## Google Code Jam 2010 Qualifying Round Solutions (Part 2)

13 May

Here’s my solution to Fair Warning Large input. It’s actually very similar to my Small input solution except that longs are replaced with BigIntegers (and it’s designed to handle cases larger than 3. Anyway, here’s my solution:

import java.util.*;
import java.io.*;
import java.math.*;

public class Fair_large
{
String filename;
ArrayList<BigInteger> list;
ArrayList<BigInteger> diffs;

public Fair_large(String f)
{
filename = f;
openFiles();
}

private void openFiles()
{
Scanner s;

try
{
s = new Scanner(new File(filename));
int ct;

ct = s.nextInt();

for (int i = 0; i < ct; i++)
{
list = new ArrayList<BigInteger>();
diffs = new ArrayList<BigInteger>();
int ct2 = s.nextInt();

for (int j = 0; j < ct2; j++)

output(i + 1, process(ct2));
}

s.close();
}
catch (Exception e)
{
System.err.println("I/O error: " + e);
System.exit(0);
}
}

private String process(int length)
{
Collections.sort(list);
BigInteger T = BigInteger.ZERO;

if (length == 2)
{
T = list.get(1).subtract(list.get(0));

if (T.equals(BigInteger.ONE))
return "0";

return distToNextMultiple(list.get(0), T).toString();
}

for (int i = 1; i < list.size(); i++)

while (diffs.size() > 1)

T = diffs.get(0);

if (T.equals(BigInteger.ONE))
return "0";

return distToNextMultiple(list.get(0), T).toString();
}

private BigInteger distToNextMultiple(BigInteger x,
BigInteger T)
{
return T.subtract(x.mod(T)).mod(T);
}

private void output(int caseNum, String result)
{
System.out.println("Case #" + caseNum + ": " + result);
}

public static void main(String args[])
{
Fair_large app = new Fair_large(args[0]);
}
}


So in this solution, I did use some simple optimizations to reduce the amount of actual computation needed to produce my solution. First, I handled the 2 case by itself since you don’t actually need to use GCD for that. I created an auxiliary array called diffs that holds the differences between consecutive numbers (in the sorted array). Thus, we know that adding some $y \geq 0$ does not change the differences between them; so, if they must all be multiples of T, then their differences must already be multiples of T. Then, the GCD of the all the numbers of the auxiliary array is calculated. In my small input solution I wrote a slightly optimized (non-recursive – to save on mem read/writes on the stack) GCD function based on Euclid’s algorithm. At some point I also used Stein’s algorithm for the GCD calculation. But for the large input, I decided to just use the built-in GCD function that comes with BigInteger.

One key property that I exploited was that:

$\gcd(x_1, x_2, \ldots, x_n) = \gcd(\cdots\gcd(\gcd(x_1, x_2), x_3)\cdots)$

Thus, I computed the GCD of two consecutive numbers (the first two of the aux array), removed both, and added in their GCD. Continuing in this way until there was only one element left, I had computed the GCD of all the numbers with reasonable speed. The other interesting function was the distToNextMultiple function.

Mathematically, this function returns:

$(T-(x \mod T))\mod T$

We know that if $x < T$ then $x \equiv x \mod T$ but otherwise, it is the remainder after division by T. Subtracting this value from T gives how much is left until the next multiple. But why the extra modulus T? If $x \equiv 0 \mod T$, then $T - (x \mod T)$ yields $T$, which is wrong. Thus, the extra modulus T takes care of this case.

And that’s pretty much it. A pretty simple solution that works quickly. On the same Core 2 Duo machine, the solution for the large input runs in 0.436s. Good enough for me.

1 Comment

Posted by on May 13, 2010 in Uncategorized

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### One response to “Google Code Jam 2010 Qualifying Round Solutions (Part 2)”

1. May 17, 2010 at 6:16 AM

beast