Search found 429 matches

by Per
Thu Aug 21, 2003 1:01 am
Forum: Volume 6 (600-699)
Topic: 643 - Bulk Mailing
Replies: 5
Views: 3513

Finally found the answer, in case anyone is interested.

We do indeed print 15 as the letter count in the above case.

Another hint: the sample output is somewhat broken. We should print a blank line between "INVALID ZIP CODES" and the first invalid zip code.
by Per
Thu Aug 21, 2003 12:23 am
Forum: Volume 105 (10500-10599)
Topic: 10521 - Continuously Growing Fractions
Replies: 17
Views: 6276

My AC solution does not handle '+' characters, but for input

Code: Select all

5 6
1 1
output is

Code: Select all

5/6 = 0+1/{1+1/{5}}
1/1 = 1
And you should put spaces around the '='.
by Per
Fri Aug 15, 2003 7:16 am
Forum: Volume 2 (200-299)
Topic: 216 - Getting in Line
Replies: 57
Views: 24719

Oriol:

Input:

Code: Select all

6
111 84
55 28
43 116
38 101
28 62
5 19
Output as in first sample input case (i.e. 305.45 feet).
by Per
Tue Aug 12, 2003 8:19 pm
Forum: Volume 105 (10500-10599)
Topic: 10535 - Shooter
Replies: 25
Views: 49034

I used integers all the way, so it's certainly possible. If I remember correctly, the maximum value of the intermediate values in my code is +- 2*10000^2, which is well within range.

Though I guess it depends on the algorithm (I just brute forced it, checking all possible directions).
by Per
Mon Aug 11, 2003 7:20 pm
Forum: Volume 105 (10500-10599)
Topic: 10535 - Shooter
Replies: 25
Views: 49034

I don't think there's any real trick to this problem, I had it accepted at second try. The order of the coordinates is x1,y1,x2,y2 (or y1,x1,y2,x2 or x2,y2,x1,y1 etc, it doesn't make any difference), and I terminate on N = 0. The only problem with my first submission was that it went wrong for input...
by Per
Sun Aug 10, 2003 3:46 pm
Forum: Volume 101 (10100-10199)
Topic: 10114 - Loansome Car Buyer
Replies: 38
Views: 13669

Ah, I had completely misunderstood the meaning of the downpayment! :oops:

Thanks a lot!
by Per
Mon Aug 04, 2003 9:24 pm
Forum: Volume 105 (10500-10599)
Topic: 10531 - Maze Statistics
Replies: 8
Views: 5026

10531 - Maze Statistics

I'm getting WA on this problem. Could somebody check this I/O and tell me if I'm way off or if I just have precision errors? Input: 4 3 3 0.5 0.5 0.5 0.5 1.0 0.5 0.2 0.5 0.5 3 3 0.5 0.5 0.5 0.5 0.9 0.5 0.2 0.5 0.5 5 5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0....
by Per
Tue Jul 29, 2003 7:37 am
Forum: Volume 105 (10500-10599)
Topic: 10523 - Very Easy !!!
Replies: 63
Views: 22786

You have mixed up the order in which the parameters come. As it is, the input is illegal, but if you swap the numbers on each line, the output is correct.
by Per
Tue Jul 29, 2003 7:33 am
Forum: Volume 102 (10200-10299)
Topic: 10221 - Satellites
Replies: 34
Views: 13353

Think about angle > pi. You have a mistake there.
by Per
Tue Jul 29, 2003 7:26 am
Forum: Volume 103 (10300-10399)
Topic: 10393 - The One-Handed Typist
Replies: 40
Views: 11078

No. The output should be the same as in the first sample output.
by Per
Tue Jul 29, 2003 7:23 am
Forum: Volume 104 (10400-10499)
Topic: 10441 - Catenyms
Replies: 15
Views: 6627

The answer is aabb.ba.aabba since '.' (by the ASCII ordering) is less than 'a'. But incidentally, note that the edge "aabb" is also lexicographically less than the edge "aabba".
by Per
Fri Jul 25, 2003 12:16 pm
Forum: Volume 3 (300-399)
Topic: 374 - Big Mod
Replies: 79
Views: 12031

Maybe not philosophically, but mathematically, if you assume that the number of bits in P (i.e. log_2(P)) is some constant C (i.e. that O(#bits in P) = O(1)), than P must be bounded by some constant 2^C, and 2^P must be bounded by some constant 2^(2^C), hence O(2^P) also becomes O(1). Ordo notation ...
by Per
Fri Jul 25, 2003 11:51 am
Forum: Volume 3 (300-399)
Topic: 343 - What Base Is This?
Replies: 72
Views: 21609

Base 2 (i.e. binary) numbers can't contain the digit 5, only 0's and 1's.
by Per
Thu Jul 24, 2003 9:07 am
Forum: Volume 103 (10300-10399)
Topic: 10307 - Killing Aliens in Borg Maze
Replies: 54
Views: 17875

The borgs can only split at aliens, not in arbitrary cells!
by Per
Thu Jul 24, 2003 8:59 am
Forum: Volume 3 (300-399)
Topic: 374 - Big Mod
Replies: 79
Views: 12031

Well, yeah sure, but if you see it that way, almost any algorithm is O(1), even if it takes O(2^P), so it's not very informative . :) I'm guessing you're talking about the same algorithm as me, the standard square-and-multiply-thingy. This algorithm uses O(#bits in P) multiplications and modular red...

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