We will hold Tokio Marine & Nichido Fire Insurance Programming Contest 2023（AtCoder Beginner Contest 304).
- Contest URL: https://atcoder.jp/contests/abc304
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20230603T2100&p1=248
- Duration: 100 minutes
- Number of Tasks: 8
- Writer: cn449, leaf1415, evima, kyopro_friends, physics0523
- Tester: kyopro_friends, math957963
- Rated range: ~ 1999
The point values will be 100-200-300-400-475-525-625-625. We are looking forward to your participation!
Is there a long queue now? 30 minutes passed and my submission is still waiting for judging.
Finally get a WA after 39 minutes...
Don't DDoS AtCoder Please !!! Please !!! Please !!!
so much long queue :(
is there any issue with the judge? my solution to D is judging for 15mins, yet not started!
IOI-OI Format :(
In fact, it isn't judging. It's still "Waiting for Judging".
Yes. My submission of F(21:08) was "WJ" for about 10 minutes.
Waiting for 35mins :(
Till how much time may I expect for the verdict? :(
What happened to the queue? Waiting for my results on F...
Mine has been waiting for 20 mins.
It was difficult to connect AtCoder until 21:06(JST), so I could't reach rated.
Submitted Question E, but it's still in the queue. Will they consider it?
extended by 20 minutes
Extended time!
How does extending the contest time helps with WJ?
what's the point of extending time when solutions are still not judged?
wait if solutions are still taking a lot of time to judge how will extending contest time help? isn't the damage already done?
In such cases, at least do 1 optimisation that if on any test case verdict is not AC, then don't judge the further test cases
This will take way less time and avoid such a long queue.
well to be honest this had to be done regardless of bad queue condition or not
Actually, earlier this issue was not there, so in such cases it is good, we get to know how many TC got AC, WA, TLE, etc.
As far as queue is small this strategy is good, but in these cases not.
You're correct in that point, but still we are not contesting in a point-per-testcase system, so I didn't really feel much of a need of that statistic. I think showing the testcase number where the failed verdict first happened (just like CF does) is already enough.
in case u got 1 WA and 1 TLE, then you will try to cover the edge case as well as optimise your code in the next submission, otherwise you will be doing at least 2 submissions.
I think it is a very good idea! There should be two modes: the normal mode and the conjestion mode. The conjestion mode should not judge further when it meets a testcase without an AC verdict. Besides would you please add some system monitors like Prometheus?
Will today's contest become unrated?This evaluation speed has had an impact on me with a score of 114514.
Does problem D require some particular Algo/data structure? I thought about a binary search sol but would have been too slow in the end. Any hints?
Binary search
mmm ok... Like u can compute the number of strwb. between two horiziontal lines in nlogn and the same for vertical lines in nlogn using bs. But then? In order to put together the two informations and get the # of strwb in each piece u need up to n^2. Which won't pass. Am I loosing something?
This contest is a compensation for the unrated ABC299.
But I still think that it would be very unreasonable for this competition to be rated.
The same company who sponsored ABC299 and ABC304 is DDoS-ed again. Both contests are made unrated.
yep, and through all the unrated contest Atcoder had decided, I lose at least 80 ratings.
void solve_(){
ll w, h; cin >> w >> h; ll n; cin >> n; map<pair<ll,ll>, ll> gather_points; vector<pair<ll,ll>> points; for(ll i = 0; i < n; i++){ ll p, q; cin >> p >> q; points.push_back({p, q}); } set<ll> x_axis; set<ll> y_axis; ll a; cin >> a; for(ll i = 0; i < a; i++) { ll x; cin >> x; x_axis.insert(x); } x_axis.insert(w); ll b; cin >> b; for(ll i = 0; i < b; i++) { ll y; cin >> y; y_axis.insert(y); } y_axis.insert(h); for(auto [p, q] : points){ ll x_point, y_point; auto it_x = x_axis.upper_bound(p); x_point = *it_x; auto it_y = y_axis.upper_bound(q); y_point = *it_y; gather_points[{x_point, y_point}]++; } ll mn = 4e18, mx = -4e18; for(auto el : gather_points){ mn = min(mn, el.second); mx = max(mn, el.second); } ll total_points = (a + 1) * (b + 1); if((ll) gather_points.size() < total_points) mn = 0; cout << mn << " " << mx << "\n";}
What's wrong in this code? Please can someone help!
Problem D?
Your solution is correct, just use
lower_boundinstead ofupper_boundand it should work as we also need to consider the points on the rectangle's boundary.No need to consider borders, the constraints discard this happening.
i used lower bound and it still fails. Also boundary checks aren't neccessay since 0 < pi < W and 0 < qi < H.
My bad you guys are correct, there is no need to check boundary points.
I think it's not working because of this:
it should be
mx = max(mx, el.second);Bruh, i have this intense anger right now, this queue really irritated me coz half of the time i was looking for the verdict instead of actually debugging my code
I don't think there exists any reason to be unrated...
I lose 65 ratings...
Problem G is interesting and the solution is reminiscent of Codeforces 1616H Keep XOR Low. Thanks!
I learned a lot from editorial of problem F. Thanks to the problem writers. Here is my solution, which is a little different from the editorial.
For example, if n=12, then we should consider t=1,2,3,4,6. But, in fact we don't need consider t=1,2,3, since for t=1,2, all the results must be included in t=4, while for t=3, they must be included in t=6. Therefore, it is enough to only consider t=4,6. In general, assuming that n has prime divisors, p1,p2,...,pk, we only need consider n/p1,n/p2,..,n/pk. Moreover, integers <=1e5 have at most k=6 distinct prime divisors.
Next, we could use the inclusion-exclusion principle to solve it. Assume that d1=n/p1,d2=n/p2,..,dk=n/pk, then the final answer is f(d1)+f(d2)+...+(-1)*f(gcd(d1,d2))+(-1)*f(gcd(d1,d3))+..+f(gcd(d1,d2,d3))+...
The total complexity is O(2^k*n).
It turns out that one of my AC submissions on problem D is wrong, and it fails on the following test.
Is it still possible to add any test now? If yes,can you add it?
Are ratings updated for ABC 304 ?
No,because the round is unrated.