結果

問題 No.2978 Lexicographically Smallest and Largest Subarray
ユーザー Magentor
提出日時 2024-12-02 00:18:44
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 231 ms / 2,000 ms
コード長 3,151 bytes
コンパイル時間 5,701 ms
コンパイル使用メモリ 316,188 KB
実行使用メモリ 25,460 KB
平均クエリ数 1501.00
最終ジャッジ日時 2024-12-02 00:19:05
合計ジャッジ時間 20,623 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
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
other AC * 57
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

#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
using namespace atcoder;
template<typename T> inline bool chmax(T &a, T b) { return ((a < b) ? (a = b, true) : (false)); }
template<typename T> inline bool chmin(T &a, T b) { return ((a > b) ? (a = b, true) : (false)); }
#define rep(i, n) for (long long i = 0; i < (long long)(n); i++)
#define rep2(i, m ,n) for (int i = (m); i < (long long)(n); i++)
#define REP(i, n) for (long long i = 1; i < (long long)(n); i++)
typedef long long ll;
#define updiv(N,X) (N + X - 1) / X
#define l(n) n.begin(),n.end()
#define YesNo(Q) Q==1?cout<<"Yes":cout<<"No"
using P = pair<int, int>;
using mint = modint;
const int MOD = 998244353LL;
const ll INF = 999999999999LL;
vector<long long> fact, fact_inv, inv;
/*  init_nCk :二項係数のための前処理
    計算量:O(n)
*/
template <typename T>
void input(vector<T> &v){
 rep(i,v.size()){cin>>v[i];}
  return;
}
void init_nCk(int SIZE) {
    fact.resize(SIZE + 5);
    fact_inv.resize(SIZE + 5);
    inv.resize(SIZE + 5);
    fact[0] = fact[1] = 1;
    fact_inv[0] = fact_inv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < SIZE + 5; i++) {
        fact[i] = fact[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        fact_inv[i] = fact_inv[i - 1] * inv[i] % MOD;
    }
}
/*  nCk :MODでの二項係数を求める(前処理 int_nCk が必要)
    計算量:O(1)
*/
long long nCk(int n, int k) {
    assert(!(n < k));
    assert(!(n < 0 || k < 0));
    return fact[n] * (fact_inv[k] * fact_inv[n - k] % MOD) % MOD;
}
long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
ll POW(ll a,ll n){
  long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a;
        a = a * a;
        n >>= 1;
    }
    return res;
}
ll q,y;
vector<string> v;
void otp(stack<ll> st){
  ll nn = st.size();
  rep(i,nn){
    cout << st.top() << endl;
    st.pop();
  }
}
ll ev(ll t){
    stack<ll> st;
    rep(i,q){
     // otp(st);
        if(v[i]=="+"){
        ll a = st.top();st.pop();
        ll b = st.top();st.pop();
        st.push(a+b);
        }
        else if(v[i]=="min"){
        ll a = st.top();st.pop();
        ll b = st.top();st.pop();
        st.push(min(a,b));  
        }
        else if(v[i]=="max"){
        ll a = st.top();st.pop();
        ll b = st.top();st.pop();
        st.push(max(a,b));      
        }
        else if(v[i]=="X"){
        st.push(t); 
        }
        else{
            st.push(stol(v[i]));
        }
    }
    return st.top();
}
int main() {
int n,q;cin>>n>>q;
vector<int> a;vector<int> b;//min-max
rep(i,n/2){
  cout << "? " << 2*i+1 << " " << n << " " << 2*i+2 << " " << n << endl;
  int t;cin>>t;
  if(t==1){a.push_back(2*i+1);b.push_back(2*i+2);}
  else{
    b.push_back(2*i+1);a.push_back(2*i+2);
  }
}
int mn = 0;
int mx = 0;
rep(i,a.size()){
  cout << "? " << a[mn] << " " << n << " " << a[i] << " " << n << endl; 
  int t;cin>>t;
  if(t==0){mn = i;}
}
rep(i,b.size()){
  cout << "? " << b[mx] << " " << n << " " << b[i] << " " << n << endl; 
  int t;cin>>t;
  if(t==1){mx = i;}
}
cout << "! " << a[mn] << " " << a[mn] << " " << b[mx] << " " << n << endl; 
}
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