UVa OJ Board

Thanks... there must be some careless mistake in my code then. I'll go check it out.

Someone said on another post somewhere that if such a case (as above) appears, then u remove all of them, not just any one of them. Perhaps someone who has solved this problem would like to say something?

The problem description says that If there had been several candidates with the fewest first choice votes, any of them, selected at random, could be selected for elimination. Consider this input data: 3 4 1 2 3 2 1 3 3 2 1 3 1 2 Candidate 1 has 1 vote, candidate 2 has 1 vote and candidate 3 has 2 vo...

all you need is the original string, which should be a character array of size 81, and a array storing the avaliable start positions. the way i do this question is to find all palindromes of length i, before moving on to find palindromes of length i + 1. It is easy to eliminate all duplicates just b...

haha... i did this question almost half a year back... dun think i still haf my source... y dun u email your code to me and i'll take a look for u?

[url]mailto:funkyboy2000@swirvemail.com[/url]

as above... int i; long total[1000], next[1000], temp[1000]; long * pTotal, pNext, pTemp; long carry; pTotal = total; pNext = next; pTemp = temp; carry = 0; for (i = 0; i < 1000; i++) { pTemp[i] += carry + pNext[i] + pTotal[i]; carry = pTemp[i] / 100000; /* assuming 5 digits stored per element */ pT...

my guess is that you are programming in windows.. probably using microsoft visual c or some variant... cos these functions starting with '_' do not exist in ANSI C, which is generally (with a few exceptions - for example, the long long int type) what the judge is running... so, use other functions t...

yes, i totally agree with you... the problem really has no mathematical basis... but anyway, the thing i stated earlier is just the solution

this problem requires a solution that takes linear time (ie. no nested loops, no recursion needed)

personally, i use pointers and long array... its much faster than using a char array, cos u can then process 9 digits at once (for addition), and 4 digits at once (for multiplication). So what you can do is to create 3 arrays, one to store your current total, one to store the next number to add, and...

well... i guess the answers are quite obvious for positive input... well... when input is -1, f(-1) = f(0) / 0 = 1 / 0 = positive infinity (overflow)... den when u divide that by a negative num... u get another answer... and when u divide that by another negative num... what do you get?... that's th...

well... apparently they just reset the system... so stuff submitted after like year 2000 all disappeared... *sigh*... i'm back down to 0 solved questions =)... hopefully they fix it soon...

well, this question has got to do with searching for negative cycles in a graph. represent each currency by a vertex in a graph, with the edge weights the negative of the log of the currency exchange value. If a negative cycle can be found, then a arbitrage cycle exists. The length of the negative c...

well... i think the online judge is okay, but the email system is down, cos the clock on the status page is counting... but no new judging is going on.