結果

問題 No.186 中華風 (Easy)
ユーザー fumiphysfumiphys
提出日時 2019-04-30 16:28:56
言語 C++14
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 3,834 bytes
コンパイル時間 1,433 ms
コンパイル使用メモリ 168,624 KB
実行使用メモリ 4,380 KB
最終ジャッジ日時 2023-09-27 00:55:06
合計ジャッジ時間 2,402 ms
ジャッジサーバーID
（参考情報） 		judge12 / judge11
このコードへのチャレンジ
（要ログイン）

テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 1 ms
4,380 KB
testcase_01 AC 2 ms
4,376 KB
testcase_02 AC 2 ms
4,376 KB
testcase_03 AC 1 ms
4,376 KB
testcase_04 AC 1 ms
4,376 KB
testcase_05 AC 1 ms
4,376 KB
testcase_06 AC 1 ms
4,376 KB
testcase_07 AC 1 ms
4,380 KB
testcase_08 AC 1 ms
4,380 KB
testcase_09 AC 2 ms
4,376 KB
testcase_10 AC 1 ms
4,380 KB
testcase_11 AC 2 ms
4,376 KB
testcase_12 AC 1 ms
4,376 KB
testcase_13 AC 1 ms
4,380 KB
testcase_14 AC 1 ms
4,376 KB
testcase_15 AC 1 ms
4,380 KB
testcase_16 AC 2 ms
4,376 KB
testcase_17 AC 2 ms
4,380 KB
testcase_18 AC 1 ms
4,380 KB
testcase_19 AC 1 ms
4,376 KB
testcase_20 AC 2 ms
4,376 KB
testcase_21 AC 1 ms
4,380 KB
testcase_22 AC 2 ms
4,376 KB
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>

// macros
#define ll long long int
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#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(st, itr) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define vrep(v, i) for(int i = 0; i < (v).size(); i++)
#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 FI first
#define SE second
#define dump(a, n) for(int i = 0; i < n; i++)cout << a[i] << "\n "[i + 1 != n];
#define dump2(a, n, m) for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)cout << a[i][j] << "\n "[j + 1 != m];
using namespace std;

//  types
typedef pair<int, int> P;
typedef pair<ll, int> Pl;
typedef pair<ll, ll> Pll;
typedef pair<double, double> Pd;
 
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1e9 + 7;

// solve
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;}

template<typename T>
T extgcd(T a, T b, T &x, T &y){
  T d = a;
  if(b != 0){
    d = extgcd(b, a % b, y, x);
    y -= (a / b) * x;
  }else{
    x = 1, y = 0;
  }
  return d;
}

template<typename T>
pair<T, T> chinese_reminder_theorem(vector<T> b, vector<T> m){
  T r = 0, M = 1;
  for(int i = 0; i < b.size(); i++){
    T x, y;
    T d = extgcd<T>(M, m[i], x, y);
    if((b[i] - r) % d != 0)return make_pair(0, -1);
    T tmp = (b[i] - r) / d * x % (m[i] / d);
    r += M * tmp;
    M *= m[i] / d;
  }
  r %= M;
  if(r < 0)r += M;
  return make_pair(r, M);
}

int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  vector<ll> x(3), y(3);
  rep(i, 3)cin >> x[i] >> y[i];
  auto p = chinese_reminder_theorem(x, y);
  if(p.FI == 0 && p.SE == -1)cout << -1 << endl;
  else cout << (p.FI == 0 ? p.FI + p.SE: p.FI) << endl;
  return 0;
}
0