C programming assignment (with arrays)

I am new to C and i just learned how to use arrays. I have to complete the task below:
There are 100 doors (numbered 1 through 100) and 100 students
(numbered 1 through 100) waiting in a line. Initially, all the doors are closed. A student numbered x passes through all the doors that are multiples of x (i.e., student numbered x goes to doors x; 2x; 3x; ... ) and changes the status of each door, i.e., opens the door if it is closed and closes the door if it is open. After all the students complete their turns, which doors remain open? Write a C program that mimics this experiment and prints the numbers of the doors that remain open at the end.

I am stuck on how to know which door is in what state. This is what I have now.

Code:

---

#include <stdio.h>


int main(void)
{
    int i,x=0,j,door,k=1,d[100],s[100];
   
    for (i=1;i<100;i++)
        d[i]=i;
    for(k=1;k<100;k++)
        s[k]=k;
    k=1;
    for(i=1;i<=100 || k==100;i++)
    {
        j=d[i]%s[k];
        for(;i==100;k++)
            i=1;
        if(j==0)   
    }
    return;
}

---

If I pasted the code wrong I sincerely apologize and I will try again. As you can see the code isn't fully finished. I think I am on the right track I just can't figure out how to know what happens to which door. Thanks for any help.

Ok... before you start pounding code that you'll just end up trashing... You need to do some preparatory work...

Turn off your computer...

1) ANAYSE the assignment, until you understand it in very small detail... For example; you might pick one student and follow him or her through a set of doors.

2) PLAN a step by step solution, on paper, to the problem based on your newfound level of understanding.

Now you can turn on your computer...

3) WRITE your code working in small sections testing as you go...

4) TEST your code to see if there are any problems that weren't obvius in the planning stage.


All programmers follow a similar process, even for the simplest of programs.
This is necessary because nobody can solve a problem they don't understand.

Quote:

---

Originally Posted by CommonTater

Ok... before you start pounding code that you'll just end up trashing... You need to do some preparatory work...

Turn off your computer...

1) ANAYSE the assignment, until you understand it in very small detail... For example; you might pick one student and follow him or her through a set of doors.

2) PLAN a step by step solution, on paper, to the problem based on your newfound level of understanding.

Now you can turn on your computer...

3) WRITE your code working in small sections testing as you go...

4) TEST your code to see if there are any problems that weren't obvius in the planning stage.


All programmers follow a similar process, even for the simplest of programs.
This is necessary because nobody can solve a problem they don't understand.

---

I feel like that is what I have done. I have analyzed the problem. If all the doors start off closed, we take one student through each of the 100 doors, if that student can be divided into the door number the state of the door changes. Once all of the students have gone through all of the doors print the doors that are in the open state. I guess I am missing the planned portion now. I just keep getting stuck. Thanks for the tips though!

You keep getting stuck because you don't yet understand the problem. Your statements are sweeping, general in nature... computers can't deal with that. Also take note that *your* description does not match that of your assignment... Not all students go through all doors.

Some things to think about...

Take any student... lets say #58 ... how many doors does that student go through?

Take any door... precisely what happens when a student goes through it?

And so on... programming is all about the minutia... you have to understand these things *at the computer's level* because it's a sure and certain thing that it's not going to understand it at ours.

Quote:

---

Originally Posted by CommonTater

You keep getting stuck because you don't yet understand the problem. Your statements are sweeping, general in nature... computers can't deal with that. Also take note that *your* description does not match that of your assignment... Not all students go through all doors.

Some things to think about...

Take any student... lets say #58 ... how many doors does that student go through?

Take any door... precisely what happens when a student goes through it?

And so on... programming is all about the minutia... you have to understand these things *at the computer's level* because it's a sure and certain thing that it's not going to understand it at ours.

---

How does my description not match the assignment? I wasn't saying all students go through all doors. I was saying all students "test" to see if they go through the door or not. Is that not the right mentality.

Student 58 would go through 1 door, door number 58.

If a student goes through the door it changes the door depending on if the door is already closed or open.

I really don't understand what is wrong with my logic. Are you saying my logic doesn't fit the computers logic or that my logic is just plain wrong?

