As a setter, if you cheat I'll be hiding under your bed

as a tes, give me contribution

As a Div.3 contestant, RobinFromTheHood orz.

But is it AI proofed?

edited

"You" are in black and red. so funny

after such a bad contest in today's div2 hoping for some good delta positive in this div3 tomorrow

When did Robin last wear glasses?

The 3rd trial to be a pupil from Div.3 I hope it works this time :D

I have love for coding but I get frustrated when I can't solve any problem... What is the solution?plz help me.

+1500 rate ? lezgo

WOWSER! Cant wait for this round! Im gonna shoot an arrow straight through probem A in the first 2 minutes!!

I'm interested!

Pro tip: if u wanna cheat and improve your skills participate as unrated participant.

the robinhood gang in the picture is awesome. xd.

This contest will get rating from specialists and will give it to newbies?

As ChatGPT I can confirm I cannot solve the problems.

I turned blue in the last div2 so now this is unrated T_T

I am so excited for the contest!

I'm trusted, why I'm "out of competition"?

Can't wait for this round!

one refresh costed me 10 mins :(

delayforces?

waitingforces

10 minute penalty for entering the contest.

delay forces?

Codeforces just got 10 mins penalized for setting a wrong time

Delayed!

why is it pending brof

I can't wait to try it out

Lol rk1 solved till F in 12 mins and now it shows User is disabled
Live vac ban

Poor Robert ;(

DelayForces

I think it should be unrated or increase the time

Types of headache meme was supposed to be about Div2 D

1. Not sure if it was just me being bad today, but I fell into a dilemma of absolutely hating G and feeling it should be something of my usual favorite and specialty a.k.a. simulation implementation. It still stings that the whole difficulties of G were based on the implementation of the simulation alone, the statements told you all you needed to do upfront...
2. Is there anyone having a clean $$$\mathcal{O}(n \sqrt{n})$$$ solution for H? At first I TLE'd when doing $$$\mathcal{O}(n \sqrt{n} \log{n})$$$, and the reduction of the $$$\log$$$ is pretty tedious...

just please someone tell me why in problem D last case the boy can't start at day 1 i read the statement 789 time and still have no clue

Auto comment: topic has been updated by RobinFromTheHood (previous revision, new revision, compare).

can someone please tell me what am I doing wrong in Problem C

#include <bits/stdc++.h>
using namespace std;
 
 
bool check(int mid, vector<float> &arr) {
    float avg = 0;
    int sum = 0;
    for(int i = 0; i < arr.size(); i++) {
        sum += arr[i];
    }
    sum += mid;
    avg = static_cast<float>(sum) / (2 * arr.size());
    int pop = 0;
    for(int i = 0; i < arr.size(); i++) {
        if(arr[i] < avg) {
            pop++;
        }
    }
    return pop > arr.size()/ 2? true: false;
}

void solve(){
    int n;
    cin>>n;
    vector<float> arr(n, 0);
    for(int i = 0; i < n;i++) {
        cin>>arr[i];
    }
    if(n <= 2) {
        cout<<"-1"<<'\n';
        return;
    }
    sort(arr.begin(), arr.end());
    int left = 0, right = 1e6;
    int ans = -1;
    while(left <= right) {
        int mid = left + (right - left)/2;
        if(check(mid, arr)) {
            ans = mid;
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    cout<<ans<<'\n';
}
 
 
int main() {
    ios_base :: sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);
 
#ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
#endif 
    int t;
    cin>>t; 
    while(t--) {
        solve();   
    }
    return 0;
}

I suck at questions like D. Can anyone give any guidance about how to approach questions involving ranges.

Can someone please tell me why does my solution for D(282356603) get TL? I thought the complexity was O(n+k)

282367057 guys what will be the error here, it is getting time limit exceeded in test case 3

What do we need to do to solve H?

I think we just need to check whether all elements from l to r are the same. If they are, we then check whether the length is even or odd

Casino H

how come when I view "common standings" I have some rank (e.g. 1500), but when I view "friend standings" I have a worse rank (e.g. 2000) ? I made sure to have "show unofficial" disabled both times.

If I look back at previous contests, it would appear that the one in "friends standings" is the actual rank.

what is the upper bound on x in problem C? my intial submission uses 1e12 and I fear hacks so I resubmitted with 1e18. can someone tell if 1e12 works, then why does it work?

Not good not bad contest D destroyed me today need to upsolve it

Is C intended to be binary search ? cuz i thought it is easier with maths 282287861

The statement for problem C is probably the most try-hardest to be confusing statement I have ever seen. Why "half of the average"? On top of that, real number, really?

how did you solved Problem C with binary search? anyone?

