DaWei

Ahhhhhhh, well, don't depend on me, then. Good luck.

__________________
Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code.
Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Java|Tera

14 hours to go...I'm pulling an all-nighter, can someone please help?

Ooble

The mode is just the one that pops up the most. Sort the data, and then loop through and find out which one(s) occur most often.

__________________
Me :: You :: Them

Java|Tera

Do I need a multi-dimensional array to do that? Or can I just use a regular array?

And how do I loop through it and find the stuff?

Mjordan2nd

You could easily do that with a single dimension array.

Loop through it with a for loop -- then use a sorting algorithm to sort out the numbers from greatest to least, or however you plan on going about finding the mode.

__________________
"Every gun that is made, every warship launched, every rocket signifies, in the final sense, a theft from those who hunger and are not fed, from those who are cold and are not clothed. The world in arms is not spending money alone. It is spending the sweat of its laborers, the genius of its scientists, the hopes of its children." - Dwight D. Eisenhower

groovicus

MODE:
To find the mode, all you need is a simple array (filled with zeroes) that is the same size as the largest possible random number that can be generated . Then you loop through your array of 1000 randomly generated numbers. If the first number is a one, then you put the number one in index one in your counting array. If your second number is a one, then you increment the count at one again, so then you have two. Whatever the random number is, increment the count in your array by one. Once you are done, then go through your counting array, see which cell has the largest number, and that is your mode.

So, assume your aray of random numbers is called rand, and your counting array is called count:

get rand[i] //get random number
count[rand]++; // increment count in cell rand


Now loop to check for biggest:

int most = 0;
if rand[i]> most
then most = rand[1]

Once you are done, most will hold the index of the most frequently generated number.


MEDIAN:
Sort your array using the built in sort, then divide by two to get the index of the median.

SHOW ALL NUMBERS:
I'm not sure why you would have a problem with that, If you know how to loop through your array, then that should be pretty simple.

__________________
HijackThis Team-SFDC

Java|Tera

I don't know what's wrong with showing all the numbers..

When I print them in the same method as they are generated in (main method) it works fine, but when I pass them to the menu method and print them, they print all of the same number...

As for median, what if the length is odd?

And for mode, do I need 2 arrays then?

groovicus

Not that it probably matters now..

You need to re-examine your code. Apparently your array is not being passed properly.

As long as it is sorted, you will get at least almost the middle. When the array is odd length, you won't ever get the exact median, because the median will always be between two numbers.. If you divide by two for your index, then it will work for both cases (even and odd lengthed arrays)

That's how I would do it, but I don't think you understand the concept, so you might want to try something else. Unfortunately, I can't think of any other good way to do it.

__________________
HijackThis Team-SFDC

DaWei

I suspect you're right -- it probably doesn't matter now. I must confess that I presume (possibly erroneously, probably not) that a 16-hour deadline was the result of procrastination. There is perhaps something to be learned by that, for the future.

__________________
Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code.
Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

groovicus

When was the last time you heard someone ask for urgent help for an assignment that was due in one week?

__________________
HijackThis Team-SFDC