Imprime los primeros k dígitos de 1/n donde n es un número entero positivo

Dado un entero positivo n, imprima los primeros k dígitos después del punto en valor de 1/n. Su programa debe evitar el desbordamiento y la aritmética de punto flotante.
Ejemplos: 
 

Input:   n = 3, k = 3
Output:  333

Input:   n = 50, k = 4
Output:  0200

Recomendamos encarecidamente minimizar el navegador y probarlo usted mismo primero.
Consideremos un ejemplo n = 7, k = 3. El primer dígito de 1/7 es ‘1’, se puede obtener haciendo el valor entero de 10/7. El resto de 10/7 es 3. El siguiente dígito es 4, que se puede obtener tomando el valor entero de 30/7. El resto de 30/7 es 2. Los siguientes dígitos son 2, que se pueden obtener tomando el valor entero de 20/7 
 

#include <iostream>
using namespace std;
 
// Function to print first k digits after dot in value
// of 1/n.  n is assumed to be a positive integer.
void print(int n, int k)
{
   int rem = 1; // Initialize remainder
 
   // Run a loop k times to print k digits
   for (int i = 0; i < k; i++)
   {
         // The next digit can always be obtained as
         // doing (10*rem)/10
         cout << (10 * rem) / n;
 
         // Update remainder
         rem = (10*rem) % n;
   }
}
 
// Driver program to test above function
int main()
{
    int n = 7, k = 3;
    print(n, k);
    cout << endl;
 
    n = 21, k = 4;
    print(n, k);
 
    return 0;
}

// Java code to Print first k
// digits of 1/n where n is a
// positive integer
import java.io.*;
 
class GFG
{
    // Function to print first
    // k digits after dot in value
    // of 1/n. n is assumed to be
    // a positive integer.
    static void print(int n, int k)
    {
        // Initialize remainder
        int rem = 1;
         
        // Run a loop k times to print k digits
        for (int i = 0; i < k; i++)
        {
            // The next digit can always be
            // obtained as doing (10*rem)/10
            System.out.print( (10 * rem) / n);
 
            // Update remainder
            rem = (10 * rem) % n;
             
        }
         
    }
     
    // Driver program
    public static void main(String []args)
    {
        int n = 7, k = 3;
        print(n, k);
        System.out.println();
         
        n = 21;
        k = 4;
        print(n, k);
         
    }
}
 
// This article is contributed by vt_m

# Python code to Print first k
# digits of 1/n where n is a
# positive integer
import math
 
# Function to print first k digits
# after dot in value of 1/n. n is
# assumed to be a positive integer.
def Print(n, k):
    rem = 1 # Initialize remainder
     
    # Run a loop k times to print
    # k digits
    for i in range(0, k):
        # The next digit can always
        # be obtained as doing
        # (10*rem)/10
        print(math.floor(((10 * rem)
                       / n)), end="")
         
        # Update remainder
        rem = (10*rem) % n
 
# Driver program to test
# above function
n = 7
k = 3
Print(n, k);
print(" ")
n = 21
k = 4
Print(n, k);
 
# This code is contributed by Sam007.

// C# code to Print first k digits of
// 1/n where n is a positive integer
using System;
 
class GFG {
     
    // Function to print first
    // k digits after dot in value
    // of 1/n. n is assumed to be
    // a positive integer.
    static void print(int n, int k)
    {
         
        // Initialize remainder
        int rem = 1;
         
        // Run a loop k times to
        // print k digits
        for (int i = 0; i < k; i++)
        {
             
            // The next digit can always be
            // obtained as doing (10*rem)/10
            Console.Write( (10 * rem) / n);
 
            // Update remainder
            rem = (10 * rem) % n;
        }
    }
     
    // Driver program
    public static void Main()
    {
        int n = 7, k = 3;
        print(n, k);
        Console.WriteLine();
         
        n = 21;
        k = 4;
        print(n, k);
    }
}
 
// This code is contributed by Sam007.

<?php
// Function to print first k digits
// after dot in value of 1/n. n is
// assumed to be a positive integer.
 
function println($n, $k)
{
    // Initialize remainder
    $rem = 1;
 
// Run a loop k times
// to print k digits
for ($i = 0; $i < $k; $i++)
{
    // The next digit can always
    // be obtained as doing
    // (10 * rem) / 10
    echo floor((10 * $rem) / $n);
 
    // Update remainder
    $rem = (10 * $rem) % $n;
}
}
 
// Driver Code
$n = 7; $k = 3;
println($n, $k);
echo "\n";
 
$n = 21; $k = 4;
println($n, $k);
 
// This code is contributed by aj_36
?>

<script>
 
// Function to print first k digits after dot in value
// of 1/n.  n is assumed to be a positive integer.
function print(n,  k)
{
   let rem = 1; // Initialize remainder
   let ans = '';
   // Run a loop k times to print k digits
   for (let i = 0; i < k; i++)
   {
         // The next digit can always be obtained as
         // doing (10*rem)/10
         ans += Math.floor(((10 * rem) / n));
         // Update remainder
         rem = (10*rem) % n;
   }
   document.write(ans)
}
 
// Driver program to test above function
let n = 7;
let k = 3;
print(n, k);
document.write("<br>");
n = 21;
k = 4;
print(n, k);
 
</script>

Producción : 

142
0476

Complejidad de tiempo: O(k)

Espacio Auxiliar: O(1)

Referencia:  
Algoritmos y programación: problemas y soluciones por Alexander Shen
Este artículo es una contribución de Sachin . Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.