ // Storing N raised to power 0 var power = []; power.push(1); // Initializing empty result var res = []; // One by one compute next powers and // add their corner digits. for (var i=1; i<=X; i++) { // Call Function that store power // in Power array. power = nextPower(N, power) ; // Store unit and last digits of // power in res. res.push(power[power.length-1]); res.push(power[0]); } for (var i=0 ; i < res.length; i++) document.write( res[i] );}// Driver Codevar N = 19 , X = 4;printPowerNumber(X, N);</script> |
Output
19316911
Time Complexity: O(XlogX)
Auxiliary Space: O(N)
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