Demlo number (Square of 11…1)

Given a number of the form 11..1 such that number of digits in it is smaller than 10, find square of the number.

Examples:

Input : 111111
Output : 12345654321

Input : 1111
Output : 1234321

Squares of 11…1 (where length is smaller than 10) are called Demlo numbers. A demlo number is number which consist of 1, 2, 3….n, n-1, n-2, …..1.
Demlo numbers are:

1 = 1
11 * 11 = 121
111 * 111 = 12321
1111 * 1111 = 1234321

To find demlo number, first we find append n numbers which are increasing and then append numbers which are decreasing.

C++

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// CPP program to print DemloNumber
#include <bits/stdc++.h>
using namespace std;
  
// To return demlo number. This function assumes
// that the length of str is smaller than 10.
string printDemlo(string str)
{
    int len = str.length();
    string res = "";
  
    // Add numbers to res upto size
    // of str and then add number
    // reverse to it
    for (int i = 1; i <= len; i++)
        res += char(i + '0');
  
    for (int i = len - 1; i >= 1; i--)
        res += char(i + '0');
  
    return res;
}
  
// Driver program to test printDemlo()
int main()
{
    string str = "111111";
    cout << printDemlo(str);
    return 0;
}

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Java

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// Java program to print DemloNumber
public class Main {
  
    // To return demlo number. This function assumes
    // that the length of str is smaller than 10.
    static String printDemlo(String str)
    {
        int len = str.length();
        String res = "";
  
        // Add numbers to res upto size
        // of str and then add number
        // reverse to it
        for (int i = 1; i <= len; i++)
            res += Integer.toString(i);
  
        for (int i = len - 1; i >= 1; i--)
            res += Integer.toString(i);
  
        return res;
    }
  
    // Driver program to test printDemlo()
    public static void main(String[] args)
    {
        String str = "111111";
        System.out.println(printDemlo(str));
    }
}

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Python

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# Python program to print Demlo Number
  
# To return demlo number
# Length of s is smaller than 10
def printDemlo(s):
    l = len(s)
    res = ""
      
    # Add numbers to res upto size
    # of s then add in reverse
      
    for i in range(1,l+1):
        res = res + str(i)
      
    for i in range(l-1,0,-1):
        res = res + str(i)
      
    return res
      
  
# Driver Code
s = "111111"    
print printDemlo(s)
  
# Contributed by Harshit Agrawal

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C#

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// C# program to print DemloNumber
using System;
public class GFG {
  
    // To return demlo number. This function assumes
    // that the length of str is smaller than 10.
    static String printDemlo(String str)
    {
        int len = str.Length;
        String res = "";
  
        // Add numbers to res upto size
        // of str and then add number
        // reverse to it
        for (int i = 1; i <= len; i++)
            res += i.ToString();
  
        for (int i = len - 1; i >= 1; i--)
            res += i.ToString();
  
        return res;
    }
  
    // Driver program to test printDemlo()
    public static void Main()
    {
        String str = "111111";
        Console.WriteLine(printDemlo(str));
    }
}
// This code is contributed by mits

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PHP

= 1; $i–)
$res .= chr($i + 48);

return $res;
}

// Driver Code
$str = “111111”;
echo printDemlo($str);

// This code is contributed by mits
?>


Output:

12345654321

This article is contributed by nuclode. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : Mithun Kumar