Project Euler

Problem 26: Reciprocal Cycles

A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that ¹/₇ has a 6-digit recurring cycle. Find the value of d<1000 for which ¹/_d contains the longest recurring cycle in its decimal fraction part.

The Catch

How to find the length of a recurring cycle of ¹/_d

The Light

Use an algorithm which uses modulus operation to find the length of a recurring cycle:

• Take the modulus 1 % d. Call the result x.
• Times x by 10.
• Take the modulus x % d.
• Keep track of all found values for x.
• Repeat step 2 - 3 until x repeats again or x = 0.

For example, to find the recurring cycle length of 1/8:
1 % 8 = 1
10 % 8 = 2
20 % 8 = 4
40 % 8 = 0 (Stop the algorithm here; 1/8 has a recurring cycle length of 3)

The Code

import java.util.*;

public class Problem26
{
  public static void main(String[] args)
  {
    int count = 1; 
    int remainder = 0;
    int tmp = 10;
    int max = 0;
    int result = 0;

    for(int d = 2; d < 1000; d++)
    {
      ArrayList list = new ArrayList();
      list.add(new Integer(1));
      while(true)
      {
        remainder = tmp % d;
        if(list.contains(remainder) || remainder == 0)
        {
          if(count > max)
          {
            max = count;
            result = d;
          }
          count = 1;
          tmp = 10;
          break;
        }
        else
        {
          list.add(new Integer(remainder));
          tmp = remainder * 10;
          count++;
        }	
      }
    }
    System.out.println(result);
  }
}