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lgarcia

Some cases, I hope it helps.

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lgarcia

It happens

lgarcia

Change "long int" to "int" and you will got AC.

lgarcia

I did exactly what you did in the first version. The only thing you didn't name was to check the fuel used between the start (or last gas station) and the goal. Other than that, I don't see any difference. Maybe add some epsilon when you print the result.

lgarcia

My critical input was this:

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1
00??0?11

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Case 1: 2

lgarcia

You can solve it with binomial coefficients and modular multiplicative inverse.

lgarcia

Ok, I did. You may want to remove your e-mail address to avoid spammers.

lgarcia

Yes, if you increase the distance between them, the radius becomes larger. But if you make it smaller, eventually the upper 'v' also makes the circle bigger. With ternary search you can get the optimum distance.

lgarcia

Almost, I actually solved with Ternary Search and I took the distance between the two opposite upper blocks as my variable, that was enough.

lgarcia

In the third move, the fourth player takes the R2 card from the draw pile and then he put it in the discard pile, he doesn't keep it.

lgarcia

Well, by now I'm 20 and I started to submit in June 2009. I have ~660 accepted problems!

lgarcia

It's more like a DP problem, think about what states you have, a robot can be destroyed or not, and for any combination of destroyed robots you can destroy others (independently of the order they were destroyed). I hope it was helpful.

lgarcia

That was the problem (the two of them). I could say it directly, but people here don't let it be so easy. By the way, you have to keep trying, I needed several months to get one accepted and I have ~350 problems accepted by now.

And yes, you should delete the code after you got it accepted.

lgarcia

Check this.

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Case 1: Escaped again! More 2D grid problems!
Case 2: Escaped again! More 2D grid problems!

lgarcia

I'm sorry, I should be more clear. We get this from the problem, ln(n) = L + ln(1 - x) => n = e^L * (1 - x) Also we have this, |x| < 1 => -1 < x < 1 => 0 < 1 - x < 2 Then, e^L * (1 - x) < 2 * e^L => n < 2 * e^L => ln(n) < ln(2) + L => ln(n) - ln(2) < L Since we don't get (or I didn't see) any other ...

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