Project Euler 64

题目

Odd period square roots

All square roots are periodic when written as continued fractions and can be written in the form:

\(\sqrt{N}=a_0+\dfrac 1 {a_1+\dfrac 1 {a_2+ \dfrac 1 {a_3+ \dots}}}\)

For example, let us consider \(\sqrt{23}:\) \(\sqrt{23}=4+\sqrt{23}-4=4+\dfrac 1 {\dfrac 1 {\sqrt{23}-4}}=4+\dfrac 1 {1+\dfrac{\sqrt{23}-3}7}\)

If we continue we would get the following expansion:

\(\sqrt{23}=4+\dfrac 1 {1+\dfrac 1 {3+ \dfrac 1 {1+\dfrac 1 {8+ \dots}}}}\)

The process can be summarised as follows:

\(a_0=4, \dfrac 1 {\sqrt{23}-4}=\dfrac {\sqrt{23}+4} 7=1+\dfrac {\sqrt{23}-3} 7\)
\(a_1=1, \dfrac 7 {\sqrt{23}-3}=\dfrac {7(\sqrt{23}+3)} {14}=3+\dfrac {\sqrt{23}-3} 2\)
\(a_2=3, \dfrac 2 {\sqrt{23}-3}=\dfrac {2(\sqrt{23}+3)} {14}=1+\dfrac {\sqrt{23}-4} 7\)
\(a_3=1, \dfrac 7 {\sqrt{23}-4}=\dfrac {7(\sqrt{23}+4)} 7=8+\sqrt{23}-4\)
\(a_4=8, \dfrac 1 {\sqrt{23}-4}=\dfrac {\sqrt{23}+4} 7=1+\dfrac {\sqrt{23}-3} 7\)
\(a_5=1, \dfrac 7 {\sqrt{23}-3}=\dfrac {7 (\sqrt{23}+3)} {14}=3+\dfrac {\sqrt{23}-3} 2\)
\(a_6=3, \dfrac 2 {\sqrt{23}-3}=\dfrac {2(\sqrt{23}+3)} {14}=1+\dfrac {\sqrt{23}-4} 7\)
\(a_7=1, \dfrac 7 {\sqrt{23}-4}=\dfrac {7(\sqrt{23}+4)} {7}=8+\sqrt{23}-4\)

It can be seen that the sequence is repeating. For conciseness, we use the notation \(\sqrt{23}=[4;(1,3,1,8)]\), to indicate that the block \((1,3,1,8)\) repeats indefinitely.

The first ten continued fraction representations of (irrational) square roots are:

\(\sqrt{2}=[1;(2)]\), period=\(1\)
\(\sqrt{3}=[1;(1,2)]\), period=\(2\)
\(\sqrt{5}=[2;(4)]\), period=\(1\)
\(\sqrt{6}=[2;(2,4)]\), period=\(2\)
\(\sqrt{7}=[2;(1,1,1,4)]\), period=\(4\)
\(\sqrt{8}=[2;(1,4)]\), period=\(2\)
\(\sqrt{10}=[3;(6)]\), period=\(1\)
\(\sqrt{11}=[3;(3,6)]\), period=\(2\)
\(\sqrt{12}=[3;(2,6)]\), period=\(2\)
\(\sqrt{13}=[3;(1,1,1,1,6)]\), period=\(5\)

Exactly four continued fractions, for \(N \le 13\), have an odd period.

How many continued fractions for \(N \le 10\,000\) have an odd period?

解决方案

本题主要以查阅资料等方式得以解决。

该计算周期性连分数的维基百科页面中，给出了一种计算连分数序列\(\{a\}\)的递推算法：

\(\begin{aligned} m_0 &=0 \\ d_0 &= 1 \\ a_0&=\lfloor\sqrt {S}\rfloor \\ m_{n+1}&=d_na_n-m_n \\ d_{n+1}&=\dfrac{S-m_{n+1}^2}{d_n}\\ a_{n+1}&=\left\lfloor\dfrac{a_0+m_{n+1}}{d_{n+1}}\right\rfloor \end{aligned}\)

其中，\(\{m\},\{d\},\{a\}\)三个数列都是整数序列。当计算到\(a_n=2a_0\)时，算法终止迭代，\(n\)就是周期本身。

依照该维基百科提供的内容，即可完成每个连分数的\(\{a\}\)序列周期计算。

代码

from tools import int_sqrt, is_square

# https://en.wikipedia.org/wiki/Periodic_continued_fraction#Canonical_form_and_repetend
N = 10000
ans = 0
for S in range(2, N + 1):
    if is_square(S):
        continue
    m = 0
    d = 1
    a0 = int_sqrt(S)
    tm = 0
    a = a0
    while a != 2 * a0:
        m = d * a - m
        d = (S - m * m) // d
        a = (a0 + m) // d
        tm += 1
    if tm & 1:
        ans += 1
print(ans)