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SRM 115
Monday, September 30, 2002

The Problems

CoinStacks  discuss it
Used as: Division-II, Level 1:
Value 300 points
Submission Rate 207 / 247 (84%)
Success Rate 90 / 207 (43%)
High Score dudedude for 275.69 points

One of the more annoying problems, this was a straight-forward simulation. Just code the rules, in order, and you'd get the right answer. Unfortunately, mildly confusing wordiage made the problem trickier than it had to be.

In all honesty, your best approach to this would be to simply avoid thinking. Don't try to be clever, don't try to save time, just write code that corresponds exactly to what they tell you to do, one coin (or five coins, depending on the step) at a time. Since the number of coins is limited to 10,000 you won't have any speed problems. Many failed solutions included attempts to speed up the execution, or avoid having to write as much code, and obviously this is a bad idea if your resulting code doesn't work.

dudedude managed his high score with some clever tricks to reduce coding time, proving that yes, it can work. Judging by the number of people whose tricks failed, I wouldn't consider it worth the risk, though kudos to him for pulling it off.


Mail  discuss it
Used as: Division-II, Level 2 & Division-I, Level 1:
Div-II stats
Value 500 points
Submission Rate 142/247 (57%)
Success Rate 76/142 (54%)
High Score UFP2161 for 395.88 points

Div-I stats
Value 250 points
Submission Rate 128/134 (96%)
Success Rate 89/128 (70%)
High Score ZorbaTHut for 233.09 points

Definitely another case of "don't try to be too clever". The easiest way to do this was first to make an array of 0's, the same size as the number of offices. Then for each letter, find the score for each office relative to that letter. Keep the best one. If it's 10 or above, add one to the array item for that office. Otherwise move on.

Finding the score, however, proved slightly trickier. A lot of potentially painful code. Some people decided to do one pass through both zip codes and just do the tests in the appropriate order. I personally passed through once for each test and changed the digits in each code to something that would never match up ('a' for the first code, 'q' for the second - 'b' wouldn't work because 'b' is one character away from 'a', so it would have matched the off-by-one test!)

In any case, once that part was working, the problem was solved.

Shaws  discuss it
Used as: Division-II, Level 3:
Value 1000 points
Submission Rate 17/247 (7%)
Success Rate 3/17 (18%)
High Score fiddlybit for 540.06 points

Definitely not the kind of problem I'd do for fun, with lots of - if you'll excuse the expression - fiddly bits. On first inspection this problem looks like pure death. Tons of special cases and things you just plain don't want to deal with. I've come up with a solution that seems relatively simple, though.

You're going to need a few variables and arrays. First off, your output string array - you're just going to be adding things on the end of this. Second, you'll need a running total cost, started at 0. Third, you'll need a running taxable total - I don't know about you, but I'd much rather not have to go back and figure this out. Fourth, you'll need a flag for whether the rewards card has been scanned yet. Fifth (yeah, I know, I'm sorry) you'll need either one array or three arrays, to store the name, reward amount, and taxability of any unclaimed rewards.

Does this sound painful to you? I know it sure sounds painful to me.

On the other hand, it turns out it's actually all quite simple.

First off you may want to parse the input arrays into something useful. I personally would end up doing a map< string, Item > with a custom Item class. Some people might find it easier to just have a set of four arrays: "name", "taxability", "cost", and "rebate". Searching through all 50 items will be fast enough, we're only going to be doing it 50 times. Not a problem. The actual mechanics of the parsing and arrays depend entirely on your language, and I'm a C++ coder - arguably the worst TopCoder language at parsing - so I'll leave the parsing up to you.

Now, as for the actual process . . .

Read an item in. If it's the rewards card, add "Rewards Card Scanned" to your output array and flip on the "rewards card" flag. If it's not the rewards card, check your arrays. Add the appropriate string to the output array ("Itemname: cost" or "Itemname(T): cost" if it's taxable), add to the total cost and the taxable total (if applicable) and, here's the interesting part, add the rebate information to the end of your rebate array. No, not to the output array!

Then, if your reward card flag is set, *now* take everything in your rebate array - no matter how many that might be - and add them one at a time to your output array, removing the appropriate amounts from your total cost and your taxable total as you go (which you have cleverly stored in your rebate array, to keep you from having to look them up again.) And then clear your rebate array. If your reward card flag is *not* set, just leave your rebate array intact and move on.

Once you're done with your array, add tax, if applicable - make sure you don't accidentally write "Tax: 0" - and then print out the total. Congratulations, you're done.

Now, as for why that thing with the rebate array works . . . If you don't have a rebate card scanned in, you'll just accumulate more and more items as you go. If you do have it scanned in, it'll end up dumping the rebate for each item immediately after the item. If you go halfway, it'll end up dumping *all* the rebates for all the previous items immediately after the rebate card. Which is exactly what you want.

So there's *that* problem. Moving on.

