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Блог пользователя dcordb

Codeforces Round #659 Editorial

Автор dcordb, 4 года назад, По-английски

We (the setters) hope you liked the problems. Here is the editorial:

Tutorial is loading...

C++ solution

#include <iostream>

#define endl '\n'

using namespace std;

void solve()
{
    int n;
    cin >> n;

    string s(200, 'a');
    cout << s << endl;

    for (int i = 0; i < n; ++i)
    {
        int u;
        cin >> u;
        s[u] = s[u] == 'a' ? 'b' : 'a';
        cout << s << endl;
    }
}

int main()
{
    int t;
    cin >> t;

    for (int i = 0; i < t; ++i)
    {
        solve();
    }

    return 0;
}

Python solution

import sys
input = sys.stdin.readline

def main():
    t = int(input())
    
    for _ in range(t):
        n = int(input())
        a = list(map(int, input().split()))
    
        mx = max(a)
        ans = [ 'a' * (mx + 1) ] * (n + 1)
    
        for i, x in enumerate(a):
            who = 'a' if ans[i][x] == 'b' else 'b'
            ans[i + 1] = ans[i][:x] + who + ans[i][x + 1:]
    
        print('\n'.join(ans))    

main()

Problem idea: dcordb

Solution idea: dcordb and Devil

Tutorial is loading...

C++ solution

#include <bits/stdc++.h>

using namespace std;

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

	int test;
	cin >> test;
	while (test--)
	{
		int n, k, l;
		cin >> n >> k >> l;
		vector<int> d(n+2, -k);
		for (int i = 1; i <= n; ++i)
			cin >> d[i];

		set<tuple<int, int, bool>> mark;
		function<bool(int, int, bool)> go = [&](int pos, int tide, bool down)
		{
			if (pos > n) return true;
			
			if (mark.find({ pos, tide, down }) != mark.end())
				return false;

			mark.insert({ pos, tide, down });

			tide += down ? -1 : +1;
			if (tide == 0) down = false;
			if (tide == k) down = true;

			if (d[pos] + tide <= l && go(pos, tide, down))
				return true;
			if (d[pos + 1] + tide <= l && go(pos + 1, tide, down))
				return true;
			return false;
		};

		if (go(0, 0, false)) cout << "Yes\n";
		else cout << "No\n";
	}

	return 0;
}

Problem idea: Devil

Solution idea: Devil

Tutorial is loading...

C++ solution

#include <bits/stdc++.h>

using namespace std;

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

	int test;
	cin >> test;
	while (test--)
	{
		int n, k, l;
		cin >> n >> k >> l;
		vector<int> d(n+1), safe = { 0 };
		for (int i = 1; i <= n; ++i)
		{
			cin >> d[i];
			if (d[i] + k <= l)
				safe.push_back(i);
		}

		safe.push_back(n+1);
		bool ok = true;
		for (size_t i = 1; i < safe.size() && ok; ++i)
		{
			int tide = k; bool down = true;
			for (int j = safe[i-1] + 1; j < safe[i]; ++j)
			{
				tide += down ? -1 : +1;

				if (down) tide -= max(0, d[j] + tide - l);
				if (tide < 0 || d[j] + tide > l) { ok = false; break; }

				if (tide == 0) down = false;
			}
		}

		if (ok) cout << "Yes\n";
		else cout << "No\n";
	}

	return 0;
}

Problem idea: Devil

Solution idea: Devil

Tutorial is loading...

C++ solution

#include <bits/stdc++.h>

using namespace std;

const int Alp = 20;

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

	int test;
	cin >> test;
	while (test--)
	{
	    int n;
		string a, b;
		cin >> n >> a >> b;

		bool bad = false;
		vector<vector<int>> adj(Alp);
		for (int i = 0; i < n; ++i)
			if (a[i] != b[i])
			{
				if (a[i] > b[i])
				{
					bad = true;
					cout << "-1\n";
					break;
				}

				adj[a[i]-'a'].push_back(b[i]-'a');
				adj[b[i]-'a'].push_back(a[i]-'a');
			}

		if (bad) continue;

		vector<bool> mark(Alp);
		function<void(int)> dfs = [&](int u)
		{
			mark[u] = true;
			for (auto v : adj[u])
				if (!mark[v])
					dfs(v);
		};

		int ans = Alp;
		for (int i = 0; i < Alp; ++i)
			if (!mark[i])
				dfs(i), --ans;
		cout << ans << "\n";
	}

	return 0;
}

Problem idea: Devil

Solution idea: mnaeraxr and Devil

Tutorial is loading...

