Score distributing looks interesting, I'll target 4 problems then if we r lucky enough to get math problem on D kekw :)

As a participant, I want to thank you for the contest!

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After my own participation in this contest, I found your statement wrong.

You know what it means right !!!!!!. It's now time to solve four questions this time. Hahaha

I was at that once. You can do it!

Do not think about it during contest. I repeat don't. Just solve problems with accuracy. It's ez.

Legendary grandmasters have more experience than anyone else. They will probably be much better at pointing out things like if a problem has appeared before, if there exists a stupid data structure brute force, if the authors' intended greedy is incorrect and the problem is actually unsolvable, etc.

orz

By testing Problems !

That's the spirit dude

Congrats on the green!

Sir you graph is so inspiring.

Transition to Expert Round? o.O

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⡀⣀⣀⣀⣀⣀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⣠⣴⣞⠉⠉⠀⠀⠀⠀⠀⠀⠀⠈⠉⢑⣶⣤⣀⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⣤⣶⣾⣿⣿⣿⣿⣧⠀⠀⠀⠀⠀⠀⠀⠀⠀⠰⣿⣿⣿⣿⣿⣶⣤⣀⢀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⢀⣤⣴⣤⣤⣤⣤⣤⣤⣤⣤⣤⣴⣶⣿⣿⣿⣿⣿⣿⣿⣿⣿⣥⣤⣤⣤⣤⣤⣤⣤⣤⣤⣤⣽⣿⣿⣿⣿⣿⣿⣿⣿⣷⣦⣤⣤⣤⣤⣤⣤⣤⣤⣤⣤⣤⣤⣄ ⠈⠻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⠋ ⠀⠀⠀⢹⣿⣿⣿⠛⠛⠛⠛⠛⠛⠛⢛⣛⣻⣟⣛⠛⠛⠛⠛⠛⠛⠛⢻⣿⣿⣿⣿⡟⠛⠛⠛⠛⠛⠛⠛⢛⣛⣿⣛⣛⠛⠛⠛⠛⠛⠛⠛⠛⣿⣿⣿⡏⠀⠀ ⠀⠀⠀⢘⣿⣿⣿⡀⠀⣠⠋⠉⣣⠾⣿⣿⣿⣿⣿⡟⠢⡀⠀⠀⠀⠀⣼⣿⣿⣿⣿⣿⠀⠀⠀⢠⠊⡩⢿⣿⣿⡀⠀⠀⠙⢦⡀⠀⠂⠀⠀⢸⣿⣿⣿⠀⠀⠀ ⠀⠀⠀⠀⢿⣿⣿⡇⠀⣇⠀⢰⣹⢁⣿⣿⣿⣿⣿⣿⣀⢹⠀⠀⠀⢠⣿⣿⣿⣿⣿⣿⣇⠀⠀⣇⡼⠁⣸⣿⣿⣷⣥⣀⠀⠀⢱⠀⠀⠀⠀⣾⣿⣿⡇⠀⠀⠀ ⠀⠀⠀⠀⢸⣿⣿⣷⠀⠈⠒⢺⣧⣼⣿⣿⣿⣿⣿⣿⣿⣿⠇⠀⠀⣼⣿⣿⣿⣿⣿⣿⣿⡀⠀⠈⣷⣿⣿⣿⣿⣿⣿⣿⣆⠀⢸⠀⠀⠀⢀⣿⣿⣿⠆⠀⠀⠀ ⠀⠀⠀⠀⠈⣿⣿⣿⣔⠀⠀⠘⢿⣿⣿⣿⣿⣿⣿⣿⣿⡟⠀⠀⢠⣿⣿⣿⣿⣿⣿⣿⣿⣧⠀⠀⠙⣿⣿⣿⣿⣿⣿⣿⣿⣧⠏⠀⠀⠀⣼⣿⣿⡿⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⢹⣿⣿⣿⣦⡠⠀⠀⠛⢿⣿⣿⣿⣿⡿⠋⠀⠀⣠⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣦⡀⠁⠘⢿⣿⣿⣿⣿⣿⠟⠁⠀⠀⢀⣼⣿⣿⡿⠃⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠙⣿⣿⣿⣿⣶⣤⣤⣤⣬⣽⣽⣁⣠⣦⣴⣾⣿⣿⣿⣿⣏⢉⣭⣽⣿⣿⣿⣿⣿⣶⣤⣤⣈⣩⣭⣭⣥⣤⣤⣴⣾⣿⣿⣿⠟⠁⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⡏⠙⠿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⢿⣿⣿⣿⣼⡏⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⢿⠇⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⢇⠀⠀⠀⣭⡙⠛⠛⠛⠛⠛⠛⠛⠛⠉⠉⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠉⠉⠛⠛⠛⠛⠛⠛⢛⣟⠛⠉⠁⠀⢸⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⢸⡀⠀⠀⠙⠁⠀⠀⠀⠸⣿⡆⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢠⣶⠀⠀⡸⠿⠀⠀⠀⠀⡏⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⢧⠀⠀⠀⣤⡀⠀⠀⠀⠈⠀⠀⠀⢀⣴⣦⣤⠤⠤⠤⠤⠤⠤⠤⢤⣤⣶⣤⡀⠀⠀⠘⠋⠀⠀⠀⠀⠀⠀⠀⣸⠁⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠈⢇⠀⠐⠙⠃⠀⠀⠀⠀⠀⠀⠀⢻⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⢹⣿⣿⡧⠀⠀⠀⠀⠀⠀⢠⡄⠀⠀⣰⠃⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⢦⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠉⠙⣇⠀⠀⠀⠀⠀⠀⠀⠀⣼⠋⠉⠀⠀⠀⠀⠀⠀⠀⠉⠀⠀⣰⠃⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠷⣄⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠋⠳⠶⠶⠶⠖⠒⠋⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⡜⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠢⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⡴⠋⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠘⠳⣄⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣠⠔⠋⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠉⠲⢤⣠⣀⣀⡀⠀⠀⣀⣠⡀⠀⡀⢀⡀⢀⣀⣦⣤⡾⠟⠋⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢰⣾⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⡖⠂⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠙⠛⠻⠿⠿⠿⣿⣿⣿⣿⣿⣿⣿⠿⠿⠿⠛⠋⠉⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

