Python – Calculates the cycle length of a cyclic decimal

I want to make a program in python (3.6.5) to tell, for example, 1/7. The output of this example should be similar to: “Length: 6, Repeat numbers: 142857”. So far I got this :

n = int(input("numerator: "))
d = int(input("denominator: "))

def t(n, d):
    x = n * 9
    z = x
    k = 1
    while z % d:
        z = z * 10 + x
        k += 1
        print ("length:", k)
        print ("repeated numbers:", t)

return k, z / d

t(n, d)

Execute print("repeated numbers:", t) to print a representation of the t function itself, not its output.

This is a fixed version of your code. I used Python 3.6+ f-string to convert repeated numbers to strings, adding zeros in front to make them the correct length.

def find_period(n, d):
    z = x = n * 9
    k = 1
    while z % d:
        z = z * 10 + x
        k += 1

digits = f"{z // d:0{k}}"
    return k, digits

# Test

num, den = 1, 7
period, digits = find_period(num, den)
print('num:', num, 'den:', den, 'period:', period, 'digits:', digits)

num, den = 1, 17
period, digits = find_period(num, den)
print('num:', num, 'den:', den, 'period:', period, 'digits:', digits)

Output

num: 1 den: 7 period: 6 digits: 142857
num: 1 den: 17 period: 16 digits: 0588235294117647

This line may be a bit mysterious:

f"{z // d:0{k}}"

It says: Find the largest integer less than or equal to z divided by d, convert it to a string, and fill it with zeros on the left (if necessary) to give it a length of k.

As Goyo points out in his review, the algorithm is not perfect. If the decimal contains any non-repeating parts, i.e. if the denominator has any factor of 2 or 5, it gets stuck in a loop. See if you can come up with a way to deal with this.