OJ Board

Problem statement: http://www.ioi2006.org/practice/cutting_a_grid.pdf My friend and I are wondering about how to do this problem. Can someone tell me if this greedy solution works? It sounds fine to me... 1. First you assume the white area is to the left, and you divide the grid turning as many time...

The third way that I know of is similar to the addition/subtraction method:

a = a * b;
b = a / b;
a = a / b;

Well then again this method is real stupid... you can get overflow and precision errors very easily. Just for fun I knowing it, but don't use it in real programs!

Whao... thanks everyone for their replies... quite surprised that so many people bothered to code test programs.

Thanks again.

Hi, I have this question that's just out of curiosity. Which of the following code is more efficient? for (i = 0; i < 10000; i++) { if (i % 5 == 0) b++; a++; } or for (i = 0; i < 10000; i++) { count++; if (count == 5) { b++; count = 0; } a++; } In short, does multiplication / division / modulus use ...

Hmm... if you implemented all the heuristics listed you should get a better runtime. Try checking through your code and see if you have anything that would take a long time to run. An example of useless redundancy could be like: for (i = 0; i < 1000000; i++) a = i; when you could just have a = 99999...

Instead of running through everything in O(n^3) time, consider using some maths to do the calculations.

This problem is quite difficult... it requires a lot of pruning... but it is possible... :D Try these suggestions: - Find all "dominating nodes", and eliminate the rest. Let A and B be the set of nodes adjacent to node a and b respectively. Let a node be connected to itself. If A is a subs...

I think it should be "1 3 2 1 sum: 7", but cyfra had a typo. Btw, the way I see it "tracks do not repeat" does not mean the length, just that you cannot use the same track more than once.

Yes your output is correct. Have you tried reading this post?

http://online-judge.uva.es/board/viewto ... highlight=

Hmm okay... now it doesn't seem as easy...

I'm thinking of a O(n^3) DP solution, but I suppose that is too slow. Am I on the right track at all (like maybe I need to optimise the DP), or you need other methods to solve this problem (maths / smart brute force search / etc)?

Thanks.

Zhao Le: I don't really know much about DP except using Matrix How do you use matrix to find factorials? Btw for this problem would there be TLE/MLE if we just precalculate all factorials and store the numbers in arrays? I haven't tried this yet but if the memory used is not too great it sounds lik...

I think I don't understand this problem. From what I see, there is no special checker for this problem, but for the sample input (31) why is the output 2 3 5 6 7 8? Can't it be 1 3 5 6 7 9 as well? I couldn't find anything in the problem that will make the solution unique. Could someone who has solv...

I don't think the statement is found anywhere. The problem setter probably thought it obvious that a species cannot consume itself. Of course test cases on the OJ rarely fit into actual real life conditions, but I guess this problem does.

My output:

1 2 3
3 2 1
total:2

The three numbers on each line should be different from each other.

Ok so that's where the odd number comes in. By the way I think I figured out why the answer must be less than half the lower bound. The question never numbered the positions, just that you have to find how many position away from Mxgobgwq. In short 1 can mean position 1 left or 1 right. Therefore, i...