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Why not thank me for participating?

Excited and hopefully the problem's statement will be short and precise just like the announcement! <3

Grey testing ?

Grey testing?

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RAMADAN contest starts 15:35UTC+2. Please

Hmmm you didn't thank us lol

do not have a point of 1900 or higher in the rating. So div 2 or div 3 ?

date is wrong

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Ramadan>contest

By the way thanks for the March 19's div3 its on the right time

why is there such a big gap (20 days) between 2 contests?? Will more contests be added in between?

UPD: Another contest added

Hopefully more contests gets added up in between.

What does trusted participant mean?

First contest during Ramadan!

I hope the tasks will be interesting

And Good Luck for every parcipicant!

Sorry Codeforces, I have to attend night-prayer at that time. so, No contest for me for the next 30 days.

спасибо за раунд

WOW! natalina,Sasha0738,vladmart wrote Codeforces Round 933 (Div. 3) and Codeforces Round 833 (Div. 3)

I hope to become specialist after this round.

"adventures of a restless mathematician"

Mathforces Incoming. Brace yourselves

In the last few days, why Div3 contest has been hosted so frequently than before?

exciting to become an Expert!

GL everybody :)))

I hope to solve as many problems as possible :DD

Hope reach pupil after this round

Good luck for everyone !!!!

as an old green all i can say is do not be afraid of losing rating

Why no thanks for gray writers/testers?

Negative delta, HERE WE GO

Negative delta, HERE WE GO

gl everyone :>

Can I use a chatbot during the exam? :))

thanks for information;

i hope i return the pupil on this div.3

Excited to give this contest on my birthday!!!

If I become pupil during Ramadan, I will convert to Muslim from Hindu

Hope to solve only A. B TO G will be so hard.

Good luck!

RAMADAN mubark :)

Expert please

Rudolf and Bernard? Game theory on problem E?

I wish this contest i could improve the score ！ because i have lose four times till now

I wish this contest i could improve the score ！ because i have lose four times till now

Problem E is one of the worst problems i have ever read

Boring problems, contest was able to be better

E has an unusual distance definition and it has not been presented in the russian version for more than half an hour

I spent more time in B than problems E and F due to my stupid mistake.

bad B, C, D, E, F, G!!!!

Binary search doesn't work for F ?

My private Screencast.. A,B,C,D,E Video Solutions here: https://www.youtube.com/watch?v=hSNp4xnn9lc

Screencast with commentary (set video quality to HD to be able to read my code)

Problem G is nice.

E seem very hard (at least for me) are there other simple solutions not involving dp with segment tree or sparse table?

Can someone explain the 4th test case in problem D.

Like how is it 2 3 5 and not 1 2 3 5?

RIP all who got WA on test 2 in F :(

I don't know which case I was missing. Can anyone debug my submission : https://codeforces.net/contest/1941/submission/250809941

Got trolled by problem F (2 wrong submissions) because (a[i] + a[i+1]) / 2 overflows int :(

Let this be a lesson to use std::midpoint... (or in Java, a[i] + (a[i+1] - a[i]) / 2)

I could only get TLA on test 3 for F, which trick was needed to reduce complexity ? FFT ?

For problem G I made a simple Dijkstra where the cost is determined by the size of a std::set which contains all the colors of the edges I've been through. Does anyone have an idea why it doesn't pass Test 3?

#include <bits/stdc++.h>
using namespace std;

#define el '\n'
#define int int

struct edge {
    int dst{0}, c{0};
};

struct pqel {
    set<int> cost{};
    int node{0};

    bool operator<(const pqel& i) const {
        return this->cost.size() > i.cost.size();
    }
};

inline void solve()
{
    int n, m; cin >> n >> m;
    vector<edge> adj[n + 1];

    for (int i = 1; i <= m; i++) {
        int u, v, c; cin >> u >> v >> c;
        adj[u].push_back({v, c});
        adj[v].push_back({u, c});
    }

    int b, e; cin >> b >> e;

    vector<bool> visited(n + 1);
    priority_queue<pqel> pq;
    pq.push({{}, b});

    while (!pq.empty()) {
        auto t = pq.top(); pq.pop();
        if (t.node == e) {
            cout << t.cost.size() << el;
            break;
        }
        visited[t.node] = 1;

        for (auto child : adj[t.node]) if (!visited[child.dst]) {
            set<int> s = t.cost; s.insert(child.c);
            pq.push({s, child.dst});
        }
    }
}

int32_t main()
{
    int t;
    cin >> t;
    while (t--)
        solve();
}

Excuse me guys, but who can help me find mistake in D?

