No xD

That gives us 1min 48sec per problem if we spend the same amount of time on each problem. Talk about the ultimate speedforces .....

If there are 67 problems,how will they be labelled.AA after Z?((((

what will happen if it really has 67 problems?

just use chinese. Letters won't end

A01, A02, ..., A67 works perfectly fine

It is 6 or 7 or 67 so it becomes 6

What does "but" mean?

вместо того чтобы блокать просто ответь

я хотел написать что у тебя нет доказательств про курсы питон,с++,2 место в олимпиаде и т.д. также если заедает enter то почему в лёгких задачах по типу A,B у тебя всё норм, а в задачах другой сложности такие сильный подозрения на ИИ. Я собрал где-то 10 посылок с 5 последних раундов и в скором времени хочу выложить пост по поводу этого. Надеюсь на бан явного читера. хаххахахах сидит на середине алгоритмов по питону и говорит полный курс питона и с++.Я знаю о тебе больше чем ты думаешь

CLIST has predictions for recent problems

contribution farming

67 problems in 2 hours is unfair. Because 2 hours=120 mins and 120/67 ~1:47 mins. we have 1:47 mins to solve each problem..

from +67 to -69 :)

++

It's basically:

How did you solved c?

For E, say you have x strings with A ones and y string with B ones

Construct any permutation, let's say 1,2,3,4...n construct the strings how many have A ones and how many have B ones? If the answer is not x and y, there is no solution

Otherwise Now we know that x of the strings are to be placed in x known positions, but we don't know how many valid ways there are to do this, and we know y strings have y places but don't know how to distribute them either

You can try any way, if it works, then all ways work, and the answer is A! * B! otherwise, no way will work and the answer is 0

If he have some strings with A ones, some with B ones, some with C ones, etc... Then the answer is either A! * B! * C! *... Or it is 0, we just repeat the same process by constructing our own strings from a random permutation and checking whether we get the same number of ones in string

To prove that if one way works then all work, and one way doesn't work then none of them work, Notice that if we swap 2 strings, let's say representing bits x and y then numbers in the permutation with both bits x and y set, or both unset, remain unchanged

Numbers with x set and y unset or vice versa, they just swap places (try to trace some examples to see this) so if a number is < 1 or > n or a duplicate, therefore making the array not a permutation it doesn't disappear by swapping around strings, so the array remains not permutation

For E, first it is clear that each of the given strings must end up at a position that has the correct number of bits (i.e. the bitcount of the string in the $$$k$$$'th position must be equal to the number of numbers at most $$$n$$$ that have the $$$k$$$'th bit set).

The key observation now is that if any such arrangement satisfying this property produces a valid permutation, then all such arrangements will also satisfy such an arrangement. To prove this, let $$$(S_1, \ldots, S_k)$$$ be a valid arrangement, and note that every other valid arrangement can be generated by repeatedly swapping $$$S_i$$$ and $$$S_j$$$ that have equal bitcount. None of these swaps will flip the validity of the arrangement (after each swap, all numbers stay distinct, and it's not hard to check that none of the numbers end up exceeding $$$n$$$). Thus, if any one arrangement works, they all work, and the answer ends up being the product of several multinomial coefficients.

Solution to $$$E$$$.

We don't know the order of numbers. Let's try to sort them from $$$1$$$ to $$$n$$$ starting from greater bits.

For the first (greatest) bit do the following. First we have to choose the row, which will correspond to it. The only criteria is the number of bits in the row: they have to be equal. For example, for $$$n = 6$$$ the greaters bit's row has to have $$$3$$$ bits: for numbers $$$4$$$, $$$5$$$ and $$$6$$$. Let's multiply the answer by number of such rows. And then swap one of these rows with the last row.

Next we need to reorder columns (numbers themselves), such that bits in newly placed row are in correct positions. Let's swap columns, so that they placed correctly.

And this was the first iteration.

Assume input:

n = 6
011001
100011
010110

In this case we need to place at bottom the row with exactly $$$3$$$ bits. All rows satisfy this. Thus multiply answer by $$$3$$$ and move one of the satisfied rows to bottom (but we don't need it here).

Next, we need to have the following picture:

xxxxxx
xxxxxx
000111

We can do this by swapping second and sixth columns:

011001
110010
000111

Then we proceed to the next bit. We do the same, but now we can't swap columns, if one of them is $$$\le 3$$$ and the other is $$$\ge 4$$$, otherwise it will break the last row. Those boundaries split row by segments. We need to find the row, which on every segment has the same number of bits, as the required segment. For example, now the second row has to be $$$011001$$$. So we need row, which on segment $$$[1,3]$$$ has two bits, and on segment $$$[4, 6]$$$ has one bit. Again multiply answer by number of such rows. The rest is the same.

ig, depends on your implementation

like? (because i was stuck on the 7th test case for an hour)

It's lowk not that bad

How you solved C?

This round's C is harder than the others' one.

bro why did my fail , Got WA on D (Goods on the Shelf) with a greedy idea and trying to understand the flaw.

My approach was:

• Count contiguous segments for each value.
• Mark values appearing in multiple segments as "bad".
• If there are too many bad values, answer NO.
• Otherwise try to repair one bad value with a single swap and verify the final array.

For example:

1 1 2 3 2 3

Here:

• 2 appears in segments [2] and [4]
• 3 appears in segments [3]and [5]

So bad values are {2,3}.

I then try swaps involving occurrences of a bad value and check whether the resulting array has every value in a single contiguous block.

This passes samples but fails hidden tests.

What is the main flaw in this reasoning? can you give counterexample that could fail here?

D is one of the worst problems in the codeforces I have seen.

I found it one of the most fascinating problems I have seen till now, although I couldn't solve it in contest.

can you try this testcase?

[1, 0, 0, 1, 0, 0, 4, 3, 2, 2]

you can transform it into

[1, 1, 0, 0, 0, 0, 4, 3, 2, 2]

I also thought it was too simple but there were so few ACs and there wasn't much time, so I thought I must be missing something and I didn't even try to submit the implementation :<

Btw, I learned today that directly using cin to input bitset has complexity $$$O(\text{max_len})$$$ thus fails on multitest.

Hey, can you explain your approach?