Our current solutions to the problems in this volume are below. If a problem is not shown here, then we have not posted a solution yet.

#11309 - Counting Chaos

Solved By:wesley
Theory Difficulty:trivial
Coding Difficulty:easy
Algorithms Used:sorting
strings
Solution Description: This problem is fairly simple, but be careful with the numbers. I just use an integer to hold the time, so for instance:

If the starting time is 05:46, then hold 546. Keep incrementing this and checking whether or not it's palindromic. Whenever it hits a value like 560, change it to be 600 instead. That is, if x%100 == 60, then x += 40.

Also, if x == 2400, then x = 0. Don't forget to wrap around to the next day!

Note that the colon (:) doesn't factor into whether or not a number is palindromic. So it's OK to say that 05:15 is palindromic.


#11310 - Delivery Debacle

Solved By:wesley
Theory Difficulty:easy
Coding Difficulty:trivial
Algorithms Used:math
Solution Description: This is a third order reccurence:

a[n] = a[n-1] + 4*a[n-2] + 2*a[n-3]

The first part means adding a column of 2 single-cell cakes to the right of the previous term.

The second part means adding an L-cake/single-cell cake combo to the right of the 2nd previous term, in four different rotations.

The third part means adding two interlocking L-cakes to the 3rd previous term, in two rotations.

With a bound of 10^18, you can use signed longs to hold the sequence.


#11313 - Gourmet Games

Solved By:wesley
Theory Difficulty:trivial
Coding Difficulty:trivial
Algorithms Used:simulation
sorting
Solution Description: You can simulate the entire process as follows:

While there are m or more contestants remaining, remove (m-1) of them. If you end up with just 1 contestant left over, then output how many rounds it took. Otherwise, it can't be done.


#11321 - Sort! Sort!! and Sort!!!

Solved By:peter
Theory Difficulty:trivial
Coding Difficulty:easy
Algorithms Used:sorting
Solution Description: Just create a comparator that follows the criteria given, and use Collections.sort or Arrays.sort to sort it.

Some things to note:

-You need to use a StringBuffer and BufferedReader to make this run in time

-Make sure when you're doing % that you're taking the ABSOLUTE values of the mods. So, when you check for evenness or oddness, make sure you catch both 1, -1, 0, and -0 (stupid Java).

-In general, use all straight comparisons (so use < and >). Don't try returning the difference between two things in your comparator since you'll usually end up with issues.

-You need to use longs, not ints.

Solved By:wesley
Theory Difficulty:trivial
Coding Difficulty:easy
Algorithms Used:sorting
Solution Description: While this *should* be a simple sorting problem, there are some points to watch out for:

First, when you are checking whether a number is even or odd, remember that (x%2) = -1 if x is a negative odd number, and not 1.

Second, you will need to use BufferedReader and StringBuffer in Java, and even then it will take about 2 seconds to run.

Contrary to what Peter has stated, you do not need to use longs. The problem statement guarantees that the numbers fit in 32-bit signed integers.

Once this has all been dealt with, the problem is quite simple. Just write your own Comparator (in Java at least) and sort with it.


#11332 - Summing Digits

Solved By:peter
Theory Difficulty:trivial
Coding Difficulty:trivial
Algorithms Used:strings
Solution Description: Get n.
if(n == 0)
---break;

While (n >= 10)
---sum = 0
---for each char in n
------sum += char-'0'
---n = sum

output n

Solved By:wesley
Theory Difficulty:trivial
Coding Difficulty:trivial
Algorithms Used:math
Solution Description: While n > 9, convert it to a string and sum (char - '0') for all characters in the string. Set n equal to this new sum. Once n <= 9, output n.


#11364 - Parking

Solved By:peter
Theory Difficulty:trivial
Coding Difficulty:trivial
Algorithms Used:math
Solution Description: Take all the store locations in, and output:

(highestLocation-lowestLocation)*2

It doesn't matter where in the range of stores that he parks, since he's going to have to walk the full range and back to his car in any case.

Solved By:wesley
Theory Difficulty:trivial
Coding Difficulty:trivial
Algorithms Used:math
Solution Description: Regardless of where you park, as long as you park somewhere between the farthest left and farthest right store, you will travel the same distance. That distance is (max-min)*2.





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