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…

将连分数计算取前四次迭代展开式分别是:

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…

接下来的三个迭代展开式分别是99/70、239/169和577/408,但是直到第八个迭代展开式1393/985,分子的位数第一次超过分母的位数。

在前一千个迭代展开式中,有多少个分数分子的位数多于分母的位数?

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

草稿:

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

其实是很早的就做了的题,觉得草稿肯定还有更多,没那么简单……

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