Solution to 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: 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.

• Java Code

package sikademy;
/**
 *
 * @author Archangel Macsika
 * Copyright (c) Sikademy. All rights reserved
 */

import java.util.HashMap;
import java.util.Map;

public class SikademyEulerSolution {
	
	public String run() {
		int bestNumber = 0;
		int bestLength = 0;
		for (int i = 1; i <= 1000; i++) {
			int len = getCycleLength(i);
			if (len > bestLength) {
				bestNumber = i;
				bestLength = len;
			}
		}
		return Integer.toString(bestNumber);
	}
	
	
	private static int getCycleLength(int n) {
		Map stateToIter = new HashMap<>();
		int state = 1;
		for (int iter = 0; ; iter++) {
			if (stateToIter.containsKey(state))
				return iter - stateToIter.get(state);
			else {
				stateToIter.put(state, iter);
				state = state * 10 % n;
			}
		}
	}

	public static void main(String[] args) {
                SikademyEulerSolution solution = new SikademyEulerSolution();
		System.out.println(solution.run());
	}	
}

• Python Code

# 
# @author Archangel Macsika
# Copyright (c) Sikademy. All rights reserved.
# 

import itertools


def compute():
	ans = max(range(1, 1000), key=reciprocal_cycle_len)
	return str(ans)


def reciprocal_cycle_len(n):
	seen = {}
	x = 1
	for i in itertools.count():
		if x in seen:
			return i - seen[x]
		else:
			seen[x] = i
			x = x * 10 % n


if __name__ == "__main__":
	print(compute())