Codeforces Round #516 (Div. 1) C. Dwarves, Hats and Extrasensory Abilities

問題ページ

解法

2^30=10^9を使ってうまく二分割できるように構成する。まずx軸上に30点置けるような間隔で順番に置いていく。異なる色がでてきたタイミングで二分割する線のうち1端点が決まる。次にもう片方の端点を決定する。x=0,y=1e9,x=1e9の線上で3e9点存在する。線の端点を置ける領域を二分割していくように黒と白の中間に置いていく。30回であれば3e9点の区間があれば領域は足りるので問題ない。

構成の一例

#include <bits/stdc++.h>
 
using namespace std;
using ll = long long;
// #define int ll
using PII = pair<int, int>;
 
#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
 
template<typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template<typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template<typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
 
template<typename T> vector<T> make_v(size_t a) { return vector<T>(a); }
template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts) { 
  return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}
template<typename T,typename V> typename enable_if<is_class<T>::value==0>::type
fill_v(T &t, const V &v) { t=v; }
template<typename T,typename V> typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t, const V &v ) { for(auto &e:t) fill_v(e,v); }
 
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')'; return out;
}
template<typename T>
istream& operator >> (istream& is, vector<T>& vec){
  for(T& x: vec) {is >> x;} return is;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'['; for(T i: a) {out<<i<<',';} out<<']'; return out;
}
 
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0}; // DRUL
const int INF = 1<<30;
const ll LLINF = 1LL<<60;
ll MOD = 1000000007;

PII trans(ll mid) {
  if(mid <= 1000000000) return {0, mid};
  else if(mid <= 2000000000) return {mid-1000000000, 1000000000};
  else return {1000000000, 3000000000-mid};
}

signed main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  constexpr ll t = 1000000000 / 30;
  ll n;
  cin >> n;
  
  cout << 0 << " " << 0 << endl;
  string first;
  cin >> first;

  ll lb = 1, ub = 3000000000-1;
  ll x1=0, y1=1, x2=1000000000, y2=1;
  bool turn = true;
  REP(i, n-1) {
    if(turn) {
      cout << (i+1)*t << " " << 0 << endl; 
      string s;
      cin >> s;
      if(s == first) {

      } else {
        x1 = i*t+1, y1 = 0;
        turn = false;
      }
    } else {
      ll mid = (lb+ub)/2;
      PII xy = trans(mid);
      cout << xy.first << " " << xy.second << endl;
      string s;
      cin >> s;
      if(s == first) {
        lb = mid+1;
        PII p = trans(mid+1);
        x2 = p.first, y2 = p.second;
      } else {
        ub = mid-1;
        PII p = trans(mid-1);
        x2 = p.first, y2 = p.second;
      }
    }
  }

  cout << x1 << " " << y1 << " " << x2 << " " << y2 << endl;

  return 0;
}