Encuentra un número de N dígitos que sea divisible por D

Dado  y  . La tarea es encontrar un número de N dígitos que sea divisible por D (2 <= D <= 10). Si no es posible, imprima Imposible. 

Ejemplos : 

Input : N = 2 and D = 2
Output : 20

Input : N = 1 and D = 10
Output : Impossible

Enfoque: Hay dos condiciones, D = 10 y D no igual a 10. Si D = 10 y N = 1, entonces la única respuesta no es posible y en todas las demás condiciones, la respuesta será posible. 

1. If D is 10,
   Print 1 followed by n-1 times zero.
2. If D is not 10
   Print D followed by n-1 times zero

A continuación se muestra la implementación del enfoque anterior:  

// CPP program to Find N digits
// number which is divisible by D
#include <bits/stdc++.h>
using namespace std;
 
// Function to return N digits
// number which is divisible by D
string findNumber(int n, int d)
{
    // to store answer
    string ans = "";
 
    if (d != 10) {
        ans += to_string(d);
        for (int i = 1; i < n; i++)
            ans += '0';
    }
    else {
        if (n == 1)
            ans += "Impossible";
        else {
            ans += '1';
            for (int i = 1; i < n; i++)
                ans += '0';
        }
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int n = 12, d = 3;
 
    cout << findNumber(n, d);
 
    return 0;
}

// Java program to Find N digits
// number which is divisible by D
 
import java.io.*;
 
class GFG {
 
 
// Function to return N digits
// number which is divisible by D
static String findNumber(int n, int d)
{
    // to store answer
    String ans = "";
 
    if (d != 10) {
        ans += Integer.toString(d);
        for (int i = 1; i < n; i++)
            ans += '0';
    }
    else {
        if (n == 1)
            ans += "Impossible";
        else {
            ans += '1';
            for (int i = 1; i < n; i++)
                ans += '0';
        }
    }
 
    return ans;
}
 
// Driver code
 
    public static void main (String[] args) {
            int n = 12, d = 3;
 
    System.out.println(findNumber(n, d));
    }
}
// This code is contributed by anuj_67..

# Python 3 program to Find N digits
# number which is divisible by D
 
# Function to return N digits
# number which is divisible by D
def findNumber(n, d):
 
    # to store answer
    ans = ""
 
    if (d != 10) :
        ans += str(d)
        for i in range(1,n):
            ans += '0'
    else :
        if (n == 1):
            ans += "Impossible"
        else :
            ans += '1'
            for i in range(1,n):
                ans += '0'
 
    return ans
 
# Driver code
if __name__ == "__main__":
    n = 12
    d = 3
 
    print(findNumber(n, d))
     
# This code is contributed by
# ChitraNayal

// C# program to Find N digits
// number which is divisible by D
using System;
 
class GFG {
 
// Function to return N digits
// number which is divisible by D
static string findNumber(int n, int d)
{
     
    // to store answer
    string ans = "";
 
    if (d != 10) {
         
        ans += d.ToString();
         
        for (int i = 1; i < n; i++)
            ans += '0';
    }
     
    else {
         
        if (n == 1)
            ans += "Impossible";
             
        else {
             
            ans += '1';
            for (int i = 1; i < n; i++)
                ans += '0';
        }
    }
 
    return ans;
}
 
// Driver code
public static void Main ()
{
     
    int n = 12, d = 3;
    Console.WriteLine(findNumber(n, d));
}
}
 
// This code is contributed by Subhadeep

<?php
// PHP program to Find N digits
// number which is divisible by D
 
// Function to return N digits
// number which is divisible by D
function findNumber($n, $d)
{
    // to store answer
    $ans = "";
 
    if ($d != 10)
    {
        $ans .= strval($d);
        for($i = 1; $i < $n; $i++)
            $ans .= '0';
    }
    else
    {
        if (n == 1)
            $ans .= "Impossible";
        else
            $ans .= '1';
            for($i = 1; $i < $n; $i++)
                $ans .= '0';
    }
 
    return $ans;
}
 
// Driver code
$n = 12;
$d = 3;
 
print(findNumber($n, $d));
     
// This code is contributed by mits

<script>
// Javascript program to Find N digits
// number which is divisible by D
     
// Function to return N digits
// number which is divisible by D
  function findNumber(n,d)
  {
        // to store answer
    let ans = "";
   
    if (d != 10) {
        ans += (d).toString();
        for (let i = 1; i < n; i++)
            ans += '0';
    }
    else {
        if (n == 1)
            ans += "Impossible";
        else {
            ans += '1';
            for (let i = 1; i < n; i++)
                ans += '0';
        }
      }
   
      return ans;
  }
     
  // Driver code
  let n = 12, d = 3;
  document.write(findNumber(n, d));
 
 
// This code is contributed by avanitrachhadiya2155
</script>

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