C++ solution

#include <bits/stdc++.h>

using namespace std;

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

	int test;
	cin >> test;
	while (test--)
	{
		int n;
		cin >> n;
		
		int x = 0;
		vector<int> a(n);
		for (auto &i : a)
		{
			cin >> i;
			x ^= i;
		}

		if (x == 0)
		{
			cout << "DRAW\n";
			continue;
		}

		for (int k = 30; k >= 0; --k)
			if (x >> k & 1)
			{
				vector<int> f(2);
				for (auto &i : a) ++f[i >> k & 1];
				if (f[1] % 4 == 3 && f[0] % 2 == 0)
					cout << "LOSE\n";
				else
					cout << "WIN\n";
				break;
			}
	}

	return 0;
}

Python solution

import sys
input = sys.stdin.readline

d = { 1: 'WIN', 0: 'LOSE', -1: 'DRAW' }

def main():
    t = int(input())

    for _ in range(t):
        n = int(input())
        a = map(int, input().split())

        f = [0] * 30

        for x in a:
            for b in range(30):
                if x >> b & 1:
                    f[b] += 1

        ans = -1
        for x in reversed(range(30)):
            if f[x] % 2 == 1:
                ans = 0 if f[x] % 4 == 3 and (n - f[x]) % 2 == 0 else 1
                break

        print(d[ans])

main()

Problem idea: Devil

Solution idea: Devil and dcordb

Tutorial is loading...

C++ solution

#include <bits/stdc++.h>

using namespace std;

const int Alp = 20;

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

	int test;
	cin >> test;
	while (test--)
	{
	    int len;
		string a, b;
		cin >> len >> a >> b;
		
		vector<int> adj(Alp);
		vector<vector<int>> G(Alp);
		for (int i = 0; i < len; ++i)
			if (a[i] != b[i])
			{
				adj[a[i]-'a'] |= 1 << (b[i]-'a');
				G[a[i]-'a'].push_back(b[i]-'a');
				G[b[i]-'a'].push_back(a[i]-'a');
			}

		vector<bool> mark(Alp);
		function<void(int)> dfs = [&](int u)
		{
			mark[u] = true;
			for (auto v : G[u])
				if (!mark[v])
					dfs(v);
		};

		int comp = 0;
		for (int i = 0; i < Alp; ++i)
			if (!mark[i])
				dfs(i), ++comp;

		int ans = 0;
		vector<bool> dp(1<<Alp);
		dp[0] = true;
		for (int mask = 0; mask < 1<<Alp; ++mask)
			if (dp[mask])
			{
				ans = max(ans, __builtin_popcount(mask));
				for (int u = 0; u < Alp; ++u)
					if ((~mask >> u & 1) && (adj[u] & mask) == 0)
						dp[mask | 1 << u] = true;
			}

		cout << 2*Alp - comp - ans << "\n";
	}

	return 0;
}

Problem idea: Devil

Solution idea: mnaeraxr

Tutorial is loading...

