OJ Board

Thanks, Sohel.

And he calls this D. What is "this"? I don't understand the difference between N and D in the problem statement. The author describes a process to get D to 1 and then asks how long it will take for N to get to 1, but there is nothing in the problem statement that describes how N changes. I'm confus...

Got it. Thanks, brianfry713. The Cantor set is cooler than I thought.

What am I missing here? The only members of the Cantor set that can be represented as a finite decimal fraction are 0 and 1. Everything else should be a "NON-MEMBER". It's true that I've just finished a bottle of wine by myself while solving this problem, but the logic above should still be sound, n...

You are welcome. I'm glad you liked it.

Yep. Same error here, independent of the browser. It is a bug on the website.

In the sample input case #1, why isn't "is" a word? It appears twice: as "disd" and as "psip". Both occurrences are overlapped by other words: "disd" is overlapped by "tdisdort", and "psip" is overlapped by "ozgpsipkelao". I think the sample output should be dyj is ortj lnsvkoalexot * * What am I un...

Oh garbage! Print the numbers after "Today is:" in fields of width 2 as well. After 5 years, I finally got rid of the Presentation Error and got AC.

Vytenis wrote:Maybe, you missed the fact that all four coins that you have to guess are of distinct colors?

Yep, that was it. Thanks.

Could somebody who got this problem please explain the precise rules for computing n1 and n2? I used Mastermind rules, and I'm getting WA. Thanks.

Thanks, helloneo.
I'm confused though. Does anyone understand what the problem statement means by "1 is a prime". When do we count 1? How many times do we count 1?

Edit: Nevermind. I'm dumb. Fixed.

What is the answer for n = 10000000? My WA program prints 3203899.

You could also set up a huge bit vector to make updates O(1), as long as you only use one bit per number, but it wasn't necessary.

pvncad wrote:use cycle detection algorithm

Or you could just remember all of the numbers you've seen so far and stop as soon as you reach a previously seen number. That's what I did.

Yes, one of those.
...and I do solve UVa problems at work quite often.