Animals  discuss it
Used as: Division-I, Level 2:
Value 450 points
Submission Rate 73/134 (54%)
Success Rate 36/73 (49%)
High Score ZorbaTHut for 440.73 points

Look at that score. So, what's your secret, you're asking? How did you do it so fast?

"Animals" is a textbook dynamic programming problem. Literally textbook. Make a big array, "feet" by "heads" large. I added a little buffer around the edge so I didn't have to worry about bounds, and even hardcoded the size to be the maximum size. I'll snag whatever data I want out of the middle. Initialize the entire thing to 0, except the 0,0 point, which should be set to 1 (after all, it's easy to have zero heads and zero feet - it's called having zero animals.)

Now, for each animal, you want two nested loops. If we could have negative-legged animals we'd want the "legs" loop to be the inner one, but we can't, so I believe either way works. Now, if we have animal "curanimal", "curheads" heads, "curlegs" legs, and animal leg counts "legcounts", we end up with the line animalcombinations[ curheads + 1 ][ curlegs + legcounts[ curanimal ] ] += animalcombinations[ curheads ][ curlegs ]. Representing adding one animal of the current type to whatever combinations we currently have.

Then we return animalcombinations[ goalheads ][ goallegs ]. And we're done.

Yes, it's possible for this to return 0 (1 head, 2 legs, but it has to be all cows - can't be done.) This solution does so easily.

Note that it's possible to do this with nested loops. Not fun, but possible. I wouldn't believe it if I hadn't seen it, but vorthys's solution in Room 1 took down four challengers before people gave up trying, and made it straight through systests.

Uniscraper  discuss it
Used as: Division-I, Level 3:
Value 1100 points
Submission Rate 21/134 (16%)
Success Rate 6/21 (29%)
High Score Yarin for 649.87 points

I don't know about you, but I get scared when I see an 1100pt problem. 1100 points is a lot. And that might have something to do with my overly-efficient solution.

Now, a solution that *works* is much simpler than mine.

In effect, it's just another dynamic programming deal. For any possible combination of board layout and current card, there may be half a dozen ways to get there, but only one "best" way. In this case, "best" is defined as "lexicographically smaller". You can throw out all the other ways to get there, they will never be better than the lexicographically smaller one. This gives us a lot of options.

The easiest way, by far, is to represent the current state as a large string. In fact, as the large string that's given to you as input. Hey, it works, and it's cacheable. Your function should check the cache for that string and return what's there if there is something. If there isn't, it should just check all possible moves and recursively call itself, choose the best of those moves, tack its own card on the beginning of that move list, and return that. If there are no valid moves it should return an empty move list.

Yes. That's all there is. I'm not joking here, and it works.

Part of the *reason* it works is a very limited number of possible board states. The card numbers are set from the beginning. The only thing that varies is whether a card is there or whether a card is not there. 36 board positions would seem to yield almost 70 billion possible states, but most of those aren't valid. For example, you can't have a solid pyramid with the tip of it removed. There's no way to remove the tip, it's still full!

It turns out, for reasons I do not pretend to have the math background to understand, that the number of states can be represented by a series called Catalan numbers. With a height of 8, we end up with the 9th Catalan number, or 4862 states. 4862 is not very many at all - consider this is *all* the possible states, including all run lengths. In fact we'd have to go up to a height of 11, with 208012 states, before I'd even begin worrying about speed, and you might even be able to manage a height of 13 - 2674440 states - before things started breaking.

In any case. 4862 is a very small number of states, and so this apparently brute-force solution will, in fact, work beautifully.

Stop here if you want to keep your sanity. If you want to read about my solution, which finishes the worst case in a hundreth of a second, read on. My solution would be great if we had an enormous height - as it is, it's rather redundant.

In my case, I did many evil things involving bitmasks. I represented the board as 36 bits (stored in a 64-bit int), one for each card. From there I made an associative array, with that bitmask and an integer ("current card value") used for finding an array of ints. Each step I'd take every item in that array, make all possible moves, and put it into a new array. Any board states I already had, I'd throw away the inferior move order and replace it. Then I'd basically swap arrays and do it again, until I ended up with zero states in the new array - in which case it was merely a matter of searching through the previous array for the best chain of that length. This is, in effect, a breadth-first search.

But wait! The process of "making all possible moves", which I've glossed over, was a bit trickier than that. I'd loop from 0 to 36 - one for each possible move. First I'd check to see if the card in that location would work with the card I had (the card locations were a global array, since they never changed.) Then I'd make sure that card actually *existed*. And then I'd refer to another global array, this one of 64-bit bitmasks, representing which cards couldn't exist for this move to be valid. Some of them were simply 0 (the bottom cards, which obviously have no requirements to use), but the rest included two bits set to 1. I'd do a bitwise-and with my bit array, and if there were any bits left 1, that meant there were cards in the important locations, and I couldn't make that move.

So there's my complex unnecessary solution. But it sure is fast!

And it sure took a long time to debug.

This competition's lesson: never make things more complicated than they have to be.

By ZorbaTHut
TopCoder Member