C++ solution

#include <bits/stdc++.h>

#define endl '\n'

using namespace std;

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

    int n, m;
    cin >> n >> m;

    vector<vector<int>> mat(n, vector<int>(m));

    for (int i = 0; i < n; ++i)
        for (int j = 0; j < m; ++j)
            cin >> mat[i][j];

    vector<int> h(n * m + 1);
    vector<int> v(n * m + 1);

    for (int i = 0; i < n; ++i)
    {
        int a = 0;
        for (int j = 0; j < m; ++j)
            a = max(a, mat[i][j]);
        h[a] = 1;
    }

    for (int i = 0; i < m; ++i)
    {
        int a = 0;
        for (int j = 0; j < n; ++j)
            a = max(a, mat[j][i]);
        v[a] = 1;
    }

    vector<vector<int>> fin(n, vector<int>(m));
    queue<pair<int, int>> q;

    int x = -1, y = -1;

    for (int u = n * m; u >= 1; --u)
    {
        x += h[u];
        y += v[u];

        if (h[u] || v[u])
        {
            fin[x][y] = u;
        }
        else
        {
            int qx, qy;
            tie(qx, qy) = q.front();
            q.pop();
            fin[qx][qy] = u;
        }

        if (h[u])
            for (int i = y - 1; i >= 0; --i)
                q.push({x, i});

        if (v[u])
            for (int i = x - 1; i >= 0; --i)
                q.push({i, y});
    }

    for (int i = 0; i < n; ++i)
        for (int j = 0; j < m; ++j)
            cout << fin[i][j] << " \n"[j + 1 == m];

    return 0;
}

Problem idea: gilcu3

Solution idea: gilcu3 and mnaeraxr

Tutorial is loading...

C++ solution

#include <bits/stdc++.h>

using namespace std;

const int mod = 1000000007;

int main()
{
	ios::sync_with_stdio(false), cin.tie(0);

	string s; cin >> s;
	int n = s.length();
	vector<int> dist(n);
	for (int i = 0; i < n; ++i)
		if (s[i] == '0') dist[i] = (i ? dist[i-1] : 0) + 1;

	vector<int> dp(n + 2), nxt(n+2, n);
	auto get = [&](int i) { return nxt[i] < n ? dp[nxt[i]] : 0; };

	for (int i = n-1; i >= 0; --i)
	{
		dp[i] = ((dist[i] <= dist.back()) + get(0) + get(dist[i] + 1)) % mod;
		nxt[dist[i]] = i;
	}

	int ans = n;
	if (nxt[0] < n)
		ans = ((long long)get(0) * (nxt[0] + 1)) % mod;
	cout << ans << "\n";

	return 0;
}

Problem idea: Devil

Solution idea: Devil and antontrygubO_o

Tutorial is loading...

C++ solution

#include <bits/stdc++.h>

using namespace std;

template<typename C, typename R = C>
struct dinic
{
	typedef C flow_type;
	typedef R result_type;

	static_assert(std::is_arithmetic<flow_type>::value, "flow_type must be arithmetic");
	static_assert(std::is_arithmetic<result_type>::value, "result_type must be arithmetic");

	static const flow_type oo = std::numeric_limits<flow_type>::max();

	struct edge
	{
		//int src; // not needed, can be deleted to save memory
		int dst;
		int rev;
		flow_type cap, flowp;

		edge(int src, int dst, int rev, flow_type cap, int flowp) :
			/*src(src),*/ dst(dst), rev(rev), cap(cap), flowp(flowp) {}
	};

	dinic(int n) : adj(n), que(n), level(n), edge_pos(n), flow_id(0) {}

	int add_edge(int src, int dst, flow_type cap, flow_type rcap = 0)
	{
		adj[src].emplace_back(src, dst, (int) adj[dst].size(), cap, flow_id++);
		if (src == dst) adj[src].back().rev++;
		adj[dst].emplace_back(dst, src,  (int) adj[src].size() - 1, rcap, flow_id++);
		return (int) adj[src].size() - 1 - (src == dst);
	}