Quote:

---

Originally Posted by COG92

How does my description not match the assignment? I wasn't saying all students go through all doors. I was saying all students "test" to see if they go through the door or not. Is that not the right mentality.

Student 58 would go through 1 door, door number 58.

If a student goes through the door it changes the door depending on if the door is already closed or open.

I really don't understand what is wrong with my logic. Are you saying my logic doesn't fit the computers logic or that my logic is just plain wrong?

---

Your logic may not be flawed... but it's still at too high a level...

Ask yourself how you knew that #58 only goes through one door...

How do you know which students go through which doors?
How do you know the state of each door before a student goes through?
How is the door changed when a student does go through?

Quote:

---

Originally Posted by CommonTater

Your logic may not be flawed... but it's still at too high a level...

How do you know which students go through which doors?
How do you know the state of each door before a student goes through?
How is the door changed when a student does go through?

---

I'm sorry, but I feel like we are just going in circles. We know which students go through which doors based on the multiples. I can't think of an easier way to find a multiple of something besides division. I guess you could take student number 58 and multiply starting at 1 and ending if the next one goes over 100. That seems more complicated to me though.

If you start with a state of closed, once a student goes trough it opens, then another student it closes, and so forth.

I feel like I am getting even more lost than I was before. I really don't know how else to think of this.

Ok... what I'm trying to get you to see is the *process* that is at work...

How do we know which students are going through which doors... We are using multiples of the student's number to know which door they go through... So student 1 goes through door 1 X 1 (1), 1 X 2 (2), 1 X 3 (3) etc. Student 2 goes through door 2 x 1 (2), 2 X 2 (4), 2 X 3 (6), etc. And so on...

If we follow this a bit further... We can also observe that no student goes through a door less than their number...

Also since there are only 100 doors no student can go through door any door numbered higher than 100...

Ok, the action of the doors is such that if we have an open door we close it... if we have a closed door we open it...

You may understand this implicitly, but the computer does not... The computer is a total flaming idiot, you have to tell it *every little thing* and you have to tell it what to do in the exact order you want it to do it... Thus you have to be able to think down to it's level...

Ok so now we have some sense of what's happening...

We obviously need to loop through all student numbers
We need to loop through all multiples of their number < 101
We need to track the status of the doors as open or closed
Then we have to make a report.

See how deeper understanding leads you to a process?

Quote:

---

Originally Posted by CommonTater

Ok... what I'm trying to get you to see is the *process* that is at work...

How do we know which students are going through which doors... We are using multiples of the student's number to know which door they go through... So student 1 goes through door 1 X 1 (1), 1 X 2 (2), 1 X 3 (3) etc. Student 2 goes through door 2 x 1 (2), 2 X 2 (4), 2 X 3 (6), etc. And so on...

If we follow this a bit further... We can also observe that no student goes through a door less than their number...

Also since there are only 100 doors no student can go through door any door numbered higher than 100...

Ok, the action of the doors is such that if we have an open door we close it... if we have a closed door we open it...

You may understand this implicitly, but the computer does not... The computer is a total flaming idiot, you have to tell it *every little thing* and you have to tell it what to do in the exact order you want it to do it... Thus you have to be able to think down to it's level...

Ok so now we have some sense of what's happening...

We obviously need to loop through all student numbers
We need to loop through all multiples of their number < 101
We need to track the status of the doors as open or closed
Then we have to make a report.

See how deeper understanding leads you to a process?

---

Thanks for all your help. I really appreciate it. I see your point. That method still seems more complicated to me, but I am going to finish the analyzing based on what you said then try some more planning. At this point I don't see how exactly this is going to work, but I am going to work on it some more and see where it gets me. Thanks again for the help.

It's a Word Problem! :p I love the boolean nature of the doors - they're open, or shut, nothing in between.

Since each student will be passing through a bunch of doors, it seems you'd want to have two for loops:

one for the students
and one nested inside that, for the doors each student will interact with.

The studentNumber could just be an int, maybe. The doors have to be an int array (probably best). Then you'll need another int variable for the "multiplier" of the student number.

The first for loop will increment the studentNumber, and the inner for loop will increment the "multiplier", which I'd call 'n'.