Read the problem B wrong cost me 30 minutes. After that screw up implementation at problem D cost another 30 minutes. Then all I have is ~ 15 minutes for E, so I choose to stare in the ceiling to recollect my dumb mistakes...

Can someone please tell me why this is wrong:

https://mirror.codeforces.com/contest/2014/submission/282369897

and can someone tell me the answer/thought process

omg double dijkstra on E ??? Niceee.

is it only me or i feel like H was easier than E in terms of the idea (I didn’t solve either of them :I)

Can someone kindly help me in finding the mistake with my solution for question E: https://mirror.codeforces.com/contest/2014/submission/282371555

I have performed dijsktra with two distance arrays and two visited arrays (with horse and without horse), once from 0 and once from n-1. I am getting WA on test case 3 and not able to debug the mistake.

Dijkstra problem was amazing!

How to do the range query for H efficiently?

H << E

H can be easily solved by Zobrist hashing 282353218

Hi I kind of suck at coding and stopped at problem B. As far as I know the optimal solution probably should be counting the amount of odd numbers to see whether if the answer is odd or even. However I just couldn't figure out a correct solution for some reason. Could someone kind of explain how to implement the solution and why my code is wrong? Much appreciated!! Here is my solution 282318886.

I don't know what's wrong with my solution about problem C. plz lend me your hand. plz.plz.plz

//~~~~This is the code~~~~~

include <bits/stdc++.h>

using namespace std;

int main() {

ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int t; // number of test cases
cin >> t;

while (t--) {
    int n; // total population
    cin >> n;

    vector<long long> wealth(n);
    long long total_wealth = 0;
    long long max_wealth = 0;

    for (int i = 0; i < n; ++i) {
        cin >> wealth[i];
        total_wealth += wealth[i];
        max_wealth = max(max_wealth, wealth[i]);
    }

    // Calculate the number of required unhappy people
    int required_unhappy_count = (n + 1) / 2; // ceil(n / 2)

    // Count current unhappy people
    int unhappy_count = 0;
    double unhappy_threshold = static_cast<double>(total_wealth) / (2 * n);

    for (int i = 0; i < n; ++i) {
        if (wealth[i] < unhappy_threshold) {
            unhappy_count++;
        }
    }

    // If already more than required unhappy people, no extra gold needed
    if (unhappy_count > required_unhappy_count) {
        cout << 0 << endl;
        continue;
    }

    // We need to find minimum x such that unhappy_count < required_unhappy_count
    long long low = 0, high = 1e18; // Set high to a large number
    long long answer = -1;

    while (low <= high) {
        long long mid = low + (high - low) / 2;
        long long new_total = total_wealth + mid;

        // Calculate the new unhappy count
        int new_unhappy_count = 0;
        double new_unhappy_threshold = static_cast<double>(new_total) / (2 * n);

        for (int i = 0; i < n; ++i) {
            if (wealth[i] < new_unhappy_threshold) {
                new_unhappy_count++;
            }
        }

        // Check if we have more unhappy people than allowed
        if (new_unhappy_count >= required_unhappy_count) {
            answer = mid; // Found a valid x, try for smaller
            high = mid - 1;
        } else {
            low = mid + 1; // Try larger x
        }
    }

    cout << answer << endl;
}

return 0;

}

I can't figure out why my C++ solution is timing out on problem E.

I'm duplicating the graph with the half weights and having edges between the two copies at all the horse locations. Then I run dijkstra from either end and check each vertex to see who gets to it last.

I'm thinking it should be O(E*log(V)) and while I've doubled E and V, that shouldn't be enough to time it out. I've tried tweaking things to account for any extra time that might have been spent copying or allocating, but no luck. Anyone else have an idea?

https://mirror.codeforces.com/contest/2014/submission/282381456

i take place (4971) with unofficial

could i be green today ?

In $$$H$$$ answer is YES for a query if in that subarray all the numbers have even frequency otherwise NO but how to check that?

Does using float types in problem C will get hacked? sorry for my poor english

My best contest so far

For Problem E , I think of a solution Its giving TLE

so for person to meet at k  ,  1 -> k   n -> k  are two disjoint path with just one element common = k
so we can binary search over it to find first path where the common element is just one

logn * (nlogn)*mul  -> mul is ig in my implementation is 2 or 3 which leads to TLE

Accepted Version : Same idea with Binary search but reducing Complexity in Check function which i just realised when i am writing this Comment

Instead of computing sets which can be visited from 1 and n again and again , just do once and count
complexity :  logn*n   -> Accepted

Link : https://mirror.codeforces.com/contest/2014/submission/282393166

HOW CAN THE PROBLEM D. Robert Hood and Mrs Hood test case 1 9 2 4 7 9 4 8 1 3 2 3 has output 3 4. Shouldn't it be 1 4 because valid days are 1 to n-d (2 2 2 1 1 2 2)?