Problem solution · Python

CCC 2011 J2 - Who Has Seen the Wind?

CCC 2011 J2 - Who Has Seen the Wind?: 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
19 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 2011 J2 - Who Has Seen the Wind?, 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

  • 19 lines of Python from the credited upstream file ccc11j2.py.
  • The implementation keeps its working state in language-native values and containers.
  • 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.

Full codeCCC 2011 J2 - Who Has Seen the Wind? · PythonPython
Use this to learn the idea, then write your own version.
# CCC 2011 Junior 2: Who has seen the wind# written by C. Robart# March 2011 # simple loop and if h = input()M = input() t = 1A = -6*t*t*t*t + h*t*t*t + 2*t*t + twhile t < M and A > 0:    t = t + 1    A = -6*t*t*t*t + h*t*t*t + 2*t*t + tif A > 0:    print "The balloon does not touch ground in the given time."else:    print "The balloon touches ground at hour:\n" , t 

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