	inline bool side_of_S(int u) { return level[u] == -1; }

	result_type max_flow(int source, int sink, vector<flow_type> &flow_e)
	{
		result_type flow = 0;
		while (true)
		{
			int front = 0, back = 0;
			std::fill(level.begin(), level.end(), -1);
			for (level[que[back++] = sink] = 0; front < back && level[source] == -1; ++front)
			{
				int u = que[front];
				for (const edge &e : adj[u])
					if (level[e.dst] == -1 && flow_e[rev(e).flowp] < rev(e).cap)
						level[que[back++] = e.dst] = 1 + level[u];
			}
			if (level[source] == -1)
				break;
			std::fill(edge_pos.begin(), edge_pos.end(), 0);
			std::function<flow_type(int, flow_type)> find_path = [&](int from, flow_type res)
			{
				if (from == sink)
					return res;
				for (int &ept = edge_pos[from]; ept < (int) adj[from].size(); ++ept)
				{
					edge &e = adj[from][ept];
					if (flow_e[e.flowp] == e.cap || level[e.dst] + 1 != level[from]) continue;
					flow_type push = find_path(e.dst, std::min(res, e.cap - flow_e[e.flowp]));
					if (push > 0)
					{
						flow_e[e.flowp] += push;
						flow_e[rev(e).flowp] -= push;
						if (flow_e[e.flowp] == e.cap)
							++ept;
						return push;
					}
				}
				return static_cast<flow_type>(0);
			};
			for (flow_type f; (f = find_path(source, oo)) > 0;)
				flow += f;
		}
		return flow;
	}

	result_type max_flow2(int source, int sink, vector<flow_type> &flow_e)
	{
		result_type flow = 0;
		std::function<flow_type(int, flow_type)> find_path = [&](int from, flow_type res)
		{
			level[from] = 1;
			if (from == sink)
				return res;
			for (int &ept = edge_pos[from]; ept < (int) adj[from].size(); ++ept)
			{
				edge &e = adj[from][ept];
				if (level[e.dst] == 1 || flow_e[e.flowp] == e.cap) continue;
				flow_type push = find_path(e.dst, std::min(res, e.cap - flow_e[e.flowp]));
				if (push > 0)
				{
					flow_e[e.flowp] += push;
					flow_e[rev(e).flowp] -= push;
					if (flow_e[e.flowp] == e.cap)
						++ept;
					return push;
				}
			}
			return static_cast<flow_type>(0);
		};
		
		for (bool ok = true; ok; )
		{
			int it = 0;
			std::fill(edge_pos.begin(), edge_pos.end(), 0);
			for (flow_type f; ; ++it)
			{
				std::fill(level.begin(), level.end(), -1);
				f = find_path(source, oo);
				if (f == 0)
				{
					if (it == 0) ok = false;
					break;
				}
				flow += f;
			}
		}
		return flow;
	}

	int flow_id;
private:
	std::vector<std::vector<edge>> adj;
	std::vector<int> que;
	std::vector<int> level;
	std::vector<int> edge_pos;

	inline edge& rev(const edge &e) { return adj[e.dst][e.rev]; }
};

const int inf = 25;

struct edge
{
	int u, v, w;
};

mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());

void relabel(int n, vector<edge> &e, int k)
{
	shuffle(e.begin() + k, e.end(), rng);

	vector<vector<pair<int, int>>> adj(n);
	for (auto &i : e)
		adj[i.u].push_back({ i.v, &i-&e[0] });

	vector<edge> ne = e;
	vector<int> id(n, -1);
	id[0] = 0;
	id[n-1] = n-1;
	int sz = 1, esz = k;

	function<void(int)> dfs = [&](int u)
	{
		for (auto &v : adj[u])
		{
			if (v.second >= k)
			{
				ne[esz++] = e[v.second];
				v.second = -1;
			}
			if (id[v.first] == -1)
			{
				id[v.first] = sz++;
				dfs(v.first);
			}
		}
	};
	dfs(0);
	
	for (auto &u : adj)
		for (auto v : u)
			if (v.second >= k)
				ne[esz++] = e[v.second];
	
	for (int i = 0; i < n; ++i)
		if (id[i] == -1)
			id[i] = sz++;
			
	e = ne;
	for (auto &i : e)
	{
		i.u = id[i.u];
		i.v = id[i.v];
	}
}

struct masks
{
	int x, cost;
	vector<int> flow_e;
	
	bool operator<(const masks &o) const
	{
		return cost < o.cost;
	}
};

typedef std::chrono::_V2::system_clock::time_point timepoint;

timepoint get_time()
{
	return std::chrono::high_resolution_clock::now();
}

int get_elapsed(timepoint t)
{
	return std::chrono::duration_cast<std::chrono::milliseconds>(get_time() - t).count();
}

