All square roots are periodic when written as continued fractions and can be written in the form:
\( \sqrt{N}= a0 + \frac{1}{a1+\frac{1}{a2+\frac{1}{a3+...}}}\)
For example, let us consider √23
\( \sqrt{23}= 4 +\sqrt{23} -4 = 4 + \frac{1}{\frac{1}{\frac{1}{\sqrt{23}-4}}} = 4+\frac{1}{1+\frac{\sqrt{23}-3}{7}}\)
If we continue we would get the following expansion:
\( \sqrt{23}= 4 + \frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{8+...}}}}\)
The process can be summarised as follows:
\( a0=4,\frac{1}{\sqrt{23}-4}=\frac{\sqrt{23}+4}{7}=1+\frac{\sqrt{23}-3}{7}\)
\( a1=1,\frac{7}{\sqrt{23}-3}=\frac{7(\sqrt{23}+3)}{14}=3+\frac{\sqrt{23}-3}{2}\)
\( a2=3,\frac{2}{\sqrt{23}-3}=\frac{2(\sqrt{23}+3)}{14}=1+\frac{\sqrt{23}-4}{7}\)
\( a3=1,\frac{7}{\sqrt{23}-4}=\frac{7(\sqrt{23}+4)}{7}=8+\sqrt{23}-4\)
\( a4=8,\frac{1}{\sqrt{23}-4}=\frac{\sqrt{23}+4}{7}=1+\frac{\sqrt{23}-3}{7}\)
\( a5=1,\frac{7}{\sqrt{23}-3}=\frac{2(\sqrt{23}+3)}{14}=3+\frac{\sqrt{23}-3}{2}\)
\( a6=3,\frac{2}{\sqrt{23}-3}=\frac{2(\sqrt{23}+3)}{14}=1+\frac{\sqrt{23}-4}{7}\)
\( a7=1,\frac{7}{\sqrt{23}-4}=\frac{7(\sqrt{23}+4)}{7}=8+\sqrt{23}-4\)
It can be seen that the sequence is repeating. For conciseness, we use the notation √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:
√3=[1;(1,2)], period=2
√5=[2;(4)], period=1
√6=[2;(2,4)], period=2
√7=[2;(1,1,1,4)], period=4
√8=[2;(1,4)], period=2
√10=[3;(6)], period=1
√11=[3;(3,6)], period=2
√12= [3;(2,6)], period=2
√13=[3;(1,1,1,1,6)], period=5
Exactly four continued fractions, for N ≤ 13, have an odd period.
How many continued fractions for N ≤ 10000 have an odd period?
把无理数拆解成连分数，连分数只有在小学的时候碰到过，那时候做的是这题的逆运算，求解连分数，利用到的信息就是continued fractions重复出现的特点，把题变成求解方程。对于这道题，我的思路正是源自于小学解方程。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 | #include <iostream> #include <cmath> #include <vector> using namespace std; class FRAC { private: int sq; int nom; int den; public: FRAC(int s, int n=0, int d=1); int get_a() const; FRAC expand() const; friend bool operator==(const FRAC f1, const FRAC f2); }; bool is_rational(int n); int get_continued_fraction_count(int n); int main() { int odd_cnt = 0; for (int i = 2; i <= 10000; i++) { if (is_rational(i)) continue; int cnt = get_continued_fraction_count(i); if (cnt % 2 == 1) odd_cnt += 1; } cout << odd_cnt << endl; return 0; } bool operator==(const FRAC f1, const FRAC f2) { if ( (f1.sq == f2.sq) && (f1.nom == f2.nom) && (f1.den == f2.den) ) return true; return false; } FRAC::FRAC(int s, int n, int d) { sq = s; nom = n; den = d; } int FRAC::get_a() const { int a = ((int) sqrt(sq) + nom)/den; return a; } FRAC FRAC::expand() const { int a = this->get_a(); int n = a * den -nom; int d = (sq - n*n)/den; FRAC a1 = FRAC(sq, n, d); return a1; } int get_continued_fraction_count(int n) { FRAC f = FRAC(n); vector<FRAC> fv; vector<int> av; int a = f.get_a(); av.push_back(a); FRAC nf = f.expand(); fv.push_back(nf); int begin; while(true) { bool flag = false; FRAC new_frac = nf.expand(); int a = nf.get_a(); for (int i = 0; i < fv.size(); i++) { if (new_frac == fv[i]) { begin = i+1; flag = true; break; } } fv.push_back(new_frac); av.push_back(a); if (flag) break; nf = new_frac; } int res = av.size() - begin; return res; } bool is_rational(int n) { double x = sqrt(n); if (floor(x) * floor(x) == n) return true; return false; } |