OJ Board

Do you find yourself debugging in your sleep? Enjoy solving tough problems and grappling with technical challenges? Then enter Google Code Jam and compete for more than $80,000 in cash prizes!


Google Code Jam is an annual competition in which programmers – both students and professionals – are ...

Do you find yourself debugging in your sleep? Enjoy solving tough problems and grappling with technical challenges? Then enter Google Code Jam and compete for more than $80,000 in cash prizes!


Google Code Jam is an annual competition in which programmers – both students and professionals – are ...

Your case 2 output looks quite off.

Hehe. I should have explained that that's how I actually did solve the problem. My not-really-optimized search took an hour (though I did remove translations and reflections by fixing a 1 in the upper-left corner, and applying two A > B restrictions to the 4 numbers adjacent to the 1). You need to ...

The only way I could see to solve this in a real contest would be to run a search for all the magic squares (eliminating translations/reflections) for an hour or so, put them into a table, and look them up for each example. This still isn't fast enough because of the 15,000 cases, though, so you'd ...

Technically, the final row of output should be
-15 -168 -221 -347
Since the "f" function is specified as being given in order of increasing y from top to bottom. But your output is what ends up accepted.

To avoid the PE, put a space after each line of numbers.

Make sure you're sorting them correctly. 35.91, for instance, has 3, and they should be in the order "0.21 + ..." "0.50 + ..." "0.76 + ...".

Here's my code, if someone wants to tweak it.

#include <cstdio>
#include <map>
using namespace std;
typedef signed long long i64;

int size;
map<i64, double> memo;
double nmat(i64 x) {
double &ret = memo[x];
if( ret != 0.0 ) return ret-1;
ret = 1;
int d = (x>>(2*size));
if( d == size*size ...

I tried a short DP algorithm (looking at the matrix as a string of length n^2, only the last 2n choices are relevant to any remaining choices.) Made it up to 11 before running out of memory. I'm not sure how you'd improve this. (I used doubles instead of BigInts, so the last answer's inexact ...

I get 35 as well. It sounds like your search is correct; my guess is that you have some minor error when outputting the winning moves? Something cosmetic, anyway.

Here's how I thought about it when I made the problem. For each prime p, we determine the _unique_ first move for A that will allow A to force taking p. The reason that it is unique is simple: suppose A could pick both 'x' and 'y' as his first move to force taking p, with x < y. Then when A picks x ...

Try the following input:
a
(blank line)
a
a
a
(blank line)

It's a little-known fact that Yarin's phrase "I noticed my program was wrong", when used in this context, ACTUALLY means "SnapDragon pointed out that my program was wrong".

I avoided using floating point whenever possible. I determine whether the ray intersects individual cubes wholly with integer arithmetic; the only doubles I use are for calculating how high up on the cube the intersection point is.

You have to remember a few important facts about fluid dynamics ...

This question is horribly badly worded... he WANTS you to choose your permutations in a certain way, but he gives you absolutely no useful description of what that way is, and the sample output he gives you isn't sufficient. As was pointed out, the sample output has gurcoth-gcruoth when gurcoth ...