Hi!
Topcoder SRM 750 is scheduled to start at 11:00 UTC -5, Feb 8, 2018. Registration is now open for the SRM in the Web Arena or Applet and will close 5 minutes before the match begins.
Editorials: https://www.topcoder.com/blog/single-round-match-750-editorials/
Problem Setter: misof
SRM 750 will award you double the TCO19 points you would generally receive.
This is the second SRM of Stage 3 of TCO19 Algorithm. This stage will also help you qualify and win reimbursements/tickets to TCO19 Regional Events.
Stage 3 TCO19 Points Leaderboard | Topcoder Java Applet | Next SRM 751 — Feb 21
All the best!
Why are the points doubled?
We had some feedback from the members during the last stages that if they are 3-4 points behind they stand no chance to make it to the finals.
Thus an opportunity like this will help give everyone a chance to cover and end up on the top or on top 10 to get to TCO19 Round 4. While those who are at the top can consolidate their position.
Also, stage 3 points are to help regional members qualify/earn tickets/reimbursements to travel to TCO19 Regional Events which are to take place in Japan, India, China, and South America. Thus an opportunity for regional members to get some extra points in for TCO19 Regional Events Leaderboard. And those who are not competing regularly can put in some extra effort to compete and get points to cover up to get to Regional Events.
Another interesting opportunity for those not on the very top currently — 2nd and 3rd position(if they haven’t qualified for the finals yet) from this Stage will get a fixed reimbursement to fly to their nearest regional event :)
If there are 3 SRMs per month I guess there will be 9 matches in Stage 3. Approximately, for how many rounds are you planning to double the points?
Right now, we are only doing it for this SRM. After some feedback and analysis around the stats, we will plan more around the next steps.
how can I register?
I made a terrible mistake by trying to participate in this round from web arena. After I fixed a bug in my 1000 (just 1 symbol) and clicked compile, it apparently didn't save the code automatically. When I looked at it one more time, the bug was still there.
(P.S. posting since my 1000 already was challenged)
Jacob whoever challeneged ur solution must be a beast who saw one rare symbol ? who was he .
1) It was tourist who challenged this solution 2) It shouldn't be really hard to see 3) I also left lot of debug output (because I was debugging until the last seconds) that must have TL'd everything in the end.
How can we see the testcases for problems?
or at least the failed test?
or at least the failed submission verdict?
You may submit in the practice room once it's up.
and then ?
I resubmit my (failed in contest )code in practice now, and just I got 499.99 points.
There is "Run System Test" in the menu. You need to click it to see real score.
Thanks
Off the top of my head, this should work: Division summary -> right click your handle and select "History" -> double-click row where solution failed. (This will show only a prefix if the I/O is large.)
This surely works: The full report can later be found in the problem archive. Go to round summary, find yourself, click the [*] next to your name and choose the problem.
Wow!! Never knew the history thing existed. Topcoder is really mysterious. But anyways, thanks for this.
Thanks
Does pretests contain only the given sample tests ? (1 question got hacked and 2 questions failed on system tests :(( )
I think there are no pretests, you can submit empty code and it'll be accepted until hacked by a pretest.
yes , no pretests are there , only sample test . even it is not checked .
Auto comment: topic has been updated by hmehta (previous revision, new revision, compare).
hmehta bhai itna fast editorial .. haha ! aisa laga bomb leke hi baithe the chorne ke liye
hey misof in problem 1 of div1 :
sample case 2 is :
{10, 11, 12, 1, 2, 3}
3
Returns: 10
but after 10 attempts we are left with 0 , 1 , 2 , 1 , 2 , 3 (we initially filled 10 bags with candies of type 1 , 2 and 3 )
now we can make more attempts like filling one bag with type 2 ,3 and 4 . so ways = 11 and we are now left with 0 , 0 , 1 , 0 , 2 , 3
again fill by type 3 , 5 , 6 . ways = 12
now array becomes 0 0 0 0 1 2
again fill bag by 1 candy of type 5 and 2 of type 6 so ways = 13 ?
Note that the bags should be identical. Candies of each index is of different type. So in each bag you create there should be same number of candies of each type. In the example you mentioned, 11th bag doesn't contain any candy of type 1 (1-indexed), while the initial 10 bags did. So, it's not a valid bag.
I found an even simpler way for div1 500, sadly due to me not copying the formula correctly for the sum from statements I couldn't submit in time: You just need to store the depths of each value in a map<int,int> D, then use D.lower_bound(S[i]) to find the two nodes S[i] is inbetween (if not already present). Due to the properties of tree_successor, either right child of smaller one or left child of higher one is free. And the one with the free child has bigger depth.
EDIT: To further clarify:
TreeSuccessor is either:
leftmost node in right child subtree we go down and that leftmost node will surely have left child null and be deeper than predecessor
if no right child we go up while we are coming from left_child, but this means right child is null currently and we are deeper than successor so we insert there