[gcj] Magic Trick problem mistake (?)

2014-04-13 Thread newbie007
Hi,

If in the second arrangement, the cards from the first chosen row are not all 
in different rows, then he IS a Bad magician!
I know that the magician may be luck and still be able to find the card 
depending on the volunteer choices, but it doesn't change the fact that he IS a 
Bad magician!

For me this is not a corner case. It is one thing that was not clear in the 
problem, so it should be covered in the sample cases.

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Re: [gcj] Magic Trick problem mistake (?)

2014-04-13 Thread Leopoldo Taravilse
The problem statement is very clear on when he is a bad magician. From the
problem statement:

If there are multiple cards the volunteer could have chosen, y should be
Bad magician!

So it doesn't matter if in real life that would be a bad magician. In the
problem it's only a bad magician only in that case.


On Sun, Apr 13, 2014 at 8:54 AM, newbie007 lescoutinh...@gmail.com wrote:

 Hi,

 If in the second arrangement, the cards from the first chosen row are not
 all in different rows, then he IS a Bad magician!
 I know that the magician may be luck and still be able to find the card
 depending on the volunteer choices, but it doesn't change the fact that he
 IS a Bad magician!

 For me this is not a corner case. It is one thing that was not clear in
 the problem, so it should be covered in the sample cases.

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Re: [gcj] Magic Trick problem mistake (?)

2014-04-13 Thread Xiongqi ZHANG
Sometimes you cannot interpret the problem using your own understanding of
Bad magician

The description is fairly clear.

---
​​
If there is a single card the volunteer could have chosen, y should be the
number on the card. If there are multiple cards the volunteer could have
chosen, y should be Bad magician!, without the quotes. If there are no
cards consistent with the volunteer's answers, y should be Volunteer
cheated!, without the quotes.
---

As you can see, you should output Bad magician! only if multiple cards
could have been chosen by volunteer.

As you said before, the result may depend on volunteer choices, but as long
as only a single card the volunteer could have chosen, you should still
output the number on the card instead of thinking that he is a bad
magicians.


2014-04-13 19:54 GMT+08:00 newbie007 lescoutinh...@gmail.com:

 Hi,

 If in the second arrangement, the cards from the first chosen row are not
 all in different rows, then he IS a Bad magician!
 I know that the magician may be luck and still be able to find the card
 depending on the volunteer choices, but it doesn't change the fact that he
 IS a Bad magician!

 For me this is not a corner case. It is one thing that was not clear in
 the problem, so it should be covered in the sample cases.

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 To view this discussion on the web visit
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