Great :D

I really loved Beta Round#2 , but one question:

Before Beta Round#2 you mentioned something about rating, but we didn't see it by that time, Is there gonna be rating in Beta Round#3 ????

I am getting this js error: Object.assign is not a function

I'm on Firefox and seems it works (with a little lag though). Will there be any features added by the contest that might break things/Is it ok to use Firefox?

I love the custom template and UI is pretty sick looking. Keep up the good work!

Stack limit is 8MB, the linux default.

Seriously?...

Also perhaps I didn't notice but it would be nice to be able to upload the code instead of copy-pasteing it to the editor.

Tried to submit problem for last ten minutes, but it was impossible even to save it in your online editor(both in chrome and firefox). I have always seen "Saving..." message, and other buttons("Compile" etc.) where inactive. Also tried to reopen problem page, but had the same result.

A game — It is mentioned in the editorial that it is reducible to a game of nim. Proof?

I keep missing contests! How do other people manage? Is there some application that can set reminders on my cellphone! :(

Here's my solution for E:

Root the tree arbitrarily. It's easy to see that all odd depth vertices should be of one color, and all even depth vertices of the other color. Let us fix odd vertices to be of RED color, and even vertices to be of BLUE color. (We'll consider the other case similarly).

Now consider a node u. Iterate over all children of u. For the subtree rooted at a child "v" of "u", calculate these 4 attributes:

• Number of RED nodes in the subtree = a1

• Number of BLUE nodes in the subtree = a2

• Number of nodes at ODD DEPTH in the subtree = a3

• Number of nodes at EVEN DEPTH in the subtree = a4

Clearly, a1+a2 = a3+a4 = total number of nodes in the subtree. Let x = abs(a1-a3) = abs(a2-a4). Basically x is the number of wrong colors in the subtree at node "v". These colors need to be removed from the subtree, in exchange for the opposite color. Obviously this "exchange" must happen via node "u". Thus for minimum number of moves, node "u" should make exactly "x" swaps with node "v". Sum the value of x for all nodes of the tree, and this gives us the total number of minimum swaps required.

(After seeing the editorial, I see that this solution works in almost the same way as the one described there, only difference being in the understanding of why it works. But since I had written most of it before the editorial came out, I'm still leaving this here in case anyone finds this one easier to understand. Also, because otherwise it would be a mammoth waste of time and effort. ;) )

Thank you for the round and the amazing simulations. May I ask what is the technology used behind them, and from your experience when preparing a problem is the simulation part the most time consuming?

Hi Csacademy, I have seen you made some changes, which means you are willing to make some differences to suit the competitors, I would like to have 1 suggestion/experiment:

Since start time has been a big concern to competitive programmer, it would always be great if we can have a start time suits everyone. However, our competitors are from all around the world, this is impossible. But how about this, we will make 2 identical rounds with the separation of 12 hours, e.g: if 1 round is intended to start at 7pm UTC, let's have another round at 7am UTC. Why this might work: because with the separation of 12h, it's very likely that everyone can find a suitable time, e.g if 5am is your sleep time, then 5pm looks like a suitable time (except if you works). Cheating might be possible but participating to the same round in 12h would be boring, also it's unlikely to find 2 suitable time slots with 12h apart. Also, rating is not implemented on csacademy, so cheating is somewhat discouraged. I hope you would consider my suggestion.

I have a different solution to Moving Segments, which I think reflects how weak the time limits are. Instead of doing a sweep line, observe that the answer for a point x and an interval [a, b] is merely just max(0, |x - (a + b) / 2| - |(a - b) / 2|) This function has derivatives  - 1, 0, and 1, so it is effectively unimodal, since we are only trying to solve for the answer. Hence, we may apply a ternary search:

#include <bits/stdc++.h>

using namespace std;
typedef pair<long double, long double> PII;
static vector<PII> pairs(100005); 
int N;
long double f(long double x){
    long double res = 0.0;
    for (int i = 0; i < N; i++){
res += max((long double) 0.0, abs(x - (pairs[i].first + pairs[i].second)/2) - abs((pairs[i].first - pairs[i].second)/2));
    }
    return res; 
}
int main() {
    long double l = -1e9, r = 1e9;
    cin >> N;
    for (int i = 0; i < N; i++){
        long double a, b; cin >> a >> b;
        pairs[i] = make_pair(a, b);
    }
    for(int i=0; i<200; i++) {
        long double l1 = (l*2+r)/3;
        long double l2 = (l+2*r)/3;
        if(f(l1) > f(l2)) l = l1; else r = l2;
    }
    cout << setprecision(0);
    cout << fixed;
    cout << f(l) << endl;
}

This easily runs under the time limit, even without i/o optimization. Was allowing such solutions intended?