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Print first k digits of 1/n where n is a positive integer
• Difficulty Level : Easy
• Last Updated : 13 Apr, 2021

Given a positive integer n, print first k digits after point in value of 1/n. Your program should avoid overflow and floating point arithmetic.
Examples :

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

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

We strongly recommend to minimize the browser and try this yourself first.
Let us consider an example n = 7, k = 3. The first digit of 1/7 is ‘1’, it can be obtained by doing integer value of 10/7. Remainder of 10/7 is 3. Next digit is 4 which can be obtained by taking integer value of 30/7. Remainder of 30/7 is 2. Next digits is 2 which can be obtained by taking integer value of 20/7

## C++

 `#include ``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

 `// 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`

## Python3

 `# 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#

 `// 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

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## Javascript

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Output :

```142
0476```

Reference:
Algorithms And Programming: Problems And Solutions by Alexander Shen
This article is contributed by Sachin. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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