The inner for loop will use that studenNumber * n as the index to the doors array: doors[studentNumber * n].

You'll probably want an if statement with an else attached, inside that inner for loop.

And I want to know which doors are open and which are closed, at the end. :p

Quote:

---

Originally Posted by Adak

It's a Word Problem! :p I love the boolean nature of the doors - they're open, or shut, nothing in between.

Since each student will be passing through a bunch of doors, it seems you'd want to have two for loops:

one for the students
and one nested inside that, for the doors each student will interact with.

The studentNumber could just be an int, maybe. The doors have to be an int array (probably best). Then you'll need another int variable for the "multiplier" of the student number.

The first for loop will increment the studentNumber, and the inner for loop will increment the "multiplier", which I'd call 'n'.

The inner for loop will use that studenNumber * n as the index to the doors array: doors[studentNumber * n].

You'll probably want an if statement with an else attached, inside that inner for loop.

And I want to know which doors are open and which are closed, at the end. :p

---

Thanks for the nice explanation! That is very similar to what I have right now, but not quite. I'll have to change a few things around. The trouble I am having is being about to print the number of the doors that are open at the end of everything. I'll keep thinking and see what happens. I will certainly let you know what the end result is :) Thanks again!

Quote:

---

Originally Posted by Adak

It's a Word Problem!

---

FWIW... I have a solution in about 20 lines here, that I can post after our friend finishes his assignment...

I couldn't resist writing one myself, but I haven't run it yet, and certainly won't post it, until he's through. I like keeping the suspense going, anyway. Maybe time for a good mystery movie!

It's a pretty short program, yeah. Maybe 25 lines or so. I wrote it in the forum editor, so I don't have a line counter on the side bar.

Cog92: use an if statement, inside your for loop that prints out the door status.

Quote:

---

Originally Posted by Adak

I couldn't resist writing one myself, but I haven't run it yet, and certainly won't post it, until he's through. I like keeping the suspense going, anyway. Maybe time for a good mystery movie!

---

:D ... just got finished watching "Imposter", nice bit of scifi with a surprise ending.

Quote:

---

It's a pretty short program, yeah. Maybe 25 lines or so. I wrote it in the forum editor, so I don't have a line counter on the side bar.
Cog92: use an if statement, inside your for loop that prints out the door status.

---

I might have gotten it under 20 lines but decided to toss in a little extra...

Anyway we can all compare when the OP finishes his...

Quote:

---

Originally Posted by Adak

I couldn't resist writing one myself, but I haven't run it yet, and certainly won't post it, until he's through. I like keeping the suspense going, anyway. Maybe time for a good mystery movie!

It's a pretty short program, yeah. Maybe 25 lines or so. I wrote it in the forum editor, so I don't have a line counter on the side bar.

Cog92: use an if statement, inside your for loop that prints out the door status.

---

I don't know why I am having such a hard time with this. I have gotten all the other assignments just fine. I think the arrays confuse me. Anyway, if I were to put an if statement inside that for loop won't it print the status of the doors after every student goes through versus the end result?

Quote:

---

Originally Posted by COG92

I don't know why I am having such a hard time with this. I have gotten all the other assignments just fine. I think the arrays confuse me. Anyway, if I were to put an if statement inside that for loop won't it print the status of the doors after every student goes through versus the end result?

---

Yes it would... which would be a good debugging tool.

Quote:

---

Originally Posted by COG92

I don't know why I am having such a hard time with this. I have gotten all the other assignments just fine. I think the arrays confuse me. Anyway, if I were to put an if statement inside that for loop won't it print the status of the doors after every student goes through versus the end result?

---

Oh, that's right, you're supposed to mimic the students walking through the doors -- oops, my bad!

I was thinking of the for loop at the end to show the final doors that remain open, and that would be sufficient.

Do you want to show the status of the doors, as they change? Little text based display to represent the doors?

This is a little tricky, because you have to use the index of the doors array, in an entirely new manner (for you). Post up your code and let's see where you are at.

Quote:

---

Originally Posted by Adak

Oh, that's right, you're supposed to mimic the students walking through the doors -- oops, my bad!

