Submission #151252

#	Time	Username	Problem	Language	Result	Execution time	Memory
151252		andrew	"The Lyuboyn" code (IZhO19_lyuboyn)	C++17	3 / 100	961 ms	2520 KiB

lyuboyn

#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>

#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")

#define fi first
#define se second
#define p_b push_back
#define pll pair<ll,ll>
#define pii pair<int,int>
#define m_p make_pair
#define all(x) x.begin(),x.end()
#define sset ordered_set
#define sqr(x) (x)*(x)
#define pw(x) (1ll << x)

using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
const ll MAXN = 1123456;
const ll N = 2e5;
mt19937_64 rnd(chrono::system_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>;

template <typename T> void vout(T s){cout << s << endl;exit(0);}

ll Val(string s){
    ll ans = 0;
    for(auto i : s){
        ans *= 2;
        ans |= i - '0';
    }
    return ans;
}

ll n;

string Fst(ll x){
    string ans;

    for(int i = n - 1; i >= 0; i--){
        if((x & pw(i))){
            ans += "1";
        }else ans += "0";
    }
    return ans;
}

ll kol(ll a, ll b){
    ll ans = 0;
    for(int i = 0; i < n; i++){
        if((a & pw(i)) != (b & pw(i)))ans++;
    }
    return ans;
}


int main(){
    ios_base :: sync_with_stdio(0);
    cin.tie(0);

    ll k, t;
    cin >> n >> k >> t;
    vector <ll> a(pw(n));
    for(int i = 0; i < pw(n); i++)a[i] = i;
    string st;
    cin >> st;
    ll s = Val(st);
    swap(a[0], a[s]);
    while(clock() / (double)CLOCKS_PER_SEC < 0.95){
        shuffle(a.begin() + 1, a.end(), rnd);
        bool bad = 0;
        for(int i = 1; i < pw(n); i++){
            if(kol(a[i], a[i - 1]) != k)bad = 1;
        }
        if(bad)continue;
        if(t && kol(a.back(), a[0]) != k)continue;

        cout << pw(n) << "\n";

        for(auto i : a){
            cout << Fst(i) << "\n";
        }

        return 0;

    }

    vout(-1);

    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