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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