I was thinking of the for loop at the end to show the final doors that remain open, and that would be sufficient.

Do you want to show the status of the doors, as they change? Little text based display to represent the doors?

This is a little tricky, because you have to use the index of the doors array, in an entirely new manner (for you). Post up your code and let's see where you are at.

---

I don't need to show the doors as they change, just the end result. For example I would want to print: The doors that remain open are: 2,6,8,57. Something along those lines. I don't have much of a code to show because I am playing around with things to get to what I want. I realize this is nothing like what I need it to be, but this is what I am troubleshooting with to understand what is going on(by using the debugger tool).

Code:

---

#include <stdio.h>


int main(void)
{
        int i,s,d[100],n;
       
        for (i=0;i<100;i++)
                d[i]=i;
        i=0;
        for(s=1;s<100;s++)
        {
                for(n=1;i<100;n++)
                {       
                        i=s*n;
                }
                i=0;
        }
        return (0);
}

---

Don't make fun of me because of how off I am please :/

It still looks like you're throwing code at the problem and hoping something will stick.

The first thing you need to do, even before opening and shutting various doors, is decide how you are going to represent the doors. How do you tell the difference between an open door and a closed one?

Actually you're not that far off at all... based on what you've posted, you mostly need to devise a way to make the door change, which you aren't doing yet... the loops are pretty close though.

I don't mean to lecture... but what is happening now is that analysis phase I discussed earlier, except you're doing it with live code... yes it can be done that way but it's not always the smartest way to do it since you can easily "code yourself into a corner". The bigger the job the more likely that becomes. (And yes, I've done it to myself moe than once.)

Quote:

---

Originally Posted by CommonTater

Actually you're not that far off at all... based on what you've posted, you mostly need to devise a way to make the door change, which you aren't doing yet... the loops are pretty close though.

I don't mean to lecture... but what is happening now is that analysis phase I discussed earlier, except you're doing it with live code... yes it can be done that way but it's not always the smartest way to do it since you can easily "code yourself into a corner". The bigger the job the more likely that becomes. (And yes, I've done it to myself moe than once.)

---

You're right, I am analyzing it with live code. I know realize that is a bad habit, but I have troubling doing it without being able to see what is happening.

As far as figuring out which doors are doing what I can't seem to get anywhere close to making it work. I was think that you could have some sort of system that says 0 for closed doors and 1 for open doors, at the end print all the doors with a 1. I can't figure out how to implement that though.

Ok, here's a hint... the door starts out closed... a student goes through and it's open... a student goes through and it closes... student open... student closed... see a pattern? What happens every second time? What is every second number?

#define Open 1

I agree, I thought it added to the clarity of the whole thing.

You need to swap your for loops. Put the studentNumbers on the outer for loop, and the door[] logic, inside the nested for loop. You don't need the third for loop, until you're ready to print the final status.

I would change your doors array size to 101, and set the value of each element of that array, to 0, instead of i. Start your for loops at 1, and stop at <= 100, instead of just < 100.

That will initialize all the doors starting to closed. (what I would call closed, anyway).

Quote:

---

Originally Posted by CommonTater

Ok, here's a hint... the door starts out closed... a student goes through and it's open... a student goes through and it closes... student open... student closed... see a pattern? What happens every second time? What is every second number?

---

Are you implying the fact the they change from even to odd numbers? If you start at 0 all closed would be considered even numbers, and odd would be considered open. Is that what you were saying?

Quote:

---

Originally Posted by Adak

#define Open 1

I agree, I thought it added to the clarity of the whole thing.

You need to swap your for loops. Put the studentNumbers on the outer for loop, and the door[] logic, inside the nested for loop. You don't need the third for loop, until you're ready to print the final status.

I would change your doors array size to 101, and set the value of each element of that array, to 0, instead of i. Then you'll have all the doors starting off as closed.

---

I didn't see you replied. Which third loop were you referring to? Is saying i=s*n the right thing to do? I wasn't sure about that one. I'll switch my loops, but for now I am going to try to analyze it on paper. :)

While it's undoubtedly true, a simple if() statement will handle that door open or door closed action.

you want s*n, but you need it as the index for the doors[].

