Why did the Rated Range is 0-2799 some days ago but 0-1999 now?

Why is the color above bule ?

Why is the colour above blue? (I don't like blue,so do anybody know how to switch it back?)

why can't I participate as a rated participant?

F is a very good expectation question.

D is a trash idea

Segmented sieve is new for me. Pretty cool approach!!

Problem D is nice. The way to solve it is so ingenious.

Someone please help me with question C :( I tried greedy with binary search, it didn't pass 12/32 cases.

I am pretty sure that problem D or something very similar has already occur somewhere with much stricter constraints.

I solved it today with MCMF on a graph: s -> v of capacity 2 cost 0, (v + n) -> t capacity 2 cost 0, and u -> (v + n) capacity 1 cost (1 if there were no such edge initially and -1 otherwise). Answer == (m * 2 + mincost) / 2.

However, this approach is kinda hard to prove, because it is not necessary that if edge (u -> (v + n)) is taken to the flow, then (v -> (u + n)) is also taken. If the graph is directed, then there is no issue, problem solved. However, it also works for undirected(i mean that the answer is correct, not that the flow is built in a way that all edges are taken in pairs). But i cannot prove it. Can anyone?

In problem F editorial I can't understand anything. Can someone, please, explain it more adequately?

About geminipromc: is he an AI?

D in $$$ O(n!.n log(n)) $$$ Link

I think E is very hard.I noticed that the prime solving,but I got a TLE...........Because my algorithm for factoring prime factors is not good.......

But I think D is an easy problem , It just use a DP

C is a problem to practice thinking,I got the solution in 18min。I use a Binary search and greedy to solve it.

can any one please explain how to cross the 4th problem barrier at atcoder.......

i am getting WA in C submission any testcase where it will fail

I haven't seen a problem which can search to get the full score for a long time.

I think I can say that, here we go again, for today's problem F.

:D https://mirror.codeforces.com/blog/entry/142558#comment-1272566

It seems that ABC really likes this kind of dp, which asks to find out max or min of some expectation.

appreciate any help i could get about my WA on C!

void solve() {
    int N; cin >> N;
    deque<int> S(N); for (auto& x : S) cin >> x;

    int curr = S.front(); S.pop_front();
    int end = S.back(); S.pop_back();

    sort(all(S));

    S.push_front(curr);
    S.push_back(end);

    int answer = 2;
    for (int i = 1; i < N; i++) {
        if (2 * curr >= end) break;

        if (2 * curr < S[i]) {
            curr = S[i - 1];
            answer += 1;

            if (2 * curr < S[i]) {
                cout << -1 << endl;
                return;
            }
        }
    }

    if (2 * curr >= end) cout << answer << endl;
    else cout << -1 << endl;
}

https://atcoder.jp/contests/abc412/submissions/67211725

thank you in advance!

In F, if two drawers have the same no. Of socks, the expectation should be the same, right??

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
constexpr int mod = 998244353;

int binpow(int a, int b, int m)
{
	int res = 1;
	a = a % m;
	while (b != 0)
	{
		if (b & 1)
		{
			res = (res * a) % m;
		}
		a = (a * a) % m;
		b = b >> 1;
	}
	return res;
}

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

	vector<int> a(n);

	int sum = 0;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
		sum += a[i];
	}
	sum++;
	sum = sum % mod;
	c--;
	a[c]++;


	sort(a.begin(), a.end());

	int pp = 1e18;
	int sum1 = 0, kk = 0;
	map<int,int> dp;
	int z = (sum - 1 + mod) % mod;
	for (int i = n - 1; i >= 0; i--)
	{
		sum1 += a[i];
		int y = (sum - sum1) % mod;
		//cout << y << " " << a[i] << " " << z << endl;
		int mm = (y * binpow(z, mod - 2, mod)) % mod;
		mm = (1 - mm + mod) % mod;
		//cout << kk << endl;
		mm = binpow(mm, mod - 2, mod);
		//cout<<((1 + kk) * mm) % mod<<endl;
		dp[a[i]] = ((1 + kk) * mm) % mod;
		int oo = (a[i]) % mod;
		oo = (oo * binpow(z, mod - 2, mod)) % mod;
		kk = (kk + (dp[a[i]] * oo) % mod) % mod;
	}

	cout << dp[a[c]] << endl;

	return;
}

signed main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int t = 1;
	// cin >> t;

	while (t-- != 0)
	{
		solve();
	}
	return 0;
}

also if someone can tell me what's wrong in this edit: got it I think I need to sleep

Can somebody explain me the problem E please i tried to follow the editorial but that didn't helped me . Somebody please explain in simpler terms Thankyou

For problem C,I wanna to use priorityqueue to solve ,but it doesn't work for 5 tests,and I can't find the mistakes...

import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while (t-- != 0) {
            solve(in);
        }

    }

    public static void solve(Scanner in) {
        int n = in.nextInt();
        int[] s = new int[n];
        for (int i = 0; i < n; i++) {
            s[i] = in.nextInt();
        }
        if (s[0] * 2 >= s[n - 1]) {
            System.out.println(2);
            return;
        }
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        for (int i = 1; i < n - 1; i++) {
            pq.offer(s[i]);
        }
        if (pq.isEmpty()) {
            System.out.println(-1);
            return;
        }
        int starth = s[0];
        int endh = s[n - 1];
        int ans = 2;
        while (!pq.isEmpty()) {
            int g = -1;
            while (!pq.isEmpty() && pq.peek() <= 2 * starth) {
                g = pq.poll();
            }
            if (g != -1) {
                ans++;
                starth = g;
                if (starth * 2 >= endh) {
                    System.out.println(ans);
                    return;
                }
            } else {
                System.out.println(-1);
                return;
            }
        }
    }
}