Обсуждение @ 1428. Джедайский ребус @ Timus Online Judge

Problem 1428 "Jedi riddle" has been rejudged (+)

Послано Vladimir Yakovlev (USU) 9 янв 2007 16:44

New tests were added. 77 authors lost AC.

Cool! We failed to foresee such kind of tests. Thanks to Vladimir Yakovlev (-)

Послано Dmitry 'Diman_YES' Kovalioff 9 янв 2007 22:07

Re: Problem 1428 "Jedi riddle" has been rejudged (+)

Послано SPIRiT 12 янв 2007 17:35

Don't know what kind of tests were added, but all I had to do is to change type of X,Y from int to __int64 in order to get AC again

Re: Problem 1428 "Jedi riddle" has been rejudged (+)

Послано Roma Labish[Lviv NU] 12 янв 2007 17:56

Only worse case is
1 1 32 -> And answer must be 2^32 > max_int

Re: Problem 1428 "Jedi riddle" has been rejudged (+)

Послано Burunduk1 13 янв 2007 12:04

My solution (and, as far as I know, author's one also) produces answers <= 2^31.

Re: Problem 1428 "Jedi riddle" has been rejudged (+)

Послано Roma Labish[Lviv NU] 14 янв 2007 00:12

When my soulution produced <= 2^31 I had Wa52. But when I changed it to <= 2^32 -> AC

Re: Problem 1428 "Jedi riddle" has been rejudged (+)

Послано Burunduk1 14 янв 2007 04:32

I think we speak about different things...
You speak about type, in which result is stored.
And I say that there is correct formula, which produces results <= 2^31.

Re: Problem 1428 "Jedi riddle" has been rejudged (+)

Послано KIRILL(ArcSTU) 14 янв 2007 13:49

May be 2<<31 ?

Re: Problem 1428 "Jedi riddle" has been rejudged (+)

Послано Burunduk1 14 янв 2007 14:53

Yes. If you are used to C style.
^ is usual designation for power.

Re: Problem 1428 "Jedi riddle" has been rejudged (+)

Послано KIRILL(ArcSTU) 14 янв 2007 19:01

yes I know
(long long)2<<31 = int64(2 shl 31) = 2^32
it's max answer of my formula for
test 1 1 32 which was added just 1 week ago

Обсуждение задачи 1428. Джедайский ребус