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

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. Then we can start this process again and find amount of the pairs in whichUnable to parse markup [type=CF_TEX]

is max value.## 215B - Olympic Medal

Let amagine we have values of

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,*p*

_{1}and

*p*

_{2}. Then:

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Now it’s easy of understand, we must take maximum value of

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and*p*

_{1}, and minimum value of

*p*

_{2}to maximize

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. 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

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andUnable to parse markup [type=CF_TEX]

— sides of bounding box. Then we can calculate value of functionUnable to parse markup [type=CF_TEX]

, and add to the answerUnable to parse markup [type=CF_TEX]

, where two lastest brackets mean amount of placing this bounding box to a field*n*×

*m*.

Now we must calculate

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. At first, ifUnable to parse markup [type=CF_TEX]

then the result of the function isUnable to parse markup [type=CF_TEX]

(you can prove it to youself).If

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then we must to cut 4 corners which are equal rectangles with squareUnable to parse markup [type=CF_TEX]

. 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

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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?