### NALP's blog

By NALP, 8 years ago, translation, , Hi to all!

## 215A - Bicycle Chain

Because of small constraints we can iterate i from 1 to n, iterate j from 1 to m, check that b j divide a i and find max value of . Then we can start this process again and find amount of the pairs in which is max value.

## 215B - Olympic Medal

Let amagine we have values of r 1, p 1 and p 2. Then:  r 2 2·(B·p 1 + A·p 2) = r 1 2·B·p 1 Now it’s easy of understand, we must take maximum value of r 1 and p 1, and minimum value of p 2 to maximize r 2. You could not use three cycles for iterating all values, but you could find property above about at least one value and use two cycles.

## 215C - Crosses

Let’s iterate n 1 = max(a, c) and m 1 = max(b, d) — sides of bounding box. Then we can calculate value of function f(n 1, m 1, s), and add to the answer f(n 1, m 1, s)·(n - n 1 + 1)·(m - m 1 + 1), where two lastest brackets mean amount of placing this bounding box to a field n × m.

Now we must calculate f(n 1, m 1, s). At first, if n 1·m 1 = s then the result of the function is 2·(⌊ n 1 / 2⌋ + 1)·(⌊ m 1 / 2⌋ + 1) - 1 (you can prove it to youself).

If n 1·m 1 > s then we must to cut 4 corners which are equal rectangles with square . We can iterate length of a one of sides, calculate another side, check all constraints and add 2 to the result of the function for all such pairs of sides.

The time of a solution is O(n 3), but it’s with very small constant, less then 1.

## 215D - Hot Days

You can use only two features about this problem: the solution is independenly for all regions and there are only 2 possible situations: all children are in exactly one bus or organizers must take minimum amount of bus such no children got a compensation. It’s bad idea to place some children to hot bus and to place some children to cool bus simultaneously.

For solving you must calculate this 2 values and get minimum.

## 215E - Periodical Numbers

Will be soon…  Comments (12)
 » Why in problem D (Hot Days) you only need to consider the two situations mentioned? Thank you.
•  » » because the cost is a linear function about the number of buses
•  » » I also can't understand the solution
•  » » » Supose that we will arrange n buses in ith city： If T[i]-t[i]>m/n , no children will get compensations.If T[i]-t[i]
•  » » » » Thanks a lot. I get it now :)
 » 8 years ago, # | ← Rev. 2 →   I didnt get the solution for Crosses. Here f(n1,m1,s) denotes the number of crosses that can be created with area s and n1 = max(a,c) & m1 = max(b,d). But why are we adding f(n1,m1,s).(n-n1+1).(m-m1+1) to the answer ?
•  » » Because we can additionally position the center of the cross in (n - n 1 + 1)·(m - m 1 + 1) ways on the given grid.
 » So will you write the solution for the last problem too?
 » Why the solution to problem E still hasn't been published?
 » alas even after 8 years the sol of E is not uploaded lol
 » I think the formula for volume in problem B here is incorrect
•  » » It correct D = m/V; so m = D*V;