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

	int n, m, k, q;
	cin >> n >> m >> k >> q;
	vector<edge> e(m);
	for (auto &i : e) cin >> i.u >> i.v >> i.w, --i.u, --i.v;

	vector<int> mask(1 << k, -1);
	
	relabel(n, e, k);
	dinic<int> d(n);
	for (int i = 0; i < m; ++i)
		d.add_edge(e[i].u, e[i].v, ((i < k) ? inf : e[i].w));

	vector<int> flow_e(d.flow_id);
	for (int i = 0; i < k; ++i)
		flow_e[2*i] = inf;

	mask[0] = d.max_flow(0, n-1, flow_e);
	
	priority_queue<masks> pq;
	pq.push({ 0, 0, flow_e });
	while (!pq.empty())
	{
		int x = pq.top().x;
		flow_e = pq.top().flow_e;
		pq.pop();
		
		for (int j = 0; j < k; ++j)
			if ((~x >> j & 1) && mask[x | 1 << j] == -1)
			{
				auto n_flow_e = flow_e;
				n_flow_e[2 * j] = 0;
				auto t = get_time();
				mask[x | 1 << j] = mask[x] + d.max_flow2(0, n-1, n_flow_e);
				pq.push({ x | 1 << j, get_elapsed(t), n_flow_e });
			}
	}

	vector<int> cut(1 << k), bit(1 << k);
	for (int i = 1; i < 1 << k; ++i)
		bit[i] = i & -i;

	const int U = (1 << k) - 1;
	while (q--)
	{
		for (int i = 0; i < k; ++i)
			cin >> cut[1 << i];

		int ans = mask[U];
		for (int i = 1; i < 1<<k; ++i)
		{
			cut[i] = cut[i ^ bit[i]] + cut[bit[i]];
			ans = min(ans, cut[i] + mask[U ^ i]);
		}
		cout << ans << "\n";
	}

	return 0;
}

Problem idea: mnaeraxr

Solution idea: mnaeraxr and Devil

Полный текст и комментарии »

Разбор задач Codeforces Round 659 (Div. 1)

Разбор задач Codeforces Round 659 (Div. 2)

#editorial, round 659, cuba contest

+125

dcordb
4 года назад
141

Can this DP be solved in less than O(N * K)

Автор dcordb, история, 6 лет назад, По-английски

Given an array with N elements you need to choose exactly K non-consecutive elements from it such that the sum of the elements will be minimum.

Constraints:

1 ≤ N ≤ 10⁵
1 ≤ K ≤ N
- 10⁹ ≤ a_i ≤ 10⁹

I know how to solve this in O(N·K) by using DP(i, m) = minimum sum that can be obtained in prefix i taking exactly m elements.

Can it be done faster than that? Thanks beforehand.

EDIT: I had an idea that was do a binary search on the answer, let's say answer should be less or equal than x, then for a prefix i the values that m can take so that DP(i, m) ≤ x are consecutive so I transformed the DP to DP(i) = range of the valid m values. But the problem with this is how to merge DP(i - 1) with DP(i - 2) in time.

Полный текст и комментарии »

dp, optimization

dcordb
6 лет назад
9

Help with this problem from NEERC 2008

Автор dcordb, 7 лет назад, По-английски

Hello.

I was doing this problem (problem G) from NEERC 2008.

The problem basically says that you are given N points and in a minute all the points connect to all the others points, forming segments, now you choose the center of each segment as new points an delete the old ones. You have to say the time where one point is generated by at least 2 segments or say is 0 if it never happens.

I realized that if N ≤ 3 the answer is 0, if two different segments cuts in the middle then the answer is 1. And by looking to the solutions I saw that the answer is 2 in other case. Why does this happen, can anybody proof this?

Thank you.

Полный текст и комментарии »

ad hoc, geometry, neerc

dcordb
7 лет назад
2

An interesting graph problem

Автор dcordb, история, 7 лет назад, По-английски

Hi.

