結果

問題 No.195 フィボナッチ数列の理解(2)
コンテスト
ユーザー fumiphys
提出日時 2019-09-29 23:16:08
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 4,130 bytes
コンパイル時間 1,451 ms
コンパイル使用メモリ 173,556 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-10-03 04:43:25
合計ジャッジ時間 2,317 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 20 WA * 2
権限があれば一括ダウンロードができます

ソースコード

diff # 

// includes
#include <bits/stdc++.h>
using namespace std;

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define bit(n) (1LL<<(n))
// functions
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(20);}
} fast_io_;

ll fib[55];
ll F(int i){
  if(i == -2)return 1;
  if(i == -1)return 0;
  return fib[i];
}

int main(int argc, char const* argv[])
{
  fib[0] = fib[1] = 1;
  FOR(i, 2, 55)fib[i] = fib[i-1] + fib[i-2];
  ll x[3]; rep(i, 3)cin >> x[i];
  sort(x, x + 3);
  if(x[0] == x[2]){
    if(x[0] == 1){
      cout << "1 1" << endl;
      return 0;
    }
    x[0] = 1;
  }else{
    if(x[0] == x[1]){
      swap(x[1], x[2]);
    }
  }
  pair<ll, ll> res = mk(inf, inf);
  for(int i = 0; i <= 46; i++){
    for(int j = i; j <= 46; j++){
      ll a = F(i-2), b = F(i-1);
      ll c = F(j-2), d = F(j-1);
      if(a * d - b * c == 0)continue;
      ll Y = (c * x[0] - a * x[1]) / (c * b - a * d);
      ll X = (d * x[0] - b * x[1]) / (a * d - b * c);
      if(X <= 0 || Y <= 0)continue;
      bool ok = true;
      if(a * X + b * Y != x[0])ok = false;
      if(c * X + d * Y != x[1])ok = false;
      bool ex = false;
      for(int k = j; k <= 46; k++){
        ll e = F(k-2), f = F(k-1);
        if(e * X + f * Y == x[2])ex = true;
      }
      if(!ex)ok = false;
      if(ok){
        res = min(res, mk(X, Y));
      }
    }
  }

  if(res == mk((ll)inf, (ll)inf)){
    cout << -1 << endl;
    return 0;
  }
  cout << res.first << " " << res.second << endl;
  return 0;
}
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