No, this contest has no theme

Please someone do help .I even registered from that account around 20 hours ago .Please help

Yeah I also initially misunderstood the question wrong so that you can change the order of the candy types. Fortunately they then clarified.

Price tags can only cover adjacent equal values. 1 and 5 are not adjacent.

Even in the example the values reached are [4,4,2,4,4] and the answer is 3 not 2

Opposite, couldn't solve D but got the idea for C fairly quickly

c is just gcd and lcm on continous segments. D is kadanes

Mine too, but it's passed

yeah, I initially thought so too, but I reread and they specifically had [4, 4, 2, 4, 4] example.

Yes and the given constraints made the problem impossible to solve if you can change the order.

Yes, I thought so too, but then the announcement help me to understand!

same here,but I didn't read announcement or examples :)

Just greedy in a somewhat stupid way. It is easy to find this way but it is not instantly obvious why it works.

If every element is equal to zero -> no answer.

Otherwise, group elements in two vectors: positive+zero and negative.

You build the final array as follows: Keep a variable sum denoting the sum of all elements used up to this point.

if sum > 0 you append the next negative value to the final array

if sum =< 0 you append the next positive or zero to the final array

Largest absolute value of a subarray sum is the largest absolute difference between any two prefix sums. Let max equal the largest value and let min equal the smallest value of the array. You can notice that if currently sum > 0, the next prefix sum will satisfy sum > min and if currently sum <= 0, the next prefix sum will satisfy sum <= max. This means that all prefix sums satisfy min < sum <= max. The largest possible prefix sum is thus max and the smallest possible is min + 1. The absolute difference of these is less than max - min which satisfies the problem statement.

Thanks all above :)

Take a look at Ticket 16778 from CF Stress for a counter example.

No.

bruh you took an array of 50,000 elements instead of 50,001. Also it is not advisable to create that big of an array for each test case, instead use maps.

Copied fom my previous comment:

If every element is equal to zero -> no answer (trivial to prove).

Otherwise, group elements in two vectors: positive+zero and negative.

You build the final array as follows: Keep a variable sum denoting the sum of all elements used up to this point.

if sum > 0 you append the next negative value to the final array

if sum =< 0 you append the next positive or zero to the final array

Largest absolute value of a subarray sum is the largest absolute difference between any two prefix sums. Let max equal the largest value and let min equal the smallest value of the array. You can notice that if currently sum > 0, the next prefix sum will satisfy sum > min and if currently sum <= 0, the next prefix sum will satisfy sum <= max. This means that all prefix sums satisfy min < sum <= max. The largest possible prefix sum is thus max and the smallest possible is min + 1. The absolute difference of these is less than max - min which satisfies the problem statement.

I didn't lol. It just kinda made sense :/

This is the Erdos-Ginzburg-Ziv Theorem.

You don't need factorization at all.