I assume that O(n^4) means the number of operations is ^4. What's the name of this notation; how can I find out more? Is there a method to assist in converting an O(n^6) solution to O(n^4) or less?
Thanks
Search found 7 matches
- Sat Jul 29, 2006 11:49 am
- Forum: Algorithms
- Topic: What is O(n^4)?
- Replies: 2
- Views: 1794
- Sat Jul 29, 2006 11:36 am
- Forum: C++
- Topic: 64 Memory and ~0 CPU, How? #pragma with online-judge?
- Replies: 2
- Views: 2459
- Fri Jul 28, 2006 4:47 pm
- Forum: Volume 1 (100-199)
- Topic: 100 - The 3n + 1 problem
- Replies: 1394
- Views: 320000
- Fri Jul 28, 2006 4:41 pm
- Forum: Bugs and suggestions
- Topic: SEARCHING THE FORUM
- Replies: 4
- Views: 2843
- Fri Jul 28, 2006 4:41 pm
- Forum: Bugs and suggestions
- Topic: 100 Spelling Mistake
- Replies: 1
- Views: 1897
100 Spelling Mistake
"You can assume that no opperation overflows a 32-bit integer."
should be operation
should be operation
- Fri Jul 28, 2006 4:29 pm
- Forum: Volume 1 (100-199)
- Topic: 100 - The 3n + 1 problem
- Replies: 1394
- Views: 320000
- Fri Jul 28, 2006 4:19 pm
- Forum: C++
- Topic: 64 Memory and ~0 CPU, How? #pragma with online-judge?
- Replies: 2
- Views: 2459
64 Memory and ~0 CPU, How? #pragma with online-judge?
Could someone advise me on how some authors achieve such good performance for the problems in C++? For example do they just write the code in embedded assembly or are there some good C++ performance tips I can use?
Also, are our submissions compiled in debug mode with no optimizations? Might it be ...
Also, are our submissions compiled in debug mode with no optimizations? Might it be ...