That's the hardest thing about this problem, imo.

Ignore the comment about the third for loop. Your code threw me off a bit.

Quote:

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Originally Posted by COG92

Are you implying the fact the they change from even to odd numbers? If you start at 0 all closed would be considered even numbers, and odd would be considered open. Is that what you were saying?

---

Sort of... think about a light switch... it's off... you click it 4 times where is it? On or off... Now you click it 7 times... on or off?

Quote:

---

Originally Posted by Adak

While it's undoubtedly true, a simple if() statement will handle that.

you want s*n, but you need it as the index for the doors[].

That's the hardest thing about this problem, imo.

Ignore the comment about the third for loop. Your code threw me off a bit.

---

Are you saying d[i]=s*n? Or what exactly do you mean when you say the index for the doors array? Thank you both for taking time to walk through this with me. I apologize that it isn't clicking faster.

Quote:

---

Originally Posted by CommonTater

Sort of... think about a light switch... it's off... you click it 4 times where is it? On or off... Now you click it 7 times... on or off?

---

Hm...I am not sure what you are getting at. I keep wanting to go back to the even odd thing with your light switch analogy.

Quote:

---

Originally Posted by COG92

Are you saying d[i]=s*n? Or what exactly do you mean when you say the index for the doors array? Thank you both for taking time to walk through this with me. I apologize that it isn't clicking faster.

---

Array[Index];

doors[index]

Your index for the doors array needs to be a multiple of studentNumber * a multiplier, in the inner for loop.

You can use just doors[studentNumber * n] = Open or 1 or Closed or 0.

Check out the pattern:

Student number is 3:

3 * 1 first door he visits -- n equals 1
3 * 2 second door he visits-n equals 2
3 * 3 third door he visits, etc.
3 * 4 fourth door he visits

etc.

Quote:

---

Originally Posted by COG92

Hm...I am not sure what you are getting at. I keep wanting to go back to the even odd thing with your light switch analogy.

---

If you know the intial state and count the clicks you can tell if it's on or off...
That is... if it starts out being off and you click it an even number of times, you know it's off... an odd number of clicks tells you it's on. (This premise is beyond easy to test, unless you don't have lights in your house...)

Now... how can you apply that to your problem?

Quote:

---

Originally Posted by CommonTater

If you know the intial state and count the clicks you can tell if it's on or off...
That is... if it starts out being off and you click it an even number of times, you know it's off... an odd number of clicks tells you it's on. (This premise is beyond easy to test, unless you don't have lights in your house...)

Now... how can you apply that to your problem?

---

Isn't that what I said? I am struggling to see how it is different then my even odd thing. Depending on the number of times the door is changed it will tell you if the door is open or closed. For example, if the door starts off closed, and you change it 4 times, it's open and even. I could have sworn that's what I said either. Unless I am wrong again, which with my current luck I probably am.

No you're not wrong... Perhaps I just read over it too quickly.

So, the question still remains... how can you apply that to your problem?


What I'm trying to do is to get you to think in smaller "blobs"... the way the computer does...
No kidding, that really is the programmer's "big trick"... the ability to take huge things and reduce them to itty bitty little steps...

Quote:

---

Originally Posted by CommonTater

No you're not wrong... Perhaps I just read over it too quickly.

So, the question still remains... how can you apply that to your problem?


What I'm trying to do is to get you to think in smaller "blobs"... the way the computer does...
No kidding, that really is the programmer's "big trick"... the ability to take huge things and reduce them to itty bitty little steps...

---

At least I am not wrong, for once. So if we count each time a particular door happens to work with s*n and that number ends up being even the door number is open. Right? Am I close? Like if door 5 shows up 3 times the door is closed. If door 6 shows up 6 times the door is still open.

With one if statement in your inner for loop, you have everything you need, and you don't need to count any door visits. If the door is closed when the student gets there, you open the door (set it to 1 let's say). Else, you close the door (because it was already open).

You're not wrong, but it's not the simplest way.

Quote:

---

Originally Posted by Adak

With one if statement in your inner for loop, you have everything you need, and you don't need to count any door visits. If the door is closed when the student gets there, you open the door (set it to 1 let's say). Else, you close the door (because it was already open).

