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Problem solution · Java

CCC 2006 S1 - Maternity

CCC 2006 S1 - Maternity: a Java 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
57 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 2006 S1 - Maternity, 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

  • 57 lines of Java from the credited upstream file ccc06s1.java.
  • The implementation visibly relies on sequence storage.
  • 2 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 2006 S1 - Maternity · JavaJava
Use this to learn the idea, then write your own version.
// CCC2006s1Maternity//// given a mothers and fathers genes, determine if a baby may// be theirs. (dominant/recessive gene ideas)//// Basic string handling and basic structures import java.awt.*;import hsa.*; public class CCC2006s1Maternity{    static Console c;     public static void main (String[] args)    {	c = new Console ();	TextInputFile f = new TextInputFile ("s1.3.in");	int x;	String mother, father, baby; 	mother = f.readString ();	father = f.readString ();	x = f.readInt ();	for (int i = 1 ; i <= x ; i++)	{	    baby = f.readString ();	    if (possibleBaby (mother, father, baby))		c.println ("Possible baby.");	    else		c.println ("Not their baby!");	}    }      // if baby has Capital, one parent needs to have one Capital    // if baby has lower, both parents need a lower    public static boolean possibleBaby (String m, String f, String b)    {	boolean okay = true;	for (int i = 0 ; i <= 4 && okay ; i++)	    if (b.charAt (i) >= 'A' && b.charAt (i) <= 'E')		okay = (m.charAt (i * 2) >= 'A' && m.charAt (i * 2) <= 'E') ||		    (m.charAt (i * 2 + 1) >= 'A' && m.charAt (i * 2 + 1) <= 'E') ||		    (f.charAt (i * 2) >= 'A' && f.charAt (i * 2) <= 'E') ||		    (f.charAt (i * 2 + 1) >= 'A' && f.charAt (i * 2 + 1) <= 'E');	    else		okay = ((m.charAt (i * 2) >= 'a' && m.charAt (i * 2) <= 'e') ||			(m.charAt (i * 2 + 1) >= 'a' && m.charAt (i * 2 + 1) <= 'e')) &&		    ((f.charAt (i * 2) >= 'a' && f.charAt (i * 2) <= 'e') ||			(f.charAt (i * 2 + 1) >= 'a' && f.charAt (i * 2 + 1) <= 'e'));	return okay;    }}

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