Programming Forums > Application Development > Java

xavier · May 15th, 2005, 12:30 PM

I'll try:
First let's say the user inputs a number : n witch is the area. Now you should see if Math.sqrt(n) is an integer. If it is an integer, then you hava on your hands a square.
Lets say int a = Math.sqrt(n); a will be one of the dimensions => Perimeter = a*4; Dimension = a;

The important part is to test wether Math.sqrt(n) is an integer or not. But you can google a bit for that one, cause it's not that hard.

Now , if Math.sqrt(n) is not an integer then the area is that of a rectangle. You have to find a and b , witch is length and width. Lets say n=12; you could find all its divisors, and see wich of them multiplied equals 12. Like : 3*4, 2*6, 1*12. I don't now if you have to get all the posibilities or only one. The perimeter = a+b.

That's just to get you started , hope u understood... so ... next time come with some code

JDStud6 · May 15th, 2005, 4:10 PM

rectangle perimeter is actually 2*(a+b)

JD

ZenMasterJG · May 15th, 2005, 4:30 PM

If i remember my math classes right, and i occasionally do, the optimum area for a miniumum dimensions will always be a sqare, not a rectangle. so you luck out. i think.

Dark Flare Knight · May 15th, 2005, 6:56 PM

Quote:

Originally Posted by xavier

I'll try:
First let's say the user inputs a number : n witch is the area. Now you should see if Math.sqrt(n) is an integer. If it is an integer, then you hava on your hands a square.
Lets say int a = Math.sqrt(n); a will be one of the dimensions => Perimeter = a*4; Dimension = a;

The important part is to test wether Math.sqrt(n) is an integer or not. But you can google a bit for that one, cause it's not that hard.

Now , if Math.sqrt(n) is not an integer then the area is that of a rectangle. You have to find a and b , witch is length and width. Lets say n=12; you could find all its divisors, and see wich of them multiplied equals 12. Like : 3*4, 2*6, 1*12. I don't now if you have to get all the posibilities or only one. The perimeter = a+b.

That's just to get you started , hope u understood... so ... next time come with some code

I sorta get what u mean but how do i test it?

xavier · May 15th, 2005, 11:28 PM

Quote:

Originally Posted by JDStud6

rectangle perimeter is actually 2*(a+b)

JD

Yes, sorry :o :o i don't know what i was thinking of :o

@Dark.
The way u should do it (i don't know of an easier way) is something like

(Toggle Plain Text)

public boolean isInteger(String s){
try{
       Integer.parseInt(s);
       return true;
}catch(NumberFormatException nfe){
       return false;
}
}

Now, this should work so try making an effort.

Dark Flare Knight · May 16th, 2005, 12:52 PM

Quote:

Originally Posted by xavier

Yes, sorry :o :o i don't know what i was thinking of :o

@Dark.
The way u should do it (i don't know of an easier way) is something like

(Toggle Plain Text)

public boolean isInteger(String s){
try{
       Integer.parseInt(s);
       return true;
}catch(NumberFormatException nfe){
       return false;
}
}

Now, this should work so try making an effort.

Haven't learned half of this stuff yet. I'm only in grade 9 so i have no idea what u wrote.

xavier · May 16th, 2005, 2:19 PM

I guess you could make an array of possible a^2 like : array = {4,9,16,25,36,81....... } and if the number given by the user is among these, then you have a square, else a rectangle and you can take it from there.

Dark Flare Knight · May 17th, 2005, 10:49 AM

(Toggle Plain Text)

import java.awt.*;
import hsa.Console;
public class probJ2
{
    static Console c = new Console ();
    public static void main (String[] args)
    {
        int num;
        c.clear ();
        c.print ("Enter number of pictures: ");
        do
        {
            num = c.readInt ();
            if (num == 0)
                c.close ();
            else
                if (num % 2 == 0)
                {
                    c.println ((int) Math.sqrt (num));
                }
                else
                {
                    c.println ();
                }
        }
        while (num < 0 || num > 65000);
    }
}

I'm done this much. Can someone help me from here on?

May 15th, 2005, 12:30 PM	#11
xavier Professional Programmer Join Date: Oct 2004 Location: .ro Posts: 381 Rep Power: 4	I'll try: First let's say the user inputs a number : n witch is the area. Now you should see if Math.sqrt(n) is an integer. If it is an integer, then you hava on your hands a square. Lets say int a = Math.sqrt(n); a will be one of the dimensions => Perimeter = a4; Dimension = a; The important part is to test wether Math.sqrt(n) is an integer or not. But you can google a bit for that one, cause it's not that hard. Now , if Math.sqrt(n) is not an integer then the area is that of a rectangle. You have to find a and b , witch is length and width. Lets say n=12; you could find all its divisors, and see wich of them multiplied equals 12. Like : 34, 26, 112. I don't now if you have to get all the posibilities or only one. The perimeter = a+b. That's just to get you started , hope u understood... so ... next time come with some code __________________ Don't take life too seriously, it's not permanent ! Last edited by xavier; May 15th, 2005 at 12:33 PM.

May 16th, 2005, 2:19 PM	#17
xavier Professional Programmer Join Date: Oct 2004 Location: .ro Posts: 381 Rep Power: 4	I guess you could make an array of possible a^2 like : array = {4,9,16,25,36,81....... } and if the number given by the user is among these, then you have a square, else a rectangle and you can take it from there. __________________ Don't take life too seriously, it's not permanent !

May 17th, 2005, 10:49 AM	#18
Dark Flare Knight Hobbyist Programmer Join Date: Mar 2005 Location: Canada Posts: 113 Rep Power: 4	(Toggle Plain Text) import java.awt.; import hsa.Console; public class probJ2 { static Console c = new Console (); public static void main (String[] args) { int num; c.clear (); c.print ("Enter number of pictures: "); do { num = c.readInt (); if (num == 0) c.close (); else if (num % 2 == 0) { c.println ((int) Math.sqrt (num)); } else { c.println (); } } while (num < 0 \|\| num > 65000); } } import java.awt.; import hsa.Console; public class probJ2 { static Console c = new Console (); public static void main (String[] args) { int num; c.clear (); c.print ("Enter number of pictures: "); do { num = c.readInt (); if (num == 0) c.close (); else if (num % 2 == 0) { c.println ((int) Math.sqrt (num)); } else { c.println (); } } while (num < 0 \|\| num > 65000); } } I'm done this much. Can someone help me from here on?

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May 15th, 2005, 4:10 PM	#12
JDStud6 Programmer Join Date: Jan 2005 Location: Charleston, SC www.wareonearth.com Posts: 57 Rep Power: 4	rectangle perimeter is actually 2*(a+b) JD

May 15th, 2005, 4:30 PM	#13
ZenMasterJG Hobbyist Programmer Join Date: Nov 2004 Location: Boston, MA Posts: 148 Rep Power: 4	If i remember my math classes right, and i occasionally do, the optimum area for a miniumum dimensions will always be a sqare, not a rectangle. so you luck out. i think.

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Programming Forums > Application Development > Java

Area and Perimeter