This problem is from a Mexican Training Camp of this year, you can find it here (spanish only). The following is the abridged statement:

Given a undirected, simple and weighted graph with N nodes and M edges you should select a set of exactly N edges (you throw away the other M - N edges) and direct each of them, such that the sum of weights will be maximum, with the restriction that each node must have out-degree equal to 1 in this new directed graph; or say that there is not solution.

Constraints:

3 ≤ N ≤ 10000
3 ≤ M ≤ 50000
The weight of edges is between - 10⁵ and 10⁵

Example:

In this case the answer is 19, you can direct edges the following way:

4 to 2 (weight = 16)
2 to 1 (weight = 1)
1 to 3 (weight = -2)
3 to 2 (weight = 4)

I thought about flow but is not clear how I would do the graph, I also thought that N - 1 edges will be from the maximum spanning tree but that is wrong as well. Anyway could you help me?

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graphs, constructive algorithms

dcordb
7 лет назад
2

Long queue on Codeforces

Автор dcordb, история, 7 лет назад, По-английски

Admins have you noticed the long queue of submissions?

Please fix it.

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bug, queue

-3

dcordb
7 лет назад
0

Codeforces not working properly

Автор dcordb, история, 7 лет назад, По-английски

I'm having problems lately with Codeforces interface, by example sometimes the editor doesn't load properly and I can't paste my code in it, also I can't do virtual participation because the form of the time is not working. I also can't write comments on blogs.

I cleaned the cache already and the problems persists. Is anyone else having the same problems?

Admins have you seen this?? I still can't create virtual contests.

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bug

dcordb
7 лет назад
2

Proof of periodic sequence in Enigmatic Device NEERC 2009

Автор dcordb, история, 7 лет назад, По-английски

Hi. I already solved this problem by noticing that the sequence x¹, x², x⁴, x⁸, ..., x^2ⁿ modulo 2010 is periodic with period 10 and pre-period 2. This means that $\text{[math]}$ .

I would like to know a proof for this. Could you help me? You can find the problem statement here. It's the problem E.

Thanks beforehand.

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number theory, eular phi, crt, cycle

dcordb
7 лет назад
1

My solution gets TLE on 712D (Memory and Scores)

Автор dcordb, история, 8 лет назад, По-английски

Hi everyone. My solution to problem 712D gets TLE, even though its complexity is $\text{[math]}$ . Where d is the difference between the score of Memory and Lexa. Note that - 2·10⁵ ≤ d ≤ 2·10⁵.

The funny thing is that in my laptop (Intel Core i5, 4gb ram) my solution runs in 1.2s. What do you think about this, is CF that slow? Anyway could you help me?

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tle, 712d, help

dcordb
8 лет назад
4

COJ Round Contest #8 + Editorial

Автор dcordb, история, 8 лет назад, По-английски

Hello everyone.

I would like to invite you to participate in the 8th COJ (Caribbean Online Judge) Round, this will be a contest with 5 problems, and three hours of duration. The problems will have Div2-like complexity and will be in English and Spanish. The contest will have ACM-ICPC format.

You can find out more about it clicking here. You can also check out the previous rounds here.

UPD1: Time of the contest has changed because of overlapping with Euro Semifinals :). The contest will start at this time.

UPD2: The contest is about to start, get in now!!.

UPD3: The contest is over. Hope you liked the problems.

Hints:

Problem A:

The last player to play wins. Note that this last player can always get what the previous player got in the previous turn plus another positive number. Watch out with n = 1 case.

Complexity: $\text{[math]}$

Problem B:

You can start from the right: i = n, decreasing k with i - 1 while k ≥ i - 1 and append to a vector the number i. Then you can append the rest of the numbers (in increasing order) to the vector.

Complexity: $\text{[math]}$

Problem C:

Note that the graph is a tree with a cycle. The probability to disconnect the graph in one step is $\text{[math]}$ and in two steps is $\text{[math]}$ . So the expected value is $\text{[math]}$ . Here c is the length of the cycle. You can do a dfs to find c.

