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WA25 | Mortus | 1286. Космолёт марки ВАЗ | 15 янв 2024 14:03 | 1 |
WA25 Mortus 15 янв 2024 14:03 |
test 16 is terribly! HELP!!!!!! | Vasya | 1286. Космолёт марки ВАЗ | 7 фев 2015 20:48 | 5 |
I am sure that my program is right, but i get WA on test 16. Do somebody know this test or can say something interesting? I think, you haven't really proved your algorithm. I guess it's much more interesting than just gcd. Sergei Try using long long and read your data like this scanf( "%I64d%I64d%I64d%I64d%I64d%I64d", &p, &q, &sx, &sy, &ex, &ey ); Thanks for all, if not you I die near my PC. |
scanf vs cin | pmartynov | 1286. Космолёт марки ВАЗ | 11 мар 2014 22:30 | 1 |
scanf("%lld%lld%lld%lld%lld%lld", &P, &Q, &SL, &SC, &X, &Y); => WA 16 cin >> P >> Q >> SL >> SC >> X >> Y; => AC Is there anyone who can explain this? |
WA 26 | Vasily Slesarev | 1286. Космолёт марки ВАЗ | 18 фев 2014 20:33 | 4 |
WA 26 Vasily Slesarev 15 июл 2009 19:40 Have somebody any things about test N 26? Thank you! I have founf a mistake. AC now. Re: WA 26 IgorKoval [PskovSU] 17 сен 2013 03:08 I had WA 26 too. Stuped mistake. =) Was: ( v + f + i*p + k*q )%2==0 && ( -v + f + i*p - k*q )%2==0 && ( u + s - i*q - k*p )%2==0 && ( -u + s - i*q + k*p )%2==0 This get AC: ( v + f + i*(p/g) + k*(q/g) )%2==0 && ( -v + f + i*(p/g) - k*(q/g) )%2==0 && ( u + s - i*(q/g) - k*(p/g) )%2==0 && ( -u + s - i*(q/g) + k*(p/g) )%2==0 i guess || instead of && ? |
WA#7 - Does anybody have some ideas? | Alexander Samal | 1286. Космолёт марки ВАЗ | 20 авг 2009 23:04 | 1 |
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Why WA#25?Please help.Thank!!! | CHIDEMYAN SERGEY | 1286. Космолёт марки ВАЗ | 15 июл 2009 19:39 | 3 |
[code deleted] Should I use long arithmetic?Thank!!! Edited by author 06.02.2008 02:47 Somebody,who solved this problem,please,answer:what 25-th test is?Thank! Edited by author 06.02.2008 02:48 I know, that on test 25 x1==x2 and right answer is NO. |
Please help,I've WA#25.Please,give me some tests.Thank! | CHIDEMYAN SERGEY | 1286. Космолёт марки ВАЗ | 24 авг 2007 04:41 | 3 |
Anybody,who know #25,please,help me!!! Edited by author 24.08.2007 04:43 Oh,need a help! Edited by author 24.08.2007 04:41 Please,give me some interesting tests. Edited by author 24.08.2007 04:43 |
Oh, my god! (+) | Dmitry 'Diman_YES' Kovalioff | 1286. Космолёт марки ВАЗ | 19 апр 2007 01:18 | 2 |
AC finally! And I am still not sure my math solution is unbeatable. Does anybody know something about test 26? So what troubles did you had with this test? because i am having wa26 too :) |
1286 Standart way is too slow | svr | 1286. Космолёт марки ВАЗ | 8 фев 2007 13:10 | 2 |
A standart way of solution has appeared too slow(TLE14)! Standard way is to make Hermit form above __int64 of the matrix A= [p q -p -q] [q p q p] or form [0 0 * *] [0 0 0 *] by Gauss operations on columns in int ring. For initial A and Hermit form answer is equel. I were very suprized what for so small matrix we have big Time. This is due to many operations need to make zeros. Excuse me! I got quick AC(0.015) by replacing __int64 by int in the same program. I think that it because weak tests |
It's cool. | theZ[home] | 1286. Космолёт марки ВАЗ | 15 янв 2007 19:08 | 1 |
It was very interesting to solve this problem. There are some tests: test 1: 1 1 -2000000000 -2000000000 2000000000 2000000000 Yes test 2: 2 3 0 0 0 1 Yes |
Please help! | Korshun | 1286. Космолёт марки ВАЗ | 12 дек 2006 17:31 | 1 |
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No subject | Korshun | 1286. Космолёт марки ВАЗ | 12 дек 2006 17:29 | 1 |
Please help! I got WA on 9 test! |
AC :) (+) | Yu Yuanming | 1286. Космолёт марки ВАЗ | 19 окт 2005 22:28 | 4 |
A maths problem...... First, let m = x2 - x1, n = y2 - y1, d:= gcd(p, q) {assume p, q > 0, this is no harm but convenient} Second, discuss four trivial case: 1. (m = 0) and (n = 0) {Y} 2. (p = 0) and (q = 0) {N} 3. (p = 0) or (q = 0) {Y / N} 4. (m mod d <> 0) or (n mod d <> 0) {N} Third, make p = p / d, q = q / d, m = m / d, n = n / d Fourth, find x and y so that p * x + q * y = 1. Let A1 = x * m , B1 = y * m A2 = y * n , B2 = x * n You job is to judge whether there exist k1 and k2 so that (A1 + A2 + k1 * p + k2 * q ) and (B1 + B2 + k1 * q + k2 * p) are both even numbers, if so, output 'YES', otherwise 'NO'. By the way, just check (k1 = 1, 2) and (k2 = 1, 2) are enough... I have proved my algo is right. Anyway, if you doubt whether it is right, you could try to prove or disprove it by yourself, too. If you have another algo,I will be pleased if you could write it down. Sorry for my bad English, hope it can help :) Edited by author 16.05.2005 21:01 Something easier. UXMRI: Sergey Baskakov, Raphail Akhmedisheff and Denis Nikonorov 19 сен 2005 08:58 I did almost the same, but... 1) I didn't care about p=0 or q=0 2) I didn't cate about p<0 or q<0 So, I assumed that (p>0) and (q>0) - it seemed so evident from the problems's situation (-: And after having divided everything by gcd, I do the following: if p mod 2 + q mod 2 = 1 then (* p is even and q is odd or vice versa *) ___Result := true else (* both p and q are odd *) ___Result := ( deltaX + deltaY ) mod 2 = 0 ______(* we can reach a point iff its coords are both even or odd at the same time *) And I got AC. http://acm.timus.ru/status.aspx?space=1&pos=903260I use: if( Abs( Abs( DI - SI ) - Abs( DJ - SJ ) ) % NOD( Abs( P - Q ), 2 * nod_pq ) != 0 ) ____Result = false; Sorry, but what does it mean: 1) DI 2) SI 3) DJ 4) SJ 5) NOD( Abs( P - Q ), 2 * nod_pq )!=Factorial(GCD(abs(p-q),2*qcd(p,q))) |
use int64 to make difference and you'll get AC! Like me ::) | Fly [Yaroslavl_SU] | 1286. Космолёт марки ВАЗ | 10 окт 2005 03:00 | 1 |
use int64 to make difference and you'll get AC! Like me ::) |
HELP, Wa 9. But I thik my program good! | Нищий Наглец | 1286. Космолёт марки ВАЗ | 30 июл 2005 20:53 | 1 |
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