Croatian Open Competition in Informatics (COCI) 2023/2024

Блог пользователя vito1036

Croatian Open Competition in Informatics (COCI) 2023/2024 — Round #2

Автор vito1036, история, 6 месяцев назад, По-английски

Hi everyone!

The second round of COCI will be held this Saturday, December 2nd at 14:00 UTC. You can access the judging system here. If you are not familiar with COCI contest format, we encourage you to read the announcement.

The round was prepared by ToniB, Agnus, fbabic, mlicul and me.

Feel free to discuss the problems in the comment section after the contest ends.

Hope to see you and good luck!

coci, contest, croatia, bitset

vito1036
6 месяцев назад
17

Комментарии (17)

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amin_2008

6 месяцев назад, # |

+19

why is there a bitset in tags ?

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Aliyyiakbar

6 месяцев назад, # ^ |

+48

They're neccesary in COCI contests

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terracottalite

6 месяцев назад, # |

← Rev. 2 →

+19

The judge and https://hsin.hr/coci/ conflict about the start time? It seems the judge is crotia time and the announcment page is UTC, but both say 14.00. Can you clarify this?

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Abito

6 месяцев назад, # ^ |

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vito1036

6 месяцев назад, # ^ |

← Rev. 2 →

+11

Thanks for the info! The judge starting time is wrong. The round will start at 14:00 UTC. UPD: it's fixed now.

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vito1036

6 месяцев назад, # |

← Rev. 2 →

+17

Reminder: the contest starts in one hour.

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WeaponizedAutist

6 месяцев назад, # |

← Rev. 2 →

Feedback:

Task 1: Meh
Task 2: Meh
Task 3: Cute task, enjoyed it.
Task 4: A bit standard, but ok. Perhaps this should have had slightly less points?
Task 5: Again, a bit standard but a decent educational problem.

Overall, problems were good quality and educational, kind of like old style cf problems. The contest would have been more enjoyable if either task 1 or task 2 was replaced by a harder problem.

Edit: Also, the tests for task 3 might be a bit weak. My first submission had a very critical bug (which was not dependent on n) but it passed all test cases of the first 3 subtasks and 13/15 test cases of the last subtask.

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NeroZein

6 месяцев назад, # |

+19

This is the first time I AK an OI contest, it feels amazing !

I feel problems are a little bit easier than usual, but I still enjoyed the contest, well done.

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Tima19

6 месяцев назад, # |

where can i upsolve?

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AndiR

5 месяцев назад, # ^ |

https://evaluator.hsin.hr/tasks/

HONI -> 2023./2024. -> 2. kolo

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Halym2007

6 месяцев назад, # |

← Rev. 2 →

Problems was very good.But I solved 1-st problem full.other problems I got some subtasks.Can you share your idea for problems?

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DinoHadzic

6 месяцев назад, # |

← Rev. 2 →

There is $$$O((n+q)\log{}n)$$$ solution for Dizalo...

Hint

Code

#include <bits/stdc++.h>

#define ll long long
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define pb push_back
#define FIO ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)

using namespace std;

typedef pair <int, int> pii;

const int N = 1e5 + 7, inf = 1e9, ofs = (1 << 17);

int n, q;
int a[N], ti[N], va[N];
bool spe[N], del[N];
vector <int> wh[N];

struct fe {
	int fen[N] = {0};
	void Add(int x, int val) {for (; x < N; x += x & -x) fen[x] += val;}
	int Query(int x) {int out = 0; for (; x; x -= x & -x) out += fen[x]; return out;}
};

int t[2*ofs];
void Update(int x, int val) {
	x += ofs; t[x] = val;
	while (x /= 2) t[x] = max(t[2*x], t[2*x+1]);
}
int Query(int x, int l, int r, int lo = 0, int hi = ofs-1) {
	if (r < lo || hi < l) return 0;
	if (l <= lo && hi <= r) return t[x];
	int mid = (lo+hi)/2;
	return max(Query(2*x, l, r, lo, mid), Query(2*x+1, l, r, mid+1, hi));
}

int main () {
	FIO;
	cin >> n >> q;
	
	fe f1, f2, f3;
	
	for (int i = 1; i <= n; i++) ti[i] = q+1;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		ti[i] = q+1;
		f1.Add(i, 1);
		f2.Add(i, 1);
	}
	for (int i = 1; i <= q; i++) {
		cin >> va[i];
		ti[va[i]] = i;
	}
	
