No.195 フィボナッチ数列の理解(2)

解法

editorialが十分分かりやすい。

fibonacci数列が$O(2^N)$で増加するというのが肝。

実装

  • $F_{0,1}(k) = F_{1,0}(k+1)$。
  • ${\mathbf A}^{-1} = \frac{1}{\det{\mathbf A}}({\rm tr}{\mathbf A}{\mathbf E} - {\mathbf A})$。
#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>
#include <cassert>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
typedef long long ll;
template <typename T> bool setmin(T & l, T const & r) { if (not (r < l)) return false; l = r; return true; }
using namespace std;
int main() {
    vector<ll> xs(3); cin >> xs[0] >> xs[1] >> xs[2];
    sort(xs.begin(), xs.end());
    xs.erase(unique(xs.begin(), xs.end()), xs.end());
    const int l = 48;
    vector<ll> fib(l); fib[0] = 1; repeat (i,fib.size()-2) fib[i+2] = fib[i] + fib[i+1];
    assert (xs.back() < fib.back());
    if (xs.size() == 1) {
        xs.push_back(1);
        swap(xs[0], xs[1]);
    }
    ll x = xs[0], y = xs[1];
    pair<ll,ll> ans = { LLONG_MAX, -1 };
    repeat (i,l-1) {
        repeat (j,l-1) {
            ll det = fib[i] * fib[j+1] - fib[i+1] * fib[j];
            if (det == 0) continue;
            ll a =   fib[j+1] * x - fib[i+1] * y;
            ll b = - fib[j]   * x + fib[i]   * y;
            if (a % det != 0 or b % det != 0) continue;
            a /= det; b /= det;
            if (a <= 0 or b <= 0) continue;
            assert (a * fib[i] + b * fib[i+1] == x);
            assert (a * fib[j] + b * fib[j+1] == y);
            if (xs.size() == 3) {
                ll z = xs[2];
                bool found = false;
                repeat (k,l-1) {
                    if (a * fib[k] + b * fib[k+1] == z) {
                        found = true;
                        break;
                    }
                }
                if (not found) continue;
            }
            setmin(ans, make_pair(a, b));
        }
    }
    if (ans.first == LLONG_MAX) {
        cout << -1 << endl;
    } else {
        cout << ans.first << ' ' << ans.second << endl;
    }
    return 0;
}