OJ Board

Hi.

I was thinking about this code fragment

if (w_ptr == &head)
{
if (head.next == &head)
insert_to_list(w_ptr, &in_node);
else
insert_to_list((w_ptr->back), &in_node);
break;
}


why not just:

if (w_ptr == &head)
insert_to_list(head -> back, &in_node)

since (&head == head.next ...

Nope, no AC yet.

my new implementation (in C++) gets your suggested input right: "0 0 W LOST". If any one of you have gotten AC, could you do me the favor of running my test input through your solution? Thanks in advance.

just in case you want to see the new code, here is it:

# include <iostream ...

To ImLazy: thanks for you reply. Unless I'm mistaken, your suggested logic is the same as mine, except that the first condition is doubly negated ('if not not ...'). Of course, I could very well be mistaken, in which case: could you point out which part of the code is wrong?

To everyone: it's ...

upper left. Says so in the problem.

try running this:
time echo "1 1000000" | java Main
that is, take time on how long it takes to solve the biggest problem instance possible. I think you'll be surprised.

To speed it up, you might want to save some of your previous results, so you don't run through a gazillion iterations for each ...

I get WA for problem 118, which frustrates me quite a bit, since it sounds deceptively simple. I think I've read the problem carefully enough, but I might be wrong. The only 'trick' I see is the fall-over-the-edge logic, which I understand as:

next.point = getmove(current.point)
if (next.point is ...

It's possible to do this problem without floating-point calculations at all: using the given area formula, we can compute twice the area as an int, then output either '50' or '00' as the decimal part, based on whether the int is even or not.