Submission #503143

#	Time	Username	Problem	Language	Result	Execution time	Memory
503143		stanislavpolyn	"The Lyuboyn" code (IZhO19_lyuboyn)	C++17	3 / 100	1 ms	332 KiB

lyuboyn

#include <bits/stdc++.h>

#define fr(i, a, b) for(int i = (a); i <= (b); ++i)
#define rf(i, a, b) for(int i = (a); i >= (b); --i)
#define fe(x, y) for(auto& x : y)

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple

#define all(x) (x).begin(), (x).end()
#define pw(x) (1LL << (x))
#define sz(x) (int)(x).size()

using namespace std;

mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<typename T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define fbo find_by_order
#define ook order_of_key

template<typename T>
bool umn(T& a, T b) { return (a > b ? (a = b, 1) : 0); }
template<typename T>
bool umx(T& a, T b) { return (a < b ? (a = b, 1) : 0); }

using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template<typename T>
using ve = vector<T>;

int n, k, t;
string s;

ve<int> ans;
bool vis[pw(18)];

int main() {
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
#else
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
#endif

    cin >> n >> k >> t;

    cout << "-1\n";
//    cin >> s;

//    vis[0] = 1;
//    ans.pb(0);
//    while(1) {
//        bool ok = 0;
//        fr(mask, 0, pw(n) - 1) {
//            if(!vis[mask] && __builtin_popcount(mask ^ ans.back()) == k) {
//                vis[mask] = 1;
//                ans.pb(mask);
//                ok = 1;
//                break;
//            }
//        }
//        if(!ok) {
//            break;
//        }
//    }
//
//    cout << sz(ans) << "\n";
//    fe(x, ans) {
//        cout << bitset<4>(x) << "\n";
//    }

    return 0;
}

• Subtask 1: 0 / 0
• Subtask 2: 0 / 5
• Subtask 3: 3 / 3
• Subtask 4: 0 / 11
• Subtask 5: 0 / 15
• Subtask 6: 0 / 18
• Subtask 7: 0 / 31
• Subtask 8: 0 / 17