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Problem solution · Python

CCC 2014 S3 - The Geneva Confection

CCC 2014 S3 - The Geneva Confection: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Direct simulation
Source
CCCSolutions
Length
64 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For CCC 2014 S3 - The Geneva Confection, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 64 lines of Python from the credited upstream file ccc14s3.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2014 S3 - The Geneva Confection · PythonPython
Use this to learn the idea, then write your own version.
# CCC Senior 3: the Geneva Confection## A Stack problem## Test case 5 runs on my machine in 4 seconds :-)# (a Pentium 4, 3.2 Ghz, 1GB RAM)## I fine tuned the file input to shave off a second or 2.# (File input seems to be the slowest part.)## For the cars on the mountain, I just use an index (mtnCar)# to keep track of where I am.# (Nothing actually is removed from it.)## For the branch, I use a list and literally add to or# pop from it. (You can't just keep track of where you are,# because sometimes things have to 'come off' literally.)# file = open ("s3.5.in", "r")entirefile = file.readlines()p = 0t = int(entirefile[p])p = p + 1 for test in range(t):    n = int(entirefile[p])    p = p + 1    mountain = []    for i in range(n):        mountain.append(int(entirefile[p]))        p = p + 1     branch = []    mtnCar = n - 1     nextCar = 1    state = "Y"     # Next car should be either on mountian or branch    # if not, move a car from mountian into branch     while state == "Y" and nextCar <= n:        # if the next car is on the mountain, take it off        if mtnCar >= 0 and nextCar == mountain[mtnCar]:            mtnCar = mtnCar - 1            nextCar = nextCar + 1        # if the nextcar is on the branch, take it off        elif len(branch) > 0 and nextCar == branch[len(branch) - 1]:            branch.pop(len(branch) - 1)            nextCar = nextCar + 1        # otherwise move a car from the mountain to the branch        # (if you can)            elif mtnCar >= 0:            branch.append(mountain[mtnCar])            mtnCar = mtnCar - 1        # otherwise you're done        else:            state = "N"     print state

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