Submission #155685

#	Time	Username	Problem	Language	Result	Execution time	Memory
155685		qkxwsm	"The Lyuboyn" code (IZhO19_lyuboyn)	C++14	100 / 100	280 ms	6472 KiB

lyuboyn

#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
	if (a > b) a = b;
}
template<class T, class U>
void ckmax(T &a, U b)
{
	if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define SZ(x) ((int) ((x).size()))
#define ALL(x) (x).begin(), (x).end()
#define MAXN 300013

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

int N, K;
bool T;
int st;
int ans[MAXN];
vi arr;
bitset<MAXN> vis;

int32_t main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	cin >> N >> K >> T;
	string temps; cin >> temps;
	reverse(ALL(temps));
	while(SZ(temps) < N)
	{
		temps += '0';
	}
	reverse(ALL(temps));
	FOR(i, 0, N)
	{
		st *= 2; st += (temps[i] == '1');
	}
	//find N linearly independent vectors
	vis[0] = true;
	FOR(i, 0, (1 << N))
	{
		if (__builtin_popcount(i) != K) continue;
		if (vis[i]) continue;
		arr.PB(i);
		FOR(j, 0, (1 << N))
		{
			if (vis[j] || vis[j ^ i])
			{
				vis[j] = true;
				vis[i] = true;
			}
		}
	}
	if (SZ(arr) != N)
	{
		cout << "-1\n";
		return 0;
	}
	// cerr << "HEY\n";
	FOR(i, 1, (1 << N))
	{
		ans[i] = __builtin_ctz(i);
	}
	FOR(i, 1, (1 << N))
	{
		ans[i] = arr[ans[i]];
	}
	// FOR(i, 0, N)
	// {
	// 	cerr << arr[i] << ' ';
	// }
	// cerr << endl;
	ans[0] = st;
	FOR(i, 1, (1 << N))
	{
		ans[i] ^= ans[i - 1];
	}
	vis.reset();
	FOR(i, 0, (1 << N))
	{
		if (vis[ans[i]])
		{
			cout << "-1\n";
			return 0;
		}
		assert(!vis[ans[i]]);
		// cerr << ans[i] << ' ';
		vis[ans[i]] = true;
	}
	// cerr << endl;
	cout << (1 << N) << '\n';
	FOR(i, 0, (1 << N))
	{
		FORD(j, N, 0)
		{
			cout << ((ans[i] & (1 << j)) ? 1 : 0);
		}
		cout << '\n';
	}
	return 0;
}

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