OJ Board

Never mind. Found my mistake in the output section. Both, your and my version, for Hamilton is OK. Equal fractions issue doesn't seem to occur. Just for the info I'll underline the correct answer for my question. And regarding output. When choosing which methods a state is favored by, should only th...

Tried it your way for Hamilton, even used long longs. Still doesn't work. I'll try to find the mistakes without sending the code for the moment, though. If I'll be ready to give up on it, then. Then use fractions in descending order to add 1 to rep until all extra representatives are distributed. Do...

The total is 3,295 so for 49 representatives, we divide by 67.24 for the true apportionment of representatives. In the Hamilton method, the left overs go to Arizo, Idaho, Calif, and Color. I divide by 63.20 for the Jefferson method. I divide by 72.60 for the Adams method. I divide by 67.50 for the ...

You mean match for that particular case only? And I agree, I can't figure out the logic behind that output. As far as I can see it, that output seems kind of an impossibility. Anyway, hopefully the test set will get switched and problem rejudged. I don't know how long it even takes Carlos (is he sti...

brianfry713 wrote:Yes the sample output is wrong. This problem does not have a correct dataset.

Can you elaborate on how you got it accepted? What should be interpreted differently to what is stated in the problem description?

I.

Without assembler, in pure C, with serious IO and data representation optimizations. No more than that. (And I won't tell what exactly.) It took me a while to come to this solution. I picked this problem up not so long ago after not touching it for at least 1.5 years. Before that my time was about 0...

I miss the times too. It was so fun to compete for being better than others. Many of the programs are still there to be taken up again Hope I can manage someday soon enough. Till that, see you on occasions in between.

I.

It isn't possible to solve this in 0.000 at all. The time I have is the result of the glitch in the system some two years ago. Else I would have time around 0.010-0.020. That is a possible time. With good IO and a trick to generation of the numbers.

Cheers,
Ivor

Heh. :) Cool guys you still remember me. :) Very pleased to see that. I've been pretty busy for the last year or so. Studies take their fair amount of time and now work too so I don't have anything to spare for at least one more term. I drop in from time to time to see how things are going though. I...

The first three times -- 0.000 and 0.002 are probably nothing more than simple io solutions, ie one of them told me he had io somewhere and sent it. I think there might be a really easy solution to this problem but I didn't find one. The solution I have is DeBrujn cycles algorithm. Wish I could reme...

a[++y]=a[y]+1; Say y = 1. ++y => y = 2 and your code becomes: a[2] = a[2] + 1 y++; a[y]=a[y-1]+1; Same starting point, say y = 1. y++ => y = 2; and assignment becomes: a[2] = a[2 - 1] + 1, ie a[2] = a[1] + 1. That's the difference. I haven't tested it but it is the way I have always understood C co...

It has been some time since I really enjoyed some insane solving and search for better algorithms. As Carlos said 495 is one of the most interesting problems and the battle with Scott and Ivan is the one I enojoyed very much. So, 623 has actually become a lot more interesting problem from the solvin...

Or then again it doesn't matter... I sent the other version with shorter words first and it was accepted too. Hmm... I wonder what it was I messed up in my first solution.

Anyway, sorry for keeping up a conversation with myself,
Ivor

1) aab 0p, 0g (0-0-0), 0gd (0-0) 2) Aac 0p, 0g (0-0-0), 0gd (0-0) 3) aA 0p, 0g (0-0-0), 0gd (0-0) 4) aa 0p, 0g (0-0-0), 0gd (0-0) 5) BBa 0p, 0g (0-0-0), 0gd (0-0) 6) bbc 0p, 0g (0-0-0), 0gd (0-0) 7) Bb 0p, 0g (0-0-0), 0gd (0-0) 8) bb 0p, 0g (0-0-0), 0gd (0-0) Anyone bothers to explain how come this...

Thought I just might drop this one. A word can consist of only a valid suffix, ie length of the root of the word is 0. I assumed otherwise, so got a bit of nasty testing.

Ivor