##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 math


def fi():
    a, b = 1, 2
    n = 0
    for i in range(999):
        if math.floor(math.log10(a + b)) > math.floor(math.log10(b)):
            n = n + 1
        a, b = b, a + 2 * b
    return n


print(fi())
