제출 #1150030

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1150030		monkey133	Mađioničar (COI22_madionicar)	C++20	0 / 100	449 ms	416 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int main()
{
	ll n;
	cin >> n;
	ll ans = 1, ans2 = 1;
	for (ll i=1; i<=n; ++i)
	{
		ll v;
		while (1 + ans*2 <= n && 1 + ans*2 <= (ans2-1)*2) ans++;
		while (1 + ans*2 <= n)
		{
			if (i - ans <= 0 || i + ans > n) break;
			cout << '?' << ' ' << i- ans << ' ' << i + ans << endl;
			cout.flush();
			cin >> v;
			if (v) ans++;
			else break;
		}
		while ( ans2*2 <= n &&  ans2*2 <= (ans-1)*2 + 1) ans2++;
		while (ans2*2 <= n)
		{
			if (i - ans2+1 <= 0 || i + ans2 > n) break;
			cout << '?'  << ' ' << i- ans2 + 1 << ' ' << i + ans2 << endl;
			cout.flush();
			cin >> v;
			if (v) ans2++;
			else break;
		}
	}
	ll ret = max((ans2-1)*2, (ans-1)*2 + 1);
	cout << ret << endl;
	return 0;
}

• Subtask 1: 0 / 13
• Subtask 2: 0 / 25
• Subtask 3: 0 / 25
• Subtask 4: 0 / 37