	for (int i = n; i; i--) {
		wh[Query(1, 0, a[i]-1)].pb(i);
		Update(a[i], ti[i]);
	}
	
	ll sum = 0;
	set <pii> s;
	s.insert({0, 0}); s.insert({n+1, n+1});
	
	for (auto y : wh[0]) {
		sum += y-a[y];
		f3.Add(y, 1);
		spe[y] = 1;
		s.insert({a[y], y});
	}
	
	cout << sum+n << " ";
	for (int j = 1; j <= q; j++) {
		int x = va[j];
		del[x] = 1;
		
		if (spe[x]) {
			f3.Add(x, -1);
			sum -= f1.Query(x) - f2.Query(a[x]);
			s.erase({a[x], x});
		}
		else sum -= f3.Query((--s.lower_bound({a[x], 0})) -> S) - f3.Query(x);
		f1.Add(x, -1);
		f2.Add(a[x], -1);
		
		for (auto y : wh[j]) {
			if (del[y]) continue;
			spe[y] = 1;
			f3.Add(y, 1);
			s.insert({a[y], y});
			sum += f1.Query(y) - f2.Query(a[y]);
		}
		
		cout << sum+n-j << " ";
	}

	return 0;
}

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ToniB

6 месяцев назад, # ^ |

← Rev. 2 →

+13

You can do it online with the same complexity.

Code

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 1e5 + 2;
const int OFF = 1 << 18;

int n, q, a[MAXN], inv[MAXN];
long long ans;
set<int> suffix_min;

struct Fenwick {
	int fen[MAXN] = {0};

	void add(int x, int val = 1) {
		for (++x; x <= n; x += x & -x) fen[x] += val;
	}
	int count(int x) {
		int ret = 0;
		for (++x; x; x -= x & -x) ret += fen[x];
		return ret;
	}
	int count(int l, int r) {
		return count(r) - count(l - 1);
	}

} removed_values, removed_indices, marked_values, marked_indices;

struct Tournament {
	int tour[OFF] = {0};

	void update(int i, int val, int x = 0, int l = 0, int r = n - 1) {
		if (l > i || r < i) return ;
		if (l == r) {
			tour[x] = val; return ;
		}
		int mid = (l + r) >> 1;
		update(i, val, x * 2 + 1, l, mid);
		update(i, val, x * 2 + 2, mid + 1, r);

		tour[x] = min(tour[x * 2 + 1], tour[x * 2 + 2]);
	}

	int query(int ql, int qr, int x = 0, int l = 0, int r = n - 1) {
		if (ql <= l && r <= qr) return tour[x];
		if (ql > r || l > qr) return n;

		int mid = (l + r) >> 1;
		return min(query(ql, qr, x * 2 + 1, l, mid), query(ql, qr, x * 2 + 2, mid + 1, r));
	}

} values;

void mark(int i) {
	suffix_min.insert(i);
	marked_indices.add(i);
	marked_values.add(a[i]);

	ans += (i - removed_indices.count(i)) - (a[i] - removed_values.count(a[i]));
}

void unmark(int i) {
	ans -= (i - removed_indices.count(i)) - (a[i] - removed_values.count(a[i]));
	marked_indices.add(i, -1);
	marked_values.add(a[i], -1);

	int L = n, R = 0;
	auto it = suffix_min.find(i);

	if (++it != suffix_min.end()) R = *it;
	else R = n;
	--it;

	if (it != suffix_min.begin()) L = (*--it) + 1;
	else L = 0;

	suffix_min.erase(suffix_min.find(i));

	while (L < R) {
		int j = values.query(L, R - 1);
		if (j == n) break;

		j = inv[j];
		if (a[j] < ((R == n) ? n : a[R])) mark(j);

		L = j + 1;
	}

}

int main() {

	cin >> n >> q;

	for (int i = 0; i < n; ++i) {
		cin >> a[i], --a[i];

		values.update(i, a[i]);
		inv[a[i]] = i;
	}

	int cur = n;
	for (int i = n - 1; i >= 0; --i) {
		cur = min(cur, a[i]);
		if (cur == a[i]) {
			mark(i);
		}
	}

	ans += n;
	cout << ans << " ";

	while (q--) {
		int i;
		cin >> i, --i;

		values.update(i, n);

		if (suffix_min.count(i)) unmark(i);

		ans += marked_values.count(a[i], n - 1);
		ans -= marked_indices.count(i, n - 1);

		removed_values.add(a[i]);
		removed_indices.add(i);

		--ans;
		cout << ans << " ";
	}

	return 0;
}

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