Complexity: $\text{[math]}$

Problem D:

A number is hyperdrome if it has at most 1 digit with odd frequence. So build a graph where each node is a bitmask representing the parity of the digits's frequence of a number. Add edges between node x (0 ≤ x < 2^b) and nodes $\text{[math]}$ .

Also add a node 2^b that represents the empty set, and add and edge to itself, and to 2¹, 2², ..., 2^b - 1.

Then you can exponentiate this graph to obtain the answer.

The answer will be $\text{[math]}$ . Where a is the adjacency matrix of the graph.

Complexity: $\text{[math]}$

Problem E:

Segment Tree. In each node save the lcp of the range, and a string that is in the range. Then to combine two nodes you can do:

lcp[x] = min (lcp[2 * x], lcp[2 * x + 1], GetLCP(str[2 * x], str[2 * x + 1]))

str[x] = str[2 * x]

Where x is a node in the segment tree and GetLCP is a function that calculates the LCP for two strings. Also you should add lazy propagation to this.

Complexity: $\text{[math]}$

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coj, round, caribbean, contest

dcordb
8 лет назад
11

Long submission queue in CF

Автор dcordb, история, 9 лет назад, По-английски

Has anyone noted that there are 5+ pages of submissions in queue?
Please admins, fix it.

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long queue, bug

dcordb
9 лет назад
0

Looking for a closed formula for summation

Автор dcordb, история, 9 лет назад, По-английски

sin 1 + sin 3 + sin 5 + sin 7 + sin 9 + ...
Any ideas? PD: The angle is in radians.

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summation, sine

dcordb
9 лет назад
6

Bug: Some fonts on CF looking bad on Firefox

Автор dcordb, история, 9 лет назад, По-английски

Hi, i wanted to report that some fonts are looking bad in firefox 43.0. Look at this image:

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bug, huge fonts

dcordb
9 лет назад
4

TLE on Timus 1989

Автор dcordb, 10 лет назад, По-английски

I'm stuck in this problem:
http://acm.timus.ru/problem.aspx?space=1&num=1989

It gives me TLE on test 15, my code uses Hashing and Fenwick Tree, so it's O(mlogn). The time limit is only 0.5s. Is there a better solution? (an O(m) solution).
Thanks in advance.

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palindrome, timus, hashing

dcordb
10 лет назад
8

Help with this apparently dp problem (from POI)

Автор dcordb, 10 лет назад, По-английски

Could you help me with this problem from the 13th POI:
http://main.edu.pl/en/user.phtml?op=showtask&task=ork&con=OI13

It seems some kind of DP, but I'm not able to get it. So far I have an idea: to keep the state of the columns and try to filling them adding a valid rectangle, but this is too slow. Could you help me? Thanks beforehand.

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dp, poi

dcordb
10 лет назад
11

Could you help me with this strings problem?

Автор dcordb, 10 лет назад, По-английски

This is a problem from COCI 2003, here I left you the link. The problem name is VOKI (it is the third in the pdf).

http://hsin.hr/2003/national/second_day/juniors/problems.pdf

So far I have a DP approach that works in O(n * m * max(n, m)) which is too slow, could you help me to improve this?

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dp, strings, subsequence

dcordb
10 лет назад
3

Could you help me with this problem?

Автор dcordb, 10 лет назад, По-английски

Given a sequence of N numbers (a1, a2, ..., an) you have to find the minimum number of operations to get a consecutive group of size K such that all the elements in this group have the same height. To achieve this you can increase or decrease the height of any element by 1, any number of times, each modification costs 1 operation.

Constraints: 5 <= N <= 10^6 3 <= K <= N 0 <= ai <= 10^6

Example: N = 6 K = 4 10 4 5 2 5 7

Here the solution is a group from position 2 (1-based) to position 5 (1-based), which costs 4 operations.

So far I have a solution in O(Nlog^2(N)) using a binary search + segment tree, but this timed out. Could you help me to improve this?

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data structures

dcordb
10 лет назад
5