이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ldb;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ldb,ldb> pdd;
#define ff(i,a,b) for(int i = a; i <= b; i++)
#define fb(i,b,a) for(int i = b; i >= a; i--)
#define trav(a,x) for(auto& a : x)
#define sz(a) (int)(a).size()
#define fi first
#define se second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k) print the k-th smallest number in os(0-based)
const int mod = 1000000007;
const int inf = 1e9 + 5;
const int mxN = 200005;
int n;
int ask(int l, int r){
cout << "? " << l << " " << r << endl;
int X;
cin >> X;
return X;
}
int main() {
cin.tie(0)->sync_with_stdio(0);
cin >> n;
int a = 1, b = 0;
ff(i,1,n){
while(i - a / 2 - 1 >= 1 && i + a / 2 + 1 <= n && ask(i - a / 2 - 1, i + a / 2 + 1) == 1){
a += 2;
}
while(i - b / 2 >= a && i + b / 2 + 1 <= n && ask(i - b / 2, i + b / 2 + 1) == 1){
b += 2;
}
}
cout << max(a, b) << endl;
return 0;
}
/*
// probati bojenje sahovski
*/
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