p057.py

#It is possible to show that the square root of two can be expressed as an infinite continued fraction.
#
# sqrt 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...
#
#By expanding this for the first four iterations, we get:
#
#1 + 1/2 = 3/2 = 1.5
#1 + 1/(2 + 1/2) = 7/5 = 1.4
#1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
#1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...
#
#The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.
#
#In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?

import logging

def main(args):
    n = [3]
    d = [2]

    for i in range(1000):
        d.append(n[-1] + d[-1])
        n.append(d[-1] + d[-2])

    num = 0
    for i in range(len(n)):
        ni = n[i]
        di = d[i]
        while (di > 0):
            ni = ni // 10
            di = di // 10
        if (ni > 0):
            num += 1

    logging.info(num)