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Обсуждение задачи 1776. Праздничный фейерверк

WA# 6
Послано Artem Khizha [DNU] 9 окт 2010 20:35
I'm again and again experiencing WA#6. Can you help me?
If you got AC, please, share the correct answers to some tests.

N = 7
N = 8
N = 9
N = 10
N = 19
N = 20
N = 399
N = 400
Re: WA# 6
Послано vgu 9 окт 2010 20:42
Your idea is not correct. I have WA#6 too.
N=7; ans = 38.0;
N=8; ans = 42.6666;
N=9; ans = 46.69841269841;
N=10; ans = 50.1746031746;
Try think about 2-parameters DP p[n][k].
Re: WA# 6
Послано r1d1 10 окт 2010 01:02
How ans for N=400?
Re: WA# 6
Послано Artem Khizha [DNU] 10 окт 2010 02:00
N   = 400
ANS = 184.652746446

Edited by author 10.10.2010 02:19
Re: WA# 6
Послано Artem Khizha [DNU] 10 окт 2010 02:04
Thank you, vgu, your tests helped a lot.
Re: WA# 6
Послано olpetOdessaONU [1 2/3] 10 окт 2010 14:48
WA6 got all the contestants, used dynamic formula
f(N) = 1/N * sum(i=0..n-1)max(f(i),f(N-i-1)) + 10
It's incorrect.
Re: WA# 6
Послано bsu.mmf.team 10 окт 2010 16:47
This formula is almost correct. The following example will illustrate it. Imagine you have two groups of rockets (A and B) and you know, that group A will launch in 10 seconds with probability 1/2, and in 20 seconds with probability 1/2 (math expectation equals 15 in this case). The group B has probabilities 1/3 and 2/3 correspondingly (math expectation = 50/3). But a probability that all rockets will launch in 10 seconds equals 1/2 * 1/3 = 1/6 (it is almost evident), and in 20 seconds: 1 - 1/6 = 5/6. Math expectation = 55/3.
So, the only mistake in this formula is M(max(A,B)) != max(M(A),M(B)). Change this, and this fomula will be correct. By the way, I got AC using it.
Re: WA# 6
Послано Proba 11 окт 2010 12:19
How ans for N=6?
Re: WA# 6
Послано bsu.mmf.team 11 окт 2010 18:47
33.333333333
Re: WA# 6
Послано Ibragim Atadjanov (Tashkent U of IT) 27 окт 2010 16:20
Which of these formulas correct for this problem
1. M(A) + M(B) - M(AB)
2. M(A)*M(B/A)
Re: WA# 6
Послано Vasily Slesarev 21 май 2011 16:02
Please give tests for N=
20
30
40
50
60
70
80
90
100
Re: WA# 6
Послано Amirbekov Artem[Ivanovo SPU] 9 ноя 2011 03:17
Could you explain what formula is for max(M(A),M(B))?
bsu.mmf.team писал(a) 10 октября 2010 16:47
This formula is almost correct. The following example will illustrate it. Imagine you have two groups of rockets (A and B) and you know, that group A will launch in 10 seconds with probability 1/2, and in 20 seconds with probability 1/2 (math expectation equals 15 in this case). The group B has probabilities 1/3 and 2/3 correspondingly (math expectation = 50/3). But a probability that all rockets will launch in 10 seconds equals 1/2 * 1/3 = 1/6 (it is almost evident), and in 20 seconds: 1 - 1/6 = 5/6. Math expectation = 55/3.
So, the only mistake in this formula is M(max(A,B)) != max(M(A),M(B)). Change this, and this fomula will be correct. By the way, I got AC using it.