Dado un número x, encuentre su número selfie palindrómico de acuerdo con la regla multiplicativa selfie. Si tal número no existe, imprima «No existe tal número».
Un número selfie palindrómico satisface la regla multiplicativa selfie tal que existe otro número y con x * dígitos_inversos_de(x) = y * dígitos_inversos_de(y) , con la condición de que el número y se obtiene ordenando los dígitos en x, es decir x e y deben tener los mismos dígitos con diferente orden.
Ejemplos:
Input : 1224 Output : 2142 Explanation : Because, 1224 X 4221 = 2142 X 2412 And all digits of 2142 are formed by a different permutation of the digits in 1224 (Note: The valid output is either be 2142 or 2412) Input : 13452 Output : 14532 Explanation : Because, 13452 X 25431 = 14532 X 23541 And all digits of 14532 are formed by a different permutation of the digits in 13452 Input : 12345 Output : No such number exists Explanation : Because, with no combination of digits 1, 2, 3, 4, 5 could we get a number that satisfies 12345 X 54321 = number X reverse_of_its_digits
Acercarse :
- La idea es descomponer el número y obtener todas las permutaciones de los dígitos del número.
- Luego, se saca el número y su palíndromo del conjunto de permutaciones obtenidas, que formarán el LHS de nuestra igualdad.
- Para verificar RHS, ahora iteramos sobre todas las demás permutaciones que igualan
LHS = current_number X palindrome(current_number)
- Tan pronto como obtenemos una coincidencia, salimos del ciclo con un mensaje afirmativo, de lo contrario imprimimos «No hay tal número disponible».
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find palindromic selfie numbers
#include <bits/stdc++.h>
using namespace std;
class palindrome_selfie {
public:
// To store all permutations of digits in the number
set<int> all_permutes;
int number; // input number
palindrome_selfie(int num) { number = num; }
// Function to reverse the digits of a number
int palindrome(int num)
{
int reversednum = 0;
int d;
while (num > 0) {
d = num % 10; // Extract last digit
// Append it at the beg
reversednum = reversednum * 10 + d;
num = num / 10; // Reduce number until 0
}
return reversednum;
}
// Function that swaps the digits i and j in the num
int swap(int num, int i, int j)
{
char temp;
string charArray = to_string(num);
// Swap the ith and jth character
temp = charArray[i];
charArray[i] = charArray[j];
charArray[j] = temp;
// Convert back to int and return
return stoi(charArray);
}
// Function to get all possible permutations
// of the digits in num
void permute(int num, int l, int r)
{
// Adds the new permutation obtained in the set
if (l == r)
all_permutes.insert(num);
else {
for (int i = l; i <= r; i++) {
// Swap digits to get a different ordering
num = swap(num, l, i);
// Recurse to next pair of digits
permute(num, l + 1, r);
num = swap(num, l, i); // Swap back
}
}
}
// Function to check palindromic selfie
void palin_selfie()
{
// Length of the number required for
// calculating all permutations of the digits
int l = to_string(number).length() - 1;
permute(number, 0, l); // Calculate all permutations
/* Remove the number and its palindrome from
the obtained set as this is the LHS of
multiplicative equality */
auto n1 = all_permutes.find(palindrome(number));
auto n2 = all_permutes.find(number);
all_permutes.erase(n1);
all_permutes.erase(n2);
bool flag = false; // Denotes the status result
// Iterate over all other numbers
for (auto it : all_permutes) {
int number2 = it;
// Check for equality x*palin(x) = y*palin(y)
if (number * palindrome(number)
== number2 * palindrome(number2)) {
cout << "Palindrome multiplicative"
<< "selfie of " << number
<< " is : " << number2 << endl;
flag = true; // Answer found
break;
}
}
// If no such number found
if (flag == false) {
cout << "Given number has no palindrome selfie."
