OJ Board

If the judge data is the same as the original contest data, I think it is wrong: The forth case is: 1 5 0 1 0 0 0 0 0 0 1 1 1 2 10 0 0 5 0 0 5 This means that before the exchange Alice has 100, Bob has 100 and Cynthia has 25 crowns. Since Alice has to pay Bob 1 crown, and Bob has to pay Cynthia 5 cr...

The previous thread on this bug is locked, probably because it's considered fixed with the given work-around. Indeed, clearing the private data from Firefox enables you to get into the OJ, but that is really annoying! I use my browser for several other sites where I need to log in (gmail, for exampl...

You're right, my greedy gives "NO WAY"

You base your greedy on earliest deadline, but that's not the correct criterion. My greedy answers "FEASIBLE" for your input. Can't remember all the details of the monkeys problem, but I think greedy is correct for this problem.

As we could hear in the news, a terrible disaster hit Bangladesh on Friday, taking thousands of lives, and many more to come.
Lots of programmers from this country are frequently using this site, so I like to wish them and their loved ones my sincere sympathy if they were in any way involved.

Hi Sohel, thanks for the interest! I guess I fell into a kind of 'UVA-blackhole' the last few of months. While things where not going so great in my personal life lately, I found myself having less and less motivation to solve problems, set problems (I contributed none to the last Warm Up contest, t...

Amazing how insight can strike suddenly. I must have looked at this problem over 20 times in the past 3 years, without any idea of how to solve it.
This must be the most obfuscated problem in the problemset, which is a big credit for its setter .

It is a lot of work to completely write it out. You could write a brute force program, as Adrian suggested, and verify it for yourself. In pseudo code: function evaluate(a,b,k): /* returns 'winning'or 'losing'; a and b are the pile sizes */ if a==0 and b==0 return 'losing'. for all positions (a',b')...

Yes, my polynomials are wrong. Thanks for pointing that out.

For this inputyour program replies "WINNING", but it's a losing position.

Yes, I think you are right in that it's unlikely that there is a better way then O(n^3) to calculate the coefficients, and it's not worthwhile to do so for so few queries. Precalculation and hardcoding is no option because there's not enough room in a 32k program (and would be considered cheating by...

OK, I see. This won't happen in my method, because after determining the minimum radius in a certain step, I remove _all_ the sides for which the intersection of the angle bisectors lie at that minimum radius in one action. This means that in case of a square, the minimum radius is determined once a...

Thanks mf and sclo for their descriptions. I have one question for Simon: More about my method: I treated the case there're 2 adjacent sides parallel as special cases. The sample input doesn't have that case. It is possible that 2 adjacent sides are parallel if we keep removing sides. Can you give a...

A string doesn't have to be zero-terminated. If you replacebyyou'll get accepted. Then, please, remove your code.

Quite strange indeed. I created an exhaustive test file containing all possible legal cases this way: #include <stdio.h> int main(){ int i1,i2,i3,j1,j2,j3; for(j1=1;j1<20;j1++){ for(j2=1;j2<20;j2++){ for(j3=1;j3<20;j3++){ for(i1=1;i1<=j1;i1++) printf("%d %d %d %d %d %d\n",i1,j2,j3,j1,j2,j3...