Yep
On 1 May 2012 14:11, "xenoky" <[email protected]> wrote:

> Thank you for your explanations.
> i think in your second sentence you mean "...C is next to the back of
> B. C and B must therefore be in different lanes..."
> So car C oblige car A & B to be in the same lane (and when B overcame
> car C, there is car D that cannot permit B to change line)
>
> Now i understand that i cannot say if t is a solution only based by
> distance covered by cars
>
>
> On 30 Apr, 19:49, Luke Pebody <[email protected]> wrote:
> > At time t=10.01, the cars are at positions:
> > A - [40.04,45.04]
> > B - [49.02,54.02]
> > C - [45.01,50.01]
> > D - [54.01,59.01]
> >
> > Note that the front of A is next to the back of C. They must be in
> different lanes, and must stay in those lanes until A fully passed C at
> time 13.3333.
> >
> > Note also that the front of C is next to the back of A. C and A must
> therefore be in different lanes, and so A and B must be in the same lane.
> >
> > Note also that the front of B is next to the back of D. They must be in
> different lanes, and must stay in those lanes until B fully passes D at
> time 20.
> >
> > So to summarise, from time 10.01 to time 13.333, cars A and B are in the
> same lane and cannot change lane. Therefore, at time 12 when A reaches B,
> maximal bang occurs.
> >
> > On 29 Apr 2012, at 21:16, xenoky <[email protected]> wrote:
> >
> >
> >
> >
> >
> >
> >
> > > Hi,
> >
> > > The 4th test case in the Cruise Control problem has the following
> > > input:
> > > 4
> > > L 4 0
> > > L 2 29
> > > L 1 35
> > > L 1 44
> >
> > > The given solution is:
> > > Case #4: 12
> >
> > > calculating the distance at t=12
> >
> > > 4*12=48
> > > 2*12+29=53
> > > 12+35=47
> > > 12+44=56
> >
> > > ordering this distance [47,48,53,56]
> >
> > > The maximum distance beetween nearest triplet is 6! -> [47,48,53]
> >
> > > I can put the car this way
> >
> > > ##### #####
> > > #####
> >
> > > without accident
> >
> > > Where am i wrong?
> >
> > > My program answer is 13...
> >
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