結果

問題 No.62 リベリオン(Extra)
ユーザー fumiphys
提出日時 2019-08-25 18:59:03
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 66 ms / 5,000 ms
コード長 5,528 bytes
コンパイル時間 1,812 ms
コンパイル使用メモリ 174,220 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-11-06 21:46:37
合計ジャッジ時間 2,642 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge4
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 * 2
other AC * 3
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 #
プレゼンテーションモードにする

// includes
#include <bits/stdc++.h>
// macros
#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(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 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 bit(n) (1LL<<(n))
#define INT(n) int n; cin >> n;
#define LL(n) ll n; cin >> n;
#define DOUBLE(n) double n; cin >> n;
using namespace std;
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>;
using Pli = pair<ll, int>;
using Pil = pair<int, ll>;
using Pll = pair<ll, ll>;
using Pdd = pair<double, double>;
using cd = complex<double>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1e9 + 7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// solve
ll w, h, d, mx, my, hx, hy, vx, vy;
template<typename T>
T gcd(T a, T b) {
  if(a > b)return gcd(b, a);
  if(a == 0)return b;
  return gcd(b % a, a);
}
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, M);
}
int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  INT(q);
  rep(i_, q){
    cin >> w >> h >> d >> mx >> my >> hx >> hy >> vx >> vy;
    if(vx < 0){
      vx *= -1;
      mx = w - mx;
      hx = w - hx;
    }
    if(vy < 0){
      vy *= -1;
      my = h - my;
      hy = h - hy;
    }
    ll g = gcd<ll>(vx, vy);
    vx /= g, vy /= g;
    d *= g;
    ll gx = gcd<ll>(vx, 2 * w);
    ll gy = gcd<ll>(vy, 2 * h);
    auto check = [&](ll x, ll y){
                   if((x - hx) % gx != 0 || (y - hy) % gy != 0)return false;
                   ll tx, kx, ty, ky;
                   extgcd(vx, 2*w, tx, kx);
                   extgcd(vy, 2*h, ty, ky);
                   tx *= (x - hx) / gx;
                   ty *= (y - hy) / gy;
                   vector<ll> b(2), m(2);
                   b[0] = tx; b[1] = ty;
                   m[0] = (2 * w) / gx; m[1] = (2 * h) / gy;
                   rep(i, 2)b[i] = (b[i] % m[i] + m[i]) % m[i];
                   // cout << b << endl;
                   // cout << m << endl;
                   auto p = chinese_reminder_theorem(b, m);
                   // cout << p << " " << d << endl;
                   ll M = p.second, r = p.first;
                   if(M == -1){
                     return false;
                   }
                   if(r <= d)return true;
                   return false;
                 };
    bool ok = false;
    if(check(mx, my))ok = true;
    if(check(mx, 2 * h - my))ok = true;
    if(check(2 * w - mx, my))ok = true;
    if(check(2 * w - mx, 2 * h - my))ok = true;
    if(ok)cout << "Hit" << endl;
    else cout << "Miss" << endl;
  }
  return 0;
}
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