# Find the number d < 1000 for which 1 / d has the longest recurring cycle
# in its decimal fraction part

# It seems like the cycle is never longer than the longest of the cycle length of any of its prime divisors
# So say we want to know the length for d and d = a * b, then the cycle for d has the same length as either the cycle for a or b, whichever is longer.

# So it seems we then only have to check the primes, so first let's get a prime finder in here

import numpy as np
import math

def maxIndex(ns):
	result = 0
	m = ns[0]

	for i in range(len(ns)):
		if ns[i] > m:
			result = i
			m = ns[i]

	return result

def getPrimes(n):
	primes = [1 for p in range(n)]
	primes[0] = 0
	primes[1] = 0

	x = 2
	while x < np.sqrt(n):
		if primes[x] == 1:

			for i in range(x+x, n, x):
				primes[i] = 0

		x += 1

	result = []
	for i in range(n):
		if primes[i]:
			result.append(i)

	return result

# Now we need to find out the recurring cycle length for any number
# As a benefit for only checking prime number, we know that the cycles will always immediately start after the .'s

def cycleLength(n):
	l = int(math.log10(n)) + 1
	y = 1
	power = pow(10,l)

	result = 1
	
	while True:

		x = (y * power) // n
		rest = y * power - x * n

		if rest == 0:
			return 0
		
		elif rest == 1:
			break
		
		else:
			y = rest

		result += 1

	return result * l

def main():
	print("Hello, this is Patrick")

	ps = getPrimes(1000)
	cyclelengths = [cycleLength(p) for p in ps]
	
	for i in range(len(ps)):
		print(f"{ps[i]} has length: {cyclelengths[i]}")

	print(f"Biggest cycle is for {ps[maxIndex(cyclelengths)]} with length {max(cyclelengths)}")

if __name__ == "__main__":
	main()