Search found 20 matches
- Sat Jul 03, 2004 6:18 am
- Forum: Off topic (General chit-chat)
- Topic: People are driving Manzoor Mad
- Replies: 29
- Views: 12634
Well! Boring problems vary from person to person. For example two other problems in this contest were boring to me (I won't say which ones because I have learnt courtesy somehow). About the cow problem I have written it long ago and it was just there to be solved by everyone. Doesn't topcoder or ...
- Sun Jun 27, 2004 12:27 am
- Forum: Off topic (General chit-chat)
- Topic: People are driving Manzoor Mad
- Replies: 29
- Views: 12634
- Mon Jun 14, 2004 9:30 pm
- Forum: Volume 106 (10600-10699)
- Topic: 10668 - Expanding Rods
- Replies: 38
- Views: 25722
Per, I had the same problem during the contest. I did binary search on the radius, which is not precise enough. I guess if you have a really small difference like 0.0000001, the radius can be huge, and with the formula I used, that meant taking the cos of something very close to 0. Searching on the ...
- Wed Jun 09, 2004 5:19 pm
- Forum: Volume 106 (10600-10699)
- Topic: 10663 - Non-Powerful Subsets
- Replies: 27
- Views: 9014
- Wed Jun 09, 2004 5:16 pm
- Forum: Volume 106 (10600-10699)
- Topic: 10663 - Non-Powerful Subsets
- Replies: 27
- Views: 9014
- Sun Jan 11, 2004 10:16 pm
- Forum: Other words
- Topic: Are we going insane?
- Replies: 12
- Views: 6353
- Tue Dec 30, 2003 6:02 am
- Forum: Volume 105 (10500-10599)
- Topic: 10594 - Data Flow
- Replies: 40
- Views: 32820
10594 - Bad judge data or problem statement
After wasting a fair bit of time trying to solve this using network flow, I discovered that a much simpler approach of repeatedly picking the shortest path, then deleting that path, gets AC. This is incorrect though, according to the problem description. Consider the following test case:
8 11
1 2 0 ...
8 11
1 2 0 ...
- Sat Oct 11, 2003 8:14 pm
- Forum: Other words
- Topic: Overloaded judge during contests
- Replies: 21
- Views: 6340
- Fri Jul 04, 2003 1:34 am
- Forum: Volume 105 (10500-10599)
- Topic: 10520 - Determine it
- Replies: 15
- Views: 11439
- Tue Jul 01, 2003 5:03 am
- Forum: Volume 105 (10500-10599)
- Topic: 10518 - How Many Calls?
- Replies: 19
- Views: 17801
ya, you can do Lucas number the same way, I believe, I don't know how other ppl do it though, because mine runs pretty fast and there are some really slow solution for some reason...
Those people used a different method, where you compute values in the sequence until the sequence begins to repeat ...
Those people used a different method, where you compute values in the sequence until the sequence begins to repeat ...
- Tue Jul 01, 2003 4:53 am
- Forum: Volume 105 (10500-10599)
- Topic: 10510 - Cactus
- Replies: 31
- Views: 28072
Re: 10510 O(n)?
What is the O(n) algorithm for strongly connected components?
I just thought you can find the reverse dfs for a node, and the regular dfs for a node, and if they both reach everyone, then there's one component.. but I get WA..
Yes, that is enough to determine if a graph is strongly connected ...
I just thought you can find the reverse dfs for a node, and the regular dfs for a node, and if they both reach everyone, then there's one component.. but I get WA..
Yes, that is enough to determine if a graph is strongly connected ...
- Tue Jul 01, 2003 4:37 am
- Forum: Algorithms
- Topic: Bisection Method
- Replies: 30
- Views: 10239
- Sat Jun 28, 2003 1:23 am
- Forum: Algorithms
- Topic: Bisection Method
- Replies: 30
- Views: 10239
- Wed Jun 25, 2003 7:09 pm
- Forum: Volume 105 (10500-10599)
- Topic: 10518 - How Many Calls?
- Replies: 19
- Views: 17801
Re: 10518 WA ?
I think the second %i should be %lli.pingus wrote:why wa ?
[c]
printf("Case %i: %i %i %i\n",cases,n,b,calls[n%(m-1)]);
[/c]
- Sat Jun 14, 2003 6:10 am
- Forum: Algorithms
- Topic: Find second min number .
- Replies: 10
- Views: 5300
Re: Find second min number .
You can't just go through linearly and keep track of the 2 smallest because that can use 2n - 3 comparisons in the worst case.
The tournament method works like this. First, assume n is a power of 2. Split all the elements up into arbitrary pairs. Do a comparison on each of the n/2 pairs, and then ...
The tournament method works like this. First, assume n is a power of 2. Split all the elements up into arbitrary pairs. Do a comparison on each of the n/2 pairs, and then ...