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
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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 <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 functionint 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 integerimport 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 integerimport 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 functionn = 7k = 3Print(n, k);print(" ")n = 21k = 4Print(n, k);# This code is contributed by Sam007. |
C#
// C# code to Print first k digits of// 1/n where n is a positive integerusing 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
<?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 $rem = 1;// Run a loop k times // to print k digitsfor ($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?> |
Javascript
<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 functionlet n = 7;let k = 3;print(n, k);document.write("<br>");n = 21;k = 4;print(n, k);</script> |
Output :
142 0476
Time Complexity: O(k)
Auxiliary Space: O(1)
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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