嵇爾的吐槽

#没事画轮子的嵇尔不定期的(W)碎(E)碎(B)念(B)和(L)吐(O)槽(G)

Problem 57 Square root convergents 2017-04-17 12:10:00

It is possible to show that the square root of two can be expressed as an infinite continued fraction.

√ 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?

平方根逼近

2的平方根可以用一个无限连分数表示：

√ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + … ))) = 1.414213…

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…

``````def process(level):
a, b = 3, 2
while level > 1:
a, b = a + 2 * b, a + b
level -= 1
return a, b

result = 0
for i in range(1, 1001):
a, b = process(i)
if len(str(b)) < len(str(a)):
result += 1
# print process(i)
print result
``````

1+1/(1+a/b) = (a+2b)/(a+b)