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…

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]<k/n , ans=cost[i]*n+{m-(T[i]-t[i])*(n-1)}*x[i] =cost[i]*n+m*x[i]-(T[i]-t[i])*n*x[i]+(T[i]-t[i])*x[i] ={cost[i]-(T[i]-t[i])*x[i]}*n+(T[i]-t[i])*x[i]

cost[i],x[i],T[i],t[i],m is known,and it's a linear function about "n".So the MinCost must appear on the endpoints.

Thanks a lot. I get it now :)

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?