When you start, you're in the bronze division. It's like B-C div2 problrms.

It has 3 divisions Bronze, Silver and Gold, each of them getting harder. If it is you first contest you need write Bronze, otherwise in division at what you was last year.

Maybe some will not be agree with me, but the level is kind of random for me... Some contests are a lot harder than others, even in the same division. This month, it seems the easiest contest since a long moment, and it's like Xellos said, "B-C" div2. But I will say it's more D div2 for third problem usually.

Really? I didn't have any. Just a slow server and one "Judgement failed".

I was able to start just 45 minutes ago and I think it is still possible to start, so you should refrain from discussing problems for about 4 more hours

It should be fine now.

For each cow C, partition the area in which visible cows can be (given by the circle and tangents to it from C, non-white in the pic) like this:

The number of cows in the gray area can be found by angle-sorting the points and doing prefix sums. The numbers of cows in colored areas are given by the tangent points on the circle and orientations — list all those tangent points, then angle-sweepline the answers for them. Ugly stuff.

I thought the problem was quite nice. Here is my solution:

Each cow forms two tangent points with the circle. Consider the minor arc on the circle formed by those two points. It turns out that two cows can see each other if and only if their arcs intersect.

Therefore, we can just calculate the range of angles possible for each cow and solve the well-known problem of counting the number of pairwise intersections in a set of ranges (we need to be careful to handle the fact that the ranges are cyclic though). The total runtime is O(NlogN) since we must sort the list of angles.

I have a solution which is a bit more motivated than the one presented by scott_wu.

For each cow, there is a region it can see: it is bounded by the silo as well as the two tangent lines. The entire half-plane above the upper tangent line is visible, and similarly, the entire half-plane below the other tangent line is also visible.

In fact, we can simply add up the number of cows above the first tangent line and the number of cows below the second tangent line. Although this doesn't include the region between the cow and the silo, it also double counts the intersection of the two half-planes. Taken together, these two imperfections cancel out.

Because each pair of seeing-cows is counted twice, you simply divide the total sum by 2 at the end.

Finally, the problem is reduced to querying the number of cows in a halfplane determined by a tangent to the silo. This can then be reduced to counting the number of pairwise intersections of a set of ranges.

What do you mean by "them"? Results?

They usually publish results one week after the end of the contest... What do they mean by "soon"?

Results of Usaco announced soon? Wake up man...

Dynamic programming: for each subset of coins calculate the maximum number of products that can be purchased.

My bad. It is a wrong solution. :)

It seemed to me like easy (1.), medium (2., although a bit too easy for my taste), hard (3., still had a nice solution that hex539 and scott_wu did). That's a good thing. A bad problemset is one with 3 easy or 3 hard problems, because it doesn't give you any learning experience.

As far, as I remember, the first contest of the year is always easier than following ones.

can you tell me where i can get all the data ?

Heh, for me it was 2 symbols. Something like this:

a.resize(n); -> a.resize(sz);

But I failed only one test, so it's a great luck for me)

Just a set and a vector shouldn't exceed the memlimit, but you can run it on your computer with all the testdata and check what it does.

You never define "m", but you use it on a for loop:

for(int i=0 ; i<m ; i++)

Ray Li is just that good

I know the Australian Ray Li in real life and I can tell you that they really are 2 different people. It is a funny coincidence :)