Imprima un número que contenga K dígitos con raíz digital D

Dada una raíz digital ‘D’ y un número de dígitos ‘K’. La tarea es imprimir un número que contenga K dígitos que tenga su raíz digital igual a D. Imprime ‘-1’ si tal número no existe.

Ejemplos:  

Input: D = 4, K = 4
Output: 4000
No. of digits is 4.
Sum of digits is also 4.

Input:  D = 0, K = 1
Output: 0

Enfoque: una observación clave para resolver este problema es que agregar cualquier número de 0 a un número no cambia su raíz digital. Por lo tanto , D seguido de (K-1) 0 es una solución simple. 
 

Caso especial cuando D es 0 y K no es 1 no tiene solución ya que el único número con raíz digital 0 es el 0 mismo. 

A continuación se muestra la implementación del enfoque anterior: 

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find a number
void printNumberWithDR(int k, int d)
{
 
    // If d is 0 k has to be 1
    if (d == 0 && k != 1)
        cout << "-1";
 
    else {
        cout << d;
        k--;
 
        // Print k-1 zeroes
        while (k--)
            cout << "0";
    }
}
 
// Driver code
int main()
{
    int k = 4, d = 4;
 
    printNumberWithDR(k, d);
 
    return 0;
}

// Java implementation of the above approach
 
import java.io.*;
 
class GFG {
 
 
// Function to find a number
static void printNumberWithDR(int k, int d)
{
 
    // If d is 0 k has to be 1
    if (d == 0 && k != 1)
        System.out.print( "-1");
 
    else {
        System.out.print(d);
        k--;
 
        // Print k-1 zeroes
        while (k-->0)
            System.out.print( "0");
    }
}
 
// Driver code
 
    public static void main (String[] args) {
            int k = 4, d = 4;
 
    printNumberWithDR(k, d);
    }
}
 
 
//This code is contributed by inder_verma..

# Python3 implementation of
# the above approach
 
# Function to find a number
def printNumberWithDR( k, d ) :
 
    # If d is 0, k has to be 1
    if d == 0 and k != 1 :
        print(-1, end = "")
 
    else :
        print(d, end = "")
        k -= 1
 
        # Print k-1 zeroes
        while k :
             
            print(0,end = "")
            k -= 1
             
 
# Driver code    
if __name__ == "__main__" :
 
    k, d = 4, 4
 
    # Function call
    printNumberWithDR( k, d )
             
# This code is contributed by
# ANKITRAI1

// C# implementation of the above approach
using System;
 
class GFG {
     
// Function to find a number
static void printNumberWithDR(int k, int d)
{
 
    // If d is 0 k has to be 1
    if (d == 0 && k != 1)
        Console.Write( "-1");
 
    else {
         
        Console.Write(d);
        k--;
 
        // Print k-1 zeroes
        while (k-->0)
            Console.Write( "0");
    }
}
 
// Driver code
static public void Main ()
{
    int k = 4, d = 4;
 
    printNumberWithDR(k, d);
}
}
 
// This code is contributed by ajit.

<?php
// PHP implementation of the above approach
 
// Function to find a number
function printNumberWithDR($k, $d)
{
 
    // If d is 0 k has to be 1
    if ($d == 0 && $k != 1)
        echo "-1";
 
    else
    {
        echo $d;
        $k--;
 
        // Print k-1 zeroes
        while ($k--)
            echo "0";
    }
}
 
// Driver code
$k = 4;
$d = 4;
 
printNumberWithDR($k, $d);
 
// This code is contributed
// by akt_mit
?>

<script>
 
// Javascript implementation of the above approach
 
// Function to find a number
function printNumberWithDR(k, d)
{
     
    // If d is 0 k has to be 1
    if (d == 0 && k != 1)
        document.write("-1");
 
    else
    {
        document.write(d);
        k--;
         
        // Print k-1 zeroes
        while (k-->0)
            document.write("0");
    }
}
  
// Driver Code
var k = 4, d = 4;
 
printNumberWithDR(k, d);
 
// This code is contributed by Ankita saini
                     
</script>

4000

Complejidad del tiempo: O(K)