| Показать все ветки Спрятать все ветки Показать все сообщения Спрятать все сообщения |
| Did anybody have WA3? | [SPb NRU ITMO] Niyaz Nigmatullin | 1046. Геометрические грёзы | 17 ноя 2023 14:44 | 3 |
I solved a system of two equations by hand at first and got WA3. Then I used cramer's rule and got AC. I guess cramer is far more precise (uses less divisions/multiplications etc). |
| No subject | sashabusse | 1046. Геометрические грёзы | 10 сен 2019 19:33 | 1 |
Edited by author 10.09.2019 21:16 |
| Can I solve it considering every point? | ძამაანთ [Tbilisi SU] | 1046. Геометрические грёзы | 9 авг 2013 19:00 | 1 |
huh?
just yes or no please : ] |
| Why WA on 10? | kell | 1046. Геометрические грёзы | 24 сен 2012 20:10 | 1 |
|
| Hint | Pong | 1046. Геометрические грёзы | 13 авг 2011 11:10 | 1 |
Hint Pong 13 авг 2011 11:10 Try to solve equations for the first point! |
| Be very accurate! Use epsilon with at least 12 digits after zero or you get WA15 | Leonid (SLenik) Andrievskiy | 1046. Геометрические грёзы | 20 май 2011 19:54 | 1 |
|
| Is the sample correct? | AkemiHomura | 1046. Геометрические грёзы | 9 апр 2011 09:21 | 1 |
The problem suggests
"The set of angles αi satisfies a condition that the sum of angles in any of its nonempty subsets is not aliquot to 360 degrees."
However the sample gives "90 90 90"... |
| About the statement | LSBG | 1046. Геометрические грёзы | 19 апр 2010 13:11 | 2 |
From the statement it is not obvious whether the triangle isosceles with AiMi = Ai+1Mi or some other legs equality. I think it is worth adding that to the statement. |
| brute force is ok! | ile | 1046. Геометрические грёзы | 30 мар 2010 05:12 | 1 |
I got AC by brute forcing coordinates.
Now I want to know a "real" solution for this one! Please someone contact me at ile-ngcmlfs@mail.ru
Edited by author 01.04.2010 05:20 |
| TLE #13 or WA #12 | icanwin | 1046. Геометрические грёзы | 20 окт 2009 18:16 | 2 |
This problem can be solved only by numerical methods, or it is possible to solve analytically?
Stupid handmade iterative methods do not work.
Really it is necessary to use a method of the quickest descent, or a method of Hook and Jeeves?? Very much I doubt... done icanwin 20 окт 2009 18:16 I have received AC without numerical methods. |
| TLE #4 | icanwin | 1046. Геометрические грёзы | 19 окт 2009 03:54 | 3 |
TLE #4 icanwin 19 окт 2009 03:00 http://paste.pocoo.org/show/145742/After each iteration I calculate a point, as average for two points received by different ways (a "counter-clockwise" way and a "clockwise" way, on half of quantity of points of a polygon everyone) Edited by author 19.10.2009 04:00 TLE #4 icanwin 19 окт 2009 03:22 But amount of points are less than 10. It causes fears. Is it weak convergence?
Edited by author 19.10.2009 04:03 In what way I can choose a good point for initial approach of a starting point? |
| Can someone explain? I don't understand the notion of aliquot. | xerxe | 1046. Геометрические грёзы | 25 авг 2008 21:45 | 2 |
Actually, the whole second paragraph doesn't make sense for me, as I don't know what aliquot means(in this context).
[quote]The set of angles ai satisfies a condition that the sum of angles in any of its nonempty subsets is not aliquot to 360 degrees.[/quote]
I search for the word aliquot, and found something like "part of something" and "divisor". But I'm not sure how these would make sense here. Any clarifications are welcome.
Thanks X is "aliqot Y" if there exists such ineger K satisfying equation X = K * Y, I suppose |
| rejudge? | StarForever | 1046. Геометрические грёзы | 29 июл 2008 17:26 | 5 |
has some new tests added? Yes, the problem is under investigation yet, and rejudge is not finished. When will these "investigations" end at last? Investigation is finished.
This problem had some troubles with output format and very weak tests.
Validator was fixed for this problem. Now 4.70, 4.7 and 4.7000001 are the same answers. New tests were added and Time Limit was decreased to 0.5 sec. After rejudge more than 200 submits lost AC, but about 90 WA got AC. WA3: print 5 digits after decimal point instead of 0 :)
WA19: decrease EPS from 1e-8 to 1e-14 (1e-10 wasn't enough)
Edited by author 29.07.2008 17:28 |
| What does this sentence suggest? | gush(gu_shenhua@yahoo.com.cn) | 1046. Геометрические грёзы | 29 июл 2008 17:15 | 3 |
The set of angles ai satisfies a condition that the sum of angles in
any of its nonempty subsets is not aliquot to 360 degrees.
Can you prove it? I think it mean that sum of angles is not equal to 360 It is either not divisible by 360 or does not divide 360
Edited by author 29.07.2008 17:16 |
| Format | Bohdan Istrashkin | 1046. Геометрические грёзы | 12 апр 2008 00:48 | 2 |
Format Bohdan Istrashkin 10 апр 2008 19:33 Could someone help me in next:
In what way should next values be formatted?
1.0001 -> x?
1.0091 -> x? (rounding up/down/none?)
-0.00001 -> x? ("0" or "-0.00" or..)
Or may i output as many digits as i can, but at most 2 of fraction should exactly match?
i just rounded values:
1.0001 -> 1.00
1.0091 -> 1.01
-0.00001 -> -0.00
and got AC. |
| What accuracy for output? How many digits after decimal point? | Mirzayanov Michael | 1046. Геометрические грёзы | 24 дек 2006 07:47 | 5 |
The answer should be with 10^-2 accuracy.
But keep in mind that we do not accept extra zeroes.
1 != 1.0 != 1.00
1.1 != 1.10 But what does 'extra zeros' mean?
For example, if the accurate value is 6.0004, shall I output 6 or 6.00?
And what if 6.000000000000004, which is just floating-point error? for C++, it would be a troublesome
cout<<setiosflags(ios::fixed)<<setprecision(2)<<result<<endl;
the program above will not work
so I have to write a void to do it myself. |
| Any Hint? | Miguel Angel | 1046. Геометрические грёзы | 16 фев 2005 00:43 | 3 |
Anyone can help me to start thinking how to solve it? :) I believe that you still need help.
If you have one vertex of a figure you can obtain second, third, N-th and N+1-st! But when N+1-st point will be same as 1-st? |
| Is there more than one output for the sample input? | chinachen | 1046. Геометрические грёзы | 24 июн 2003 07:39 | 2 |
(2/3,2/3) (3,2) (2,3) is also ok
Is anyone solution is ok? No.There is only one output for every input. |
| Anyone have idea how to solve this problem... | Dejan Kolundzija | 1046. Геометрические грёзы | 8 апр 2002 21:32 | 2 |
I've found that I have to memorize two coeficients ... But I can't
evaluate coeficients after rotation ...
I want to know how to compute coordinates after rotation around point
M(x, y) by angle alpha ??
faith respect,
..::dejan::.. > I've found that I have to memorize two coeficients ... But I can't
> evaluate coeficients after rotation ...
>
> I want to know how to compute coordinates after rotation around
point
> M(x, y) by angle alpha ??
>
> faith respect,
> ..::dejan::..
I've already solved it .. and got AC ... I've used the method that
I've mentioned earlier ...
..dejan.. |
