Will there be an editorial after the contest?

Will there be a live scoreboard?

[Deleted]

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omg IOI rules official round

you are right

We can't see the standings during the contest?

GL & HF for you all guys!

please upvote this comment, thanks.

Goodluck everybody. Hope you're doing well.

After the contest, solutions and statements will be available in this site: https://inf-open.ru/?lang=en

And you can also see the Qualification stage of Open Olympiad 2023-2024 in that site.

If you want to upsolve problem for Qualification round go here: https://codeforces.com/gym/104922

And for editorial visit this link: https://inf-open.ru/2023-24/zaoch-materials/

Although i could only solve one problem,i had a wonderful time.

It was a good contest, so could you open standings?

How to solve D? I have a naive solution O(N^2*W) solution , which will fail eventually.

How to solve D? I have a naive O(N^2*W) solution , which will fail eventually.

I need editorial for problem D :(, I really liked the problem but I only came up with a $$$O(n^2W)$$$ solution and I got 56 pts.

A is an enjoyable problem :)

btw when will the standings be public

Can anyone explain why this solution for D gives WA on group 2 test 12?

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

const int N = 205, M = 505;
int n, sm, ans, a[N];
vector<vector<vector<int>>> f[2];

int dfs(int L, int R, int Al, int Bl) {
    if (L > R) return 0;
    if (Al && Bl) return dfs(L, R, Al - min(Al, Bl), Bl - min(Al, Bl));
    int& ret = f[!Al ? 0 : 1][L][R][!Al ? Bl : Al];
    if (~ret) return ret;
    ret = 0;
    if (!Al && Bl) {
        ret = 0;
        ret = max(ret, dfs(L + 1, R, a[L], Bl) + a[L]);
        ret = max(ret, dfs(L, R - 1, a[R], Bl) + a[R]);
    } else if (Al && !Bl) {
        ret = 1e9;
        ret = min(ret, dfs(L + 1, R, Al, a[L]));
        ret = min(ret, dfs(L, R - 1, Al, a[R]));
    } else {
        if (L == R) ret = a[L];
        else {
            int LL = dfs(L + 2, R, a[L], a[L + 1]) + a[L];
            int LR = dfs(L + 1, R - 1, a[L], a[R]) + a[L];
            int RL = dfs(L + 1, R - 1, a[R], a[L]) + a[R];
            int RR = dfs(L, R - 2, a[R], a[R - 1]) + a[R];
            ret = max(ret, min(LL, LR));
            ret = max(ret, min(RL, RR));
        }
    }
    return ret;
}

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    int mx = *max_element(a + 1, a + n + 1);
    f[0].assign(n + 5, vector<vector<int>>(n + 5, vector<int>(mx + 5, -1)));
    f[1].assign(n + 5, vector<vector<int>>(n + 5, vector<int>(mx + 5, -1)));
    sm = accumulate(a + 1, a + n + 1, 0);
    ans = dfs(1, n, 0, 0);
    cout << ans << " " << sm - ans << endl;
    return 0;
}

Can anybody explain the solution of problem D? I thought if they play optimally then the minimum difference of their sum will be the answer. So following this I used dynamic programming on O(n*W) but it is showing wrong answer from group 3 to 5

HOW TO SOLVE THAT TREE PROBLEM B? ANY HINTS PLEASE....

Can someone suggest a case for this? link

Second test group fails, but the worst case i could make runs locally in 0.6 seconds.

how to solve C?

Can someone share their full AC code for D? Is your complexity better than $$$ O(n*W) $$$.

how to solve day2 C?

D is too short code this is my code : https://ideone.com/HwO0wc

May someone help me with C question Day 1. With my solution I was able to get only 14 points by passing group1 testcases and on rest testcases some are passing some are giving WR and time limit exceed. Also I can't think of an idea to make it fast. Here is my Code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <stdbool.h>
int main(){
    // int p;
    // scanf("%d", &p);

    // while(p--) {

        int n, k, t;
        scanf("%d %d %d", &n, &k, &t);

        int temps[n];
        for(int i = 0; i < n; i++) scanf("%d", &temps[i]);

        int arr[n * k];
        int count = 0;
        for(int i = 0; i < k; i++) {
            for(int i = 0; i < n; i++) arr[count++] = temps[i];
        }

        int i = 0;

        int ans = 0, prev = -1;
        int temp[k];
        memset(temp, 0, sizeof(temp));
        int cnt = 0;
        int len = n * k;
        int numGift = t;

        while(i < len) {
            while(i < len && numGift > 0) {
                // printf("i : %d\n", arr[i]);
                bool flag = true;
                for(int j = 0; j < cnt; j++) {
                    if(arr[i] == temp[j]) {
                        flag = false;
                        break;
                    }
                }

                if(flag) {
                    temp[cnt++] = arr[i];
                    --numGift;
                }

                prev = arr[i];

                while(prev == arr[i]) ++i;
            
                if(numGift == 0) {
                    ++ans;
                    memset(temp, 0, sizeof(temp));
                    numGift = t;
                    cnt = 0;
                }
            }

        }

        if(numGift != 0 && numGift < t) ++ans;

        printf("%d", ans);

        printf("\n");
    // }

}

B is Dp ?

Will we have a merged scoreboard?

the scoreboard of day2 is not fully rejudged, please fix this sir Ormlis

edit: fixed

How to solve B day2 ?

Will there be official scoreboard and editorials in the near future?

is there any editorial? Ormlis