My answer is different with correct answer in only one number, and I coulnd't find mistake.

Here is my code:

void solve() {
	int n, m, x;
	cin >> n >> m >> x;
	vector<int> ans;
	int r; char c;
	cin >> r >> c;
	if (c == '0') {
		ans.push_back((x + r)%n + 1);
	} else if (c == '1') {
		int frm = (x-r+n)%n;
		ans.push_back(frm + 1);
	} else if (c == '?') {
		ans.push_back((x + r)%n + 1);
		int frm = (x-r+n)%n;
		ans.push_back(frm + 1);
	}
	for (int i = 0; i < m-1; i ++) {
		cin >> r >> c;
		if (c == '0') {
			for (auto el : ans) {
				el = (el + r)%n + 1;
			}
		} else if (c == '1') {
			for (auto el : ans) {
				int frm = (el-r+n)%n + 1;
				el = frm;
			}
		} else if (c == '?') {
			vector<int> v;
			for (auto el : ans) {
				v.push_back((el + r)%n + 1);
				int frm = (el-r+n)%n + 1;
				v.push_back(frm);
			}
			ans = v;
		}
	}
	set<int> res;
	for (auto i : ans) {
		i --;
		if (i == 0) {
			res.insert(n);
		} else {
			res.insert(i);
		}
	}
	cout << len(res) << nl;
	for (auto i : res) {
		cout << i << " ";
	}
}

I will be very grateful to anyone who offers a helping hand

Can anyone give me some insights over the Solution of Problem — F Rudolf and Imbalance

Is there anyone ignore the "consecutive" in problem E like me ._.

I skipped E as F seemed easier yet failed on test2 , I have no clue where.

here's my submission — click here

The code is clean, any help appreciated

G was a nice problem

Can't help but point out that C has more ACs than B tho..

WT compilation error !!

Can't wait to become Expert!

I couldn't implement it in time, but I just wanted to make sure my strategy is correct for E.

Let row_ans[r] represent the minimum cost for some row. We can use prefix sums to calculate some contiguous row_ans and choose the minimum in O(n-k) time. The minimum will be the answer.

As for calculating row_ans[r], we use dp to calculate the minimum cost from dp[r][c] to dp[r][m]. row_ans[r] = dp[r][1].

Simmply apply dfs over dp[r][c] and check the minimum cost between dp[r][c] and dp[r][c+1], dp[r][c+2], ..., dp[r][c+d+1].

Is there anything more/wrong?

someone tell the approach of PROBLEM D

Again missed cyan by 2 points. Rip my life

Again missed cyan by 2 points. Rip my life

(haven't given this contest but) wanted to ask which topics do i need to learn to solve div 3/4's F and g?

I tried F instead of E as seemed easier but failed on test2.

here's the submission (code is clean)
any help is appreciated.

"Note that the penalty for the wrong submission in this round is 10 minutes." Does it mean I can't submit for the next 10 min? or is it anything else?

How to do F??Any hints??

Is there anyway to solve problem E with memoization? Just curious.

Problem A — Simple sort and binary search.

Problem B — Note that 1st element can only be modified by operations on element 2. Accordingly iterate over the array, simulate the operations on each, and check the last 2 elements for 0. In any case, element value goes < 0, answer is not possible.

Problem C — Simple substring search. First, count for mapie (requires only 1 operation). Then check for rest of the possiblities (pie, map).

Problem D — Let dp(i,j) represents whether all the m turns result in player j having the ball, if we are at move i and currently player j has the ball. Do a simple recurive dp, and final answer is the count of dp[i][m] == 1 over all i players.

Problem E — For each row, let dp(i) represent the min. cost to place supports, such that we place a support at a[row][j]. dp(i) can be calculated as min j from i-d-1 to i-1. Use segment tree to speed up this calculation. For every row, cost for that is dp(m). Final answer is a subarray of size k with minimum sum, that can be done using simple prefix sums.