Calculate period length of recurring decimal fraction

问题

I want to do a program in python (3.6.5) that tell the length of e.g. 1/7. The output should be for this example something like: "length: 6, repeated numbers: 142857". I got this so far:

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)

回答1:

Doing print ("repeated numbers:", t) prints the representation of the t function itself, not its output.

Here's a repaired version of your code. I use a Python 3.6+ f-string to convert the repeating digits to a string, and add zeros to the front to make it 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 pad it on the left with zeroes (if necessary) to give it a length of k.

As Goyo points out in the comments, this algorithm is not perfect. It gets stuck in a loop if the decimal contains any non-repeating part, that is, if the denominator has any factors of 2 or 5. See if you can figure out a way to deal with that.

来源：https://stackoverflow.com/questions/52383948/calculate-period-length-of-recurring-decimal-fraction

标签

python

python-3.6

fractions

recurring