OJ Board

hehe~~

I got AC la..... XD

Thanks Cho~ Thanks everyone~~~

Thanks again~~
It really help me a lot..
I know what is the problem now....
But i think it is quite hard to solve it.....

Thanks very much for Cho~~~

I do fix some bugs with your test case....... and get all your case correct now.
However, I still get WA... T_T....

Both correct for the above case~~~
Sigh~~~ :cry:

Any other special case?

I have try:

1
P
At most 0 rose(s) trampled.
13
RRR.RRRRRRRRR
RR....R..RRRR
RRRRRRRR..R.R
RR.R.RRRRRR.R
...R.R....R..
R.RR.RRRRRR.R
RRRR.RPR.RR.R
R.RRRRRR.RR.R
R.R....R.R..R
R.RRRRRR.R.RR
R.R..RRRRRRRR
RRRR....R..RR
RRRRRRR ...

I got 3 WA~~~

does anyone have good test case?

No.

Consider, for example, the test case
9
####.#.##
####.#.##
#..#.#.##
####.#.##
####P####
##.######
##.##....
##.######
##.######
There is only one minimum weight exit path, but applying your second step to that path would give you the answer "4", whereas the real answer is "2" (take two ...

What I want to ask is What is the limit of n?

I can't imagain how people can do it such quick and use 64k only? :o

If I am not wrong. To find the inverse of a matrix,
It must first find the determinants of the matrix, and then find the element of the inverse by found out the cofactor of the ...

I am not crazy, but very crazy.... 8)

You,
[cpp]
if(me==prof) printf("0 %c\n",me);
else {
:
}
[/cpp]
So, I gress you will have error when more then one point take 0 energy. If you are right with the test case, I think you are ok.

I think you are right about the second problem. I misunderstand ...

I got accepted!!

try this:

6
Y U A Z 0
Y U C A 0
Y U A Y 0
M U D Z 0
M B C D 0
M B C Y 0
A D

0 Y Z

Besides, there may have problem too.
[cpp]
close!!
[/cpp]
The 'min' you have found may not be the real minmum. if you count it together with the real one, it is wrong!

Hello, first time to reply you here. I just want more people to reply, as I got WA too.

However, I found some error in your code.
First. what you have print out when:

if(me==prof)
printf("0\n");


It should be

if(me==prof)
printf("0 %c\n",me);


Second, I am not sure am I right( I got ...

The problem say n is always >0, unless the end, so it would no be 0. The is to say the answer of that problem would never be 0. As the guy at least can pick up 1 ring.
So there would not be 0 ring/rings problem over there.

Hello, every body! First post!!!
Don't be misled by Joe, the answer of that question would never be 0;
There will be at least one ring, the boy can pick up. Unless n is Zero, but if so, the question is meaningless.