441A - Valera and Antique Items

Problem author gridnevvvit

You need to implement what written in statement. You could act like that: let's calculate q_i — minimum item price from seller i. Then if q_i < v, we can make a deal with seller i, otherwise we can't.

Jury's solution: 6850474

441B - Valera and Fruits

Problem author gridnevvvit

Let's start counting days from 1 to 3001. Let current day be i. Additionally, we'll have cur variable — number of fruit we didn't collect previous days. Suppose now fruit is ripen current day. If now + cur ≤ v, we need to add now + cur to answer and update cur value (cur = 0). Otherwise we add v to answer, but cur value need to be updated as follows. Let tv = max(v - cur, 0). Then cur = now - tv. In other words, we try to collect fruits that will not be collectable next day.

Additionally, problem could be solved with , but this is not required.

Jury's solution: 6850502

Bonus. Suppose fruit can be collected at days a_i, a_i + 1, ..., a_i + T_i, where T_i — some number for each tree. How to solve this task optimally?

Additionaly, for every day there will be its own v (maximum number of fruit collected).

441C - Valera and Tubes

Problem author gridnevvvit

The solution is pretty simple. First we need to make such route that visits every cell exactly one time. It is not difficult:

1. Initially we stay in (1, 1) cell. Moving from left to right, we should reach (1, m) cell.
2. Move to the next line, in (2, m) cell. Moving from right to left, we should reach the most left sell of 2nd line, (2, 1).
3. Move to the next line. Repeat 1. and 2. while we have not all cells visited.

After that, we can easily find the solution: you can make first (k - 1) tubes length be 2, and the last k tube will consist from cells left.

Jury's solution: 6850508

441D - Valera and Swaps

Problem author danilka.pro

In this task you should represent permutation as graph with n vertexes, and from every vertex i exists exactly one edge to vertex p[i]. It's easy to understand that such graph consists of simple cycles only.

If we make swap (i, j), edges and will become edges and respectively. Then if i and j is in the same cycle, this cycle will break:

but if they are in different cycles, these cycles will merge into one:

this means that every swap operation increases number of cycles by one, or decreases it by one.

Assuming all above, to get permutation q from permutation p, we need to increase (or decrease) number of cycles in p to n - m. Let c — number of cycles in p. Then k always equals |(n - m) - c|.

For satisfying lexicographical minimality we will review three cases:

1) n - m < c

It's easy to understand, that in this case you must decrease cycles number by merging cycles one by one with cycle containing vertex 1. This way every swap has form (1, v), where v > 1. Because every cycle vertex is bigger than previous cycle vertex, this case can be solved with O(n).

2) n - m > c

In this case you should break cycle for every vertex, making swap with smallest possible vertex (it should be in this cycle too). This could be done if represent cycle by line . As soon as every cycle is broken with linear asymptotics, this case solution works with O(n²).

Bonus: this way of representing cycle lets us optimize solution to asymptotics, you may think how.

3) n - m = с

Besause in this case k = 0, there is nothing need to be swapped.

It's highly recommended to inspect jury's solution: 6850515

441E - Valera and Number

Problem author gridnevvvit

We will solve the task by calculating dynamic d[i][mask][last][cnt] — possibility of getting v which 8 last bits equals mask, 9th bit equals last, cnt — number of consecutive bits (following 9th bit) and equal to last, after i steps.

Good, but why we left other bits? It's clear, that using operation  +  = 1 we can change only first 0 bit with index  ≥ 9.

Transitions is pretty obvious: we add 1 or multiply by 2 (it's recommended to see them in jury's solution). Perhaps, you should ask following question. For example, we have number x = 1011111111 in binary representation.

And at this moment, we make  +  = 1. According to all above, we must go to d[1][0][1][2] condition, but we can't do that because we don't have any information about 1 in 10th position. But, as we can not change any bit with index  ≥ 9 (mask = 0) we make transition to d[1][0][1][1].

Jury's solution: 6850523

Bonus. Let us have other pseudocode.

// input x, k, p
 
for(i = 0; i < k; i += 1) {
   if (x is even) {
     rnd = random number from interval [1, 100]
     if (rnd <= p)
       x *= 2;
     else
       x += 1;
   } else {
      x *= 2;
   }
}
 
s = 0;
 
while (x is even) {
  x /= 2;
  s += 1;
}
 

As before, you must find expected value of s.

How effectively you can solve this problem? Can you prove your solution?

Your corrections of my bad English are welcome, thank you.