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Basically, its easy to count the number of triples you can form with tree numbers between X and Y considering triples as sets (i mean considering (1,2,3) the same as (2,1,3) ). Then you only have to count the number of those triples that can not form a triangle. This means the numbers of triples x ...

I havent code it yet, but I think its also solvable ( althougth not too efficient ) with max = min...

If you have <= 5000 bottles ( each with no empty space, just giving you the integer amount of wine they can take ), and you consider a matrix mat[5000], you can get
in O(n^2) which amounts of wine ...

Hi!, its not the first time i see this problem, and its not the first time i dont see how to solve it... any hints? thanks! Eric.

Well, i think many of us had the same problem. Someone can say that there was a clarification, but as it was written ( "There's an empty line after each test case." ), i misunderstood the clarification, and i thought that there was a line after each case in the input. Anyway, it was a nice contest ...

Probably yes, map is too slow for this problem, it will pass the TL if you use sorted array's and binary search. gl!, Eric.

When there are two points, redefine the dist between then as max(diffx/2, diffy/2). This is the size of the largest square.

Very near, the redefined distance ( the one should give us the size of the largest square ) betwen two points p1 = (x1,y1) and p2 = (x2,y2) is dist(p1,p2) = max(abs(x1-x2 ...

Thank you very much!. Eric.

Hi, this problems seems like a Min-Cost-Max-Flow problem, but the cost of the sources complicate everything, can anyone give me a hint? Thanks, Eric.


PD: if it is not right to post this problem here i will delete it.
PD2: where there any tricky cases for the problem 11403 ( Binary Multiplication ).

I dont know tricky cases... but my program now gives:
gl! Eric.

My AC program gives the same output. gl! Eric.

Hi joshi!, im not sure, but i think there may be leading blank spaces. gl! Eric.

Another hint?? i cant figure it out, i just see that you can get an acyclic graph with the component digraph.... :'(... Thanks!, Eric.

Well, i just wanna ask something, i did the "correct" solution ( at least the idea ) during the contest, but i thought the answer for input 1 was "YES" because you can have 3 empty lists of edges for each subgraph and ( as i see ) everything will be ok with the definitions given at the problem ...

I did an O((n^2)log(n)) solution, and i didnt have problems with the Time Limit. gl! Eric.

jvimal wrote: What is the answer if the input is "55"? Is it 2 or 4?

My Ac program gives 2.....

gl! Eric.