<< endl;
}
}
};
// Driver Function
int main()
{
// First example, input = 145572
palindrome_selfie example1(145572);
example1.palin_selfie();
// Second example, input = 19362
palindrome_selfie example2(19362);
example2.palin_selfie();
// Third example, input = 4669
palindrome_selfie example3(4669);
example3.palin_selfie();
return 0;
}
// This code is contributed by phasing17
Java
// Java program to find palindromic selfie numbers
import java.util.*;
public class palindrome_selfie {
// To store all permutations of digits in the number
Set<Integer> all_permutes = new HashSet<Integer>();
int number; // input number
public palindrome_selfie(int num)
{
number = num;
}
// Function to reverse the digits of a number
public int palindrome(int num)
{
int reversednum = 0;
int d;
while (num > 0) {
d = num % 10; // Extract last digit
// Append it at the beg
reversednum = reversednum * 10 + d;
num = num / 10; // Reduce number until 0
}
return reversednum;
}
// Function to check palindromic selfie
public void palin_selfie()
{
// Length of the number required for
// calculating all permutations of the digits
int l = String.valueOf(number).length() - 1;
this.permute(number, 0, l); // Calculate all permutations
/* Remove the number and its palindrome from
the obtained set as this is the LHS of
multiplicative equality */
all_permutes.remove(palindrome(number));
all_permutes.remove(number);
boolean flag = false; // Denotes the status result
// Iterate over all other numbers
Iterator it = all_permutes.iterator();
while (it.hasNext()) {
int number2 = (int)it.next();
// Check for equality x*palin(x) = y*palin(y)
if (number * palindrome(number) ==
number2 * palindrome(number2)) {
System.out.println("Palindrome multiplicative" +
"selfie of "+ number + " is : "
+ number2);
flag = true; // Answer found
break;
}
}
// If no such number found
if (flag == false) {
System.out.println("Given number has no palindrome selfie.");
}
}
// Function to get all possible permutations
// of the digits in num
public void permute(int num, int l, int r)
{
// Adds the new permutation obtained in the set
if (l == r)
all_permutes.add(num);
else {
for (int i = l; i <= r; i++) {
// Swap digits to get a different ordering
num = swap(num, l, i);
// Recurse to next pair of digits
permute(num, l + 1, r);
num = swap(num, l, i); // Swap back
}
}
}
// Function that swaps the digits i and j in the num
public int swap(int num, int i, int j)
{
char temp;
// Convert int to char array
char[] charArray = String.valueOf(num).toCharArray();
// Swap the ith and jth character
temp = charArray[i];
charArray[i] = charArray[j];
charArray[j] = temp;
// Convert back to int and return
return Integer.valueOf(String.valueOf(charArray));
}
// Driver Function
public static void main(String args[])
{
// First example, input = 145572
palindrome_selfie example1 = new palindrome_selfie(145572);
example1.palin_selfie();
// Second example, input = 19362
palindrome_selfie example2 = new palindrome_selfie(19362);
example2.palin_selfie();
// Third example, input = 4669
palindrome_selfie example3 = new palindrome_selfie(4669);
example3.palin_selfie();
}
}
C#
// C# program to find palindromic selfie numbers
using System;
using System.Collections.Generic;
public class palindrome_selfie
{
// To store all permutations of digits in the number
HashSet<int> all_permutes = new HashSet<int>();
int number; // input number
public palindrome_selfie(int num)
{
number = num;
}
// Function to reverse the digits of a number
public int palindrome(int num)
{
int reversednum = 0;
int d;
while (num > 0)
{
d = num % 10; // Extract last digit
// Append it at the beg
reversednum = reversednum * 10 + d;
num = num / 10; // Reduce number until 0
}
return reversednum;
}
// Function to check palindromic selfie
public void palin_selfie()
{
// Length of the number required for
// calculating all permutations of the digits
int l = String.Join("",number).Length - 1;
this.permute(number, 0, l); // Calculate all permutations
/* Remove the number and its palindrome from
the obtained set as this is the LHS of
multiplicative equality */
all_permutes.Remove(palindrome(number));
all_permutes.Remove(number);
bool flag = false; // Denotes the status result
// Iterate over all other numbers
foreach (var number2 in all_permutes)
{
// Check for equality x*palin(x) = y*palin(y)
if (number * palindrome(number) ==
number2 * palindrome(number2))
{
Console.WriteLine("Palindrome multiplicative" +
"selfie of "+ number + " is : "
+ number2);
flag = true; // Answer found
break;
}
}
// If no such number found
if (flag == false)
{
Console.WriteLine("Given number has "+
"no palindrome selfie.");
}
}
// Function to get all possible
// permutations of the digits in num
public void permute(int num, int l, int r)
{
// Adds the new permutation obtained in the set
if (l == r)
all_permutes.Add(num);
else
{
for (int i = l; i <= r; i++)
{
// Swap digits to get a different ordering
num = swap(num, l, i);
// Recurse to next pair of digits
permute(num, l + 1, r);
num = swap(num, l, i); // Swap back
}
}
}
// Function that swaps the
// digits i and j in the num
public int swap(int num, int i, int j)
{
char temp;
// Convert int to char array
char[] charArray = String.Join("",num).ToCharArray();
// Swap the ith and jth character
temp = charArray[i];
charArray[i] = charArray[j];
charArray[j] = temp;
// Convert back to int and return
return int.Parse(String.Join("",charArray));
}
// Driver code
public static void Main(String []args)
{
// First example, input = 145572
palindrome_selfie example1 = new palindrome_selfie(145572);
example1.palin_selfie();
// Second example, input = 19362
palindrome_selfie example2 = new palindrome_selfie(19362);
example2.palin_selfie();
// Third example, input = 4669
palindrome_selfie example3 = new palindrome_selfie(4669);
example3.palin_selfie();
}
}
// This code contributed by Rajput-Ji
Producción :
Palindrome multiplicative selfie of 145572 is : 157452 Given number has no palindrome selfie. Palindrome multiplicative selfie of 4669 is : 6496
Publicación traducida automáticamente
Artículo escrito por Sudarshan Khasnis y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA