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Discussion of Problem 1193. Queue at the Exam

I don't understand problem 1193. Please help me!
Posted by raxtinhac 2 Mar 2002 20:00
What change if shift beginning time of the exam to more early
time ? Please explain to me. Thank!
Re: I don't understand problem 1193. Please help me!
Posted by hajime 2 Mar 2002 23:49
> What change if shift beginning time of the exam to more early
> time ? Please explain to me. Thank!

70  +  40  =  110 // first student (70 40 150)
110 +  15  = 125  //second one      (99 15 400)
125 +  10  = 135  // third one (100 10 120),
  he is 15 min late, (he's to be free at 120)

i hope it helps.
Re: I don't understand problem 1193. Please help me!
Posted by MadPsyentist/Sam 3 Mar 2002 00:09
> > What change if shift beginning time of the exam to more early
> > time ? Please explain to me. Thank!
>
> 70  +  40  =  110 // first student (70 40 150)
> 110 +  15  = 125  //second one      (99 15 400)
> 125 +  10  = 135  // third one (100 10 120),
>   he is 15 min late, (he's to be free at 120)
>
> i hope it helps.
>
does this mean third should start at 110 and second start at
110+10=120?
Re: I don't understand problem 1193. Please help me!
Posted by hajime 3 Mar 2002 00:45
> > > What change if shift beginning time of the exam to more early
> > > time ? Please explain to me. Thank!
> >
> > 70  +  40  =  110 // first student (70 40 150)
> > 110 +  15  = 125  //second one      (99 15 400)
> > 125 +  10  = 135  // third one (100 10 120),
> >   he is 15 min late, (he's to be free at 120)
> >
> > i hope it helps.
> >
> does this mean third should start at 110 and second start at
> 110+10=120?

students should start the exam as fast as they are ready (by the order
of T1)
so second student cannot enter the room after third one ( i suppose ;)
)
arrghh i'm sorry. i still not understand the problem
Posted by MadPsyentist/Sam 3 Mar 2002 02:18
if the input is
4
100 10 120
70 40 150
99 15 400
101 50 160
what's to output and why
Re: arrghh i'm sorry. i still not understand the problem
Posted by Vinicius Fortuna 3 Mar 2002 02:43
Solution:

1st) #2 starting at 70 ending at 110
2nd) #3 starting at 110 ending at 125
3rd) #1 starting at 125 ending at 135  (a +15 shift)
4th) #4 starting at 135 ending at 185  (a +25 shift)

That way the answer should be 25

> if the input is
> 4
> 100 10 120
> 70 40 150
> 99 15 400
> 101 50 160
> what's to output and why
Re: arrghh i'm sorry. i still not understand the problem
Posted by hajime 3 Mar 2002 02:48
> if the input is
> 4
> 100 10 120
> 70 40 150
> 99 15 400
> 101 50 160
> what's to output and why

let's see

70  + 40 = 110 (70 40 150)
110 + 15 = 125 (99 15 400)
125 + 10 = 135 (100 10 120) 15 min late
135 + 50 = 185 (101 50 160) 25 min late

if examenator shifts the beginning of the exam by 25 minutes then
every student will pass the exam before T3.





Re: I don't understand problem 1193. Please help me!
Posted by raxtinhac 3 Mar 2002 07:04
Thank you hajime, I got AC.
thank you very much
Posted by MadPsyentist/Sam 3 Mar 2002 15:43
> if the input is
> 4
> 100 10 120
> 70 40 150
> 99 15 400
> 101 50 160
> what's to output and why
you are wrong
Posted by Oleg 29 Dec 2002 09:25
70  +  40  =  110 // first student (70 40 150)
110 +  10  = 120  //second one      (100 10 120)
129 +  10  = 139  // third one (99 15 400),
 he is free at 139?
 wait =0;
Re: you are wrong
Posted by Locomotive 13 Jan 2003 18:38
> 70  +  40  =  110 // first student (70 40 150)
> 110 +  10  = 120  //second one      (100 10 120)
> 129 +  10  = 139  // third one (99 15 400),
>  he is free at 139?
>  wait =0;
Oleg i think
thirs student comes in queue before second
as his t1 is 99 and thirds is 100
agree?
Re: you are wrong
Posted by ZhengYuhang 10 Aug 2007 18:32
Thank all the people above!
Re: you are wrong
Posted by Vo Minh Hien 15 Aug 2010 14:01
Thanks all. I got AC