Since 6+10+5+9=30, and since "Contestants with at least 30 points at the end of this round will advance to Round 1." then yes, I think so :)

EDIT: I had misread sample as visible test cases, so my comment was completely wrong. Apologies. :(

No.

As stated in Code Jam Rule:

4.3 Judging and Scoring.
...
"Visible Test Sets" are test sets for which the outcome of judging will be reported during a Round.

Visible test set can contain testcases other than sample.

It's enough to qualify after solving visible test sets for problem A and B.

Read the editorial.

Here's my solution:

The key idea I used was to think of the queries as sending $$$N$$$ $$$F$$$-bit numbers rather than $$$F$$$ $$$N$$$-bit strings. So, for example if you sent the following $$$3$$$ queries:

$$$1010$$$

$$$0100$$$

$$$1101$$$

Think of it like you are sending $$$4$$$ numbers, $$$5$$$, $$$6$$$, $$$1$$$, $$$4$$$ in that order. Here each number is represented by a vertical column, where the $$$i$$$th row determines the bit in the number $$$i$$$th from the right. So, first row corresponds to $$$2^{0}$$$, second row corresponds to $$$2^{1}$$$ and so on. Read the judge's response in the same way.

This makes the solution for $$$F = 10$$$ quite simple, as now you can send $$$N$$$ $$$F$$$-bit numbers, you just make each worker output its ID, that is, output $$$0$$$, $$$1$$$, $$$\ldots$$$, $$$1023$$$ in that order. Now, the judge will return some $$$N-B$$$ among these numbers and you just output the $$$B$$$ missing numbers as the answer.

Now, notice you haven't used $$$2$$$ details yet: 1. The numbers returned are returned in the order they were given. 2. $$$B$$$ cannot be more than $$$15$$$.

This allows us to let each worker output its ID modulo $$$16$$$ rather than the exact ID. So, your output looks like:

$$$0$$$, $$$1$$$, $$$\ldots$$$, $$$15$$$, $$$0$$$, $$$1$$$, $$$\ldots$$$

And the judge will return these numbers in the same order with atmost $$$15$$$ numbers missing. Now, notice each group of $$$0$$$, $$$1$$$, $$$\ldots$$$, $$$15$$$ contains $$$16$$$ numbers, so no group can go completely missing. So, in the returned numbers, if a number is less than or equal to the number before it, that means a we are in the next group. In this way, we can uniquely determine the ID's of the workers returned and thus output the ID's of the remaining workers. As, we output only $$$4$$$-bit numbers, we can solve this with $$$F = 4$$$.

Implementation

Bonus: it's possible to solve D2 with $$$F = 4$$$.

(AwakeAnay have really nice solution, my idea is way more complicated)

In the new platform, the editorial for each problem is in a tab above the problem statement and right below the last updated label

Download 2 scripts provided and run the command like the following

python3 interactive_runner.py python3 testing_tool.py 1 -- ./D

In most cases you can write your own emulator. Look: https://pastebin.com/SGGz9uza

Here is a scoreboard parser: gcj-2018-scoreboard-api It's a node application. Works pretty good for my usage.

Google Code Jam 2019 Qualification Round Stats

I just used BigInt implementation in C++ , here is the Code

My Friend Handle in CF : julianfernando Submissions : His submissions in Qual GCJ 19 He solve using string :)

Click on the contestant's name to see their all submissions.

"we do not even have to know a bevy of DP flux algorithms, whatever those are!". Google's Words!

link

Well, it is not correct to say that they don't have a usable site.

The last year's website was more or less fine, and the two years ago and below site was even better.

Nah. According to Pablo Heiber, whether a site is usable or not is subjective. Also, the Google UI people don't think usable sites are good for newcomers. /sarcasm

I decided to just not participate. Distributed isn't there, interface changes are a pain and it simply isn't worth my time.

Who cares about adding some Clojure, Dart, PHP and whatever

*outdated versions of Clojure and PHP (not to mention C++14, Java 8, python 3.5, etc)