Or don't use floating point numbers at all. When computing a/b (a,b positive integers), you can use:

a/b - if rounding down

(a+b-1)/b - if rounding up

(2*a+b)/(2*b) - if rounding to the nearest integer (with 0.5 up)

Where / is the integer division.

## Search found 30 matches

- Sun Mar 06, 2005 3:09 pm
- Forum: Volume 108 (10800-10899)
- Topic: 10823 - Of Circles and Squares
- Replies:
**50** - Views:
**10913**

- Sat Mar 05, 2005 11:53 pm
- Forum: Volume 108 (10800-10899)
- Topic: 10820 - Send a Table
- Replies:
**34** - Views:
**14313**

- Sat Feb 12, 2005 11:48 pm
- Forum: Other words
- Topic: Programming Contest for Newbies 2005
- Replies:
**34** - Views:
**5833**

- Tue Nov 16, 2004 12:21 am
- Forum: Volume 106 (10600-10699)
- Topic: 10607 - Siege
- Replies:
**21** - Views:
**6633**

- Mon Nov 15, 2004 2:02 am
- Forum: Volume 106 (10600-10699)
- Topic: 10607 - Siege
- Replies:
**21** - Views:
**6633**

Hello guys,

What is your output to this?

My AC program outputs 2, but the answer should be obviously 3 (B and C must be conquered, but they are not bordering, unless I misunderstood the statement).

What is your output to this?

Code: Select all

```
5 4
DDDD
DBBD
DBAC
DBBD
DDDD
0 0
```

- Sun Oct 24, 2004 11:59 pm
- Forum: Algorithms
- Topic: Does STL use Boyer-Moore?
- Replies:
**2** - Views:
**1416**

- Sun Oct 24, 2004 11:28 pm
- Forum: Algorithms
- Topic: calculate n choose k
- Replies:
**5** - Views:
**1871**

### Re: calculate n choose k

You can use GCD. When computing (A*B)/C, let G=GCD(A,C) thenManiac wrote:but this could overflow because I multiply before I divide. Is there a way to avoid an overflow without factorizing n and m?

(A*B)/C = (A/G) * (B/(C/G)).

- Sun Oct 17, 2004 9:42 pm
- Forum: Volume 107 (10700-10799)
- Topic: 10747 - Maximum Subsequence
- Replies:
**15** - Views:
**8998**

- Sun Oct 17, 2004 8:22 pm
- Forum: Volume 107 (10700-10799)
- Topic: 10745 - Dominant Strings
- Replies:
**38** - Views:
**17191**

marian: Thanks for your explanation. As far as I understand, you look at the strings as multisets of characters. And in your trie you do not distinguish between strings that give rise to the same multiset. Is that correct? Exactly. For example you can sort the string lexicographically. And for a gi...

- Sun Oct 17, 2004 5:24 pm
- Forum: Volume 107 (10700-10799)
- Topic: 10745 - Dominant Strings
- Replies:
**38** - Views:
**17191**

..: Aren't you checking O(n^2) pairs in the worst case (ie. no dominating strings)? Unless I didn't get your algorithm. BiK: I was just saying that one string of length K can dominate at most 2^K-2 strings. Since K<=10, one string can dominate at most 1022 strings. So it's cheaper to generate all po...

- Sun Oct 17, 2004 1:59 pm
- Forum: Volume 107 (10700-10799)
- Topic: 10745 - Dominant Strings
- Replies:
**38** - Views:
**17191**

- Wed May 12, 2004 3:18 pm
- Forum: Volume 2 (200-299)
- Topic: 209 - Triangular Vertices
- Replies:
**51** - Views:
**6993**

This problem (or at least it's input) is crap. You are right, there are integers <=0, but you should not always return false. It's weird, but I think you should return 0 1 2 are the vertices of a triangle 0 1 2 3 are the vertices of a parallelogram -1 1 4 6 are the vertices of a parallelogram At lea...

- Wed May 12, 2004 12:11 pm
- Forum: Volume 102 (10200-10299)
- Topic: 10207 - The Unreal Tournament
- Replies:
**23** - Views:
**6373**

- Mon May 10, 2004 10:26 pm
- Forum: Volume 106 (10600-10699)
- Topic: 10613 - Mushroom Misery
- Replies:
**14** - Views:
**6278**

- Mon May 10, 2004 7:14 pm
- Forum: Volume 106 (10600-10699)
- Topic: 10613 - Mushroom Misery
- Replies:
**14** - Views:
**6278**

Hello, I have totally stucked with this problem. The only solution I could think of, is to cover the circles with rectangles, and then solve simpler problem with rectangles in O(N log N) (N is number of rectangles). We can trivially cover one circle with 2*r rectangles (r is circle radius) , horizon...