Dados dos enteros N y X. Haz un número de tal manera que el número contenga el primer y el último dígito que aparecen en 
.
Ejemplos:
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.
Recomendado: Resuelva primero en » PRÁCTICA «, antes de pasar a la solución.
Enfoque:
1. Calcular todos los poderes de N de 1 a X uno por uno.
2. Almacene la salida en la array power[].
3. Almacene potencia[0], es decir, último_dígito y potencia[tamaño_potencia – 1], es decir, dígito_unidad desde la array potencia[] hasta la array resultado[].
4. Imprima la array resultado[].
A continuación se muestra la implementación del enfoque anterior:
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 power
void 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 Code
int 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 0 power = [] # Find next power by multiplying # N with current power def 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 Code N = 19 X = 4 printPowerNumber(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 power
function 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 power
function 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 Code
var N = 19 , X = 4;
printPowerNumber(X, N);
</script>
Producción :
19316911