Dado un número, comprueba si es un número misterioso o no. Un número misterioso es un número que se puede expresar como la suma de dos números y esos dos números deben ser el reverso uno del otro.
Ejemplos:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Finds reverse of given num x.
int reverseNum(int x)
{
string s = to_string(x);
reverse(s.begin(), s.end());
stringstream ss(s);
int rev = 0;
ss >> rev;
return rev;
}
bool isMysteryNumber(int n)
{
for (int i=1; i <= n/2; i++)
{
// if found print the pair, return
int j = reverseNum(i);
if (i + j == n)
{
cout << i << " " << j;
return true;
}
}
cout << "Not a Mystery Number";
return false;
}
int main()
{
int n = 121;
isMysteryNumber(n);
return 0;
}
Java
// Java implementation of above approach
class GFG
{
// Finds reverse of given num x.
static int reverseNum(int x)
{
String s = Integer.toString(x);
String str="";
for(int i=s.length()-1;i>=0;i--)
{
str=str+s.charAt(i);
}
int rev=Integer.parseInt(str);
return rev;
}
static boolean isMysteryNumber(int n)
{
for (int i=1; i <= n/2; i++)
{
// if found print the pair, return
int j = reverseNum(i);
if (i + j == n)
{
System.out.println( i + " " + j);
return true;
}
}
System.out.println("Not a Mystery Number");
return false;
}
public static void main(String []args)
{
int n = 121;
isMysteryNumber(n);
}
}
// This code is contributed by ihritik
Python3
# Python3 implementation of above approach
# Finds reverse of given num x.
def reverseNum(x):
s = str(x)
s = s[::-1]
return int(s)
def isMysteryNumber(n):
for i in range(1, n // 2 + 1):
# if found print the pair, return
j = reverseNum(i)
if i + j == n:
print(i, j)
return True
print("Not a Mystery Number")
return False
# Driver Code
n = 121
isMysteryNumber(n)
# This code is contributed by
# Mohit Kumar 29 (IIIT gwalior)
C#
// C# implementation of above approach
using System;
class GFG
{
// Finds reverse of given num x.
static int reverseNum(int x)
{
string s = x.ToString();
string str="";
for(int i=s.Length-1;i>=0;i--)
{
str=str+s[i];
}
int rev=Int32.Parse(str);
return rev;
}
static bool isMysteryNumber(int n)
{
for (int i=1; i <= n/2; i++)
{
// if found print the pair, return
int j = reverseNum(i);
if (i + j == n)
{
Console.WriteLine( i + " " + j);
return true;
}
}
Console.WriteLine("Not a Mystery Number");
return false;
}
public static void Main()
{
int n = 121;
isMysteryNumber(n);
}
}
// This code is contributed by ihritik
PHP
<?php
// PHP implementation of above approach
// Finds reverse of given num x.
function reverseNum($x)
{
$s = (string)$x;
$s = strrev($s);
$rev = (int)$s;
return $rev;
}
function isMysteryNumber($n)
{
for ($i=1; $i <= $n/2; $i++)
{
// if found print the pair, return
$j = reverseNum($i);
if ($i + $j == $n)
{
echo $i . " ".$j;
return true;
}
}
echo "Not a Mystery Number";
return false;
}
$n = 121;
isMysteryNumber($n);
return 0;
// This code is contribute by Ita_c.
?>
Javascript
<script>
// Javascript implementation of above approach
// Finds reverse of given num x.
function reverseNum(x)
{
let s = x.toString();
let str="";
for(let i=s.length-1;i>=0;i--)
{
str=str+s[i];
}
let rev=parseInt(str);
return rev;
}
function isMysteryNumber(n)
{
for (let i=1; i <= Math.floor(n/2); i++)
{
// if found print the pair, return
let j = reverseNum(i);
if (i + j == n)
{
document.write( i + " " + j+"<br>");
return true;
}
}
document.write("Not a Mystery Number<br>");
return false;
}
let n = 121;
isMysteryNumber(n);
// This code is contributed by avanitrachhadiya2155
</script>
Publicación traducida automáticamente
Artículo escrito por Kushdeep_Mittal y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA