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Problem solution · Java

CCC 2009 S1 - Cool Numbers

CCC 2009 S1 - Cool Numbers: 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
53 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 2009 S1 - Cool Numbers, 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

  • 53 lines of Java from the credited upstream file ccc09s1.java.
  • The implementation visibly relies on sequence storage.
  • 1 loop block 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 2009 S1 - Cool Numbers · JavaJava
Use this to learn the idea, then write your own version.
// CCC 2009//// S1: Cool Numbers//// For the first question this is TRICKY. Shame on the CCC. :-(//// The trick is you can't just try all the numbers from a to b// to see if they are perfect squares and cubes. ITS TOO SLOW!!!//// You need to walk thru the cubes, between a and b, checking// if any are perfect squares as well. // (To do that find the cuberoot of a and walk thru the main loop// incrementing the cuberoot by 1 until the cube exceeds b.)//  import java.awt.*;import hsa.*; public class CCC2009S1CoolNumbers{     public static void main (String[] args)    {	Console c;	int a, b;	int cubeRoot, cube, squareRoot, square;	int count; 	c = new Console ();	TextInputFile f = new TextInputFile ("s1.4.in");	a = f.readInt ();	b = f.readInt ();	count = 0;	cubeRoot = (int) (Math.pow (a, 1.0 / 3));	cube = cubeRoot * cubeRoot * cubeRoot;	while (cube <= b)	{	    if (cube >= a)   // because of truncation, 			     // I might be starting below a, hence the if	    {		squareRoot = (int) (Math.sqrt (cube));		square = squareRoot * squareRoot;		if (square == cube)		    count++;	    }	    cubeRoot++;	    cube = cubeRoot * cubeRoot * cubeRoot;	}	c.println (count);    }}

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