You're not wrong, but it's not the simplest way.

---

Logically that makes completely sense. Now, I am just trying to figure out what my for loops are supposed to look like since you said to swap them. I am having trouble figuring out how to make s go up by 1 and then go back to the outer for loop after that.

Quote:

---

Originally Posted by COG92

At least I am not wrong, for once. So if we count each time a particular door happens to work with s*n and that number ends up being even the door number is open. Right? Am I close? Like if door 5 shows up 3 times the door is closed. If door 6 shows up 6 times the door is still open.

---

Other way round... if the doors are initially closed an even number of students going through will mean it's closed...

off ... click on...click off...click on... click off ...

Even numbers always return the intial state.

Quote:

---

Originally Posted by COG92

Logically that makes completely sense. Now, I am just trying to figure out what my for loops are supposed to look like since you said to swap them. I am having trouble figuring out how to make s go up by 1 and then go back to the outer for loop after that.

---

Enlarge doors[] to 101, and your 1 off problems go away.

Add <= 100, if you don't have it already, for your for loop stop test.

Code:

---

First for loop, take studNumber from 1 to 100 {

  Second for loop takes n from 1 to 100, but stops when studNumber * n > 100 {
 
      if the door is open close it, else open it (it was closed).
  }
}

---

Each door inside the second for loop, will be doors with the index studNumber * n.

Quote:

---

Originally Posted by Adak

Enlarge doors[] to 101, and your 1 off problems go away.

Add <= 100, if you don't have it already, for your for loop stop test.

Code:

---

First for loop, take studNumber from 1 to 100 {

  Second for loop takes n from 1 to 100, but stops when studNumber * n > 100 {
 
      if the door is open close it, else open it (it was closed).
  }
}

---

Each door inside the second for loop, will be doors with the index studNumber * n.

---

This is what I have right now. Am I on the right track at least?

Code:

---

#include <stdio.h>
 
 
int main(void)
{
    int i,s,d[101],n=1;
   
    for (i=0;i<=100;i++)
        d[i]=0;
   
    for(s=1;s<=100;s++)
    {
            d[s*n]=1;
        for(n=1;(s*n)<=100;n++)
        { 
            if(d[s*n]==1)
                    d[s*n]=0;
            else
                    d[s*n]=1;
        }
       
    }
    return (0);
}

---

Yes... that is one way to solve it... and it will work...

However line 13 may sabotage you.

Now you just need to write the report and see if it gives you the right answers.

(FWIW... I did it a slightly different way...)

With the current code I have the value of d[i] never really changes, which means I can't print d[i]. And the values for d[s*n] are either one or zero. Needless to say I am not sure what to print.

Quote:

---

Originally Posted by COG92

With the current code I have the value of d[i] never really changes, which means I can't print d[i]. And the values for d[s*n] are either one or zero. Needless to say I am not sure what to print.

---

You need to look at the status, after calculation, of each door... if d[i] == 1, it's open... The reason you don't see the doors changing perminently in the array is that every time through the loop you set it so the door is open on line 13, you need to get that out of there so the program operates on the actual array's status as it changes.

I agree. Delete line 13. A last simple for loop outside of all the other for loops, and you're done.

I guess I don't fully understand it then. If I delete line 13 how will d[s*n] ever change? Won't it just stay 0 since it never gets set to anything else?

Quote:

---

Originally Posted by COG92

I guess I don't fully understand it then. If I delete line 13 how will d[s*n] ever change? Won't it just stay 0 since it never gets set to anything else?

---

That's what lines 16 through 19 do for you...
They test the state of the door and flip it over from 0 to 1 or 1 to 0...

No insinuation here... but one might expect you to have known that, having written the code yourself...

Quote:

---

Originally Posted by CommonTater

That's what lines 16 through 19 do for you...
They test the state of the door and flip it over from 0 to 1 or 1 to 0...

No insinuation here... but one might expect you to have known that, having written the code yourself...

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Oh, duh! I can't believe I missed that. I realize you said no insults, but now I feel like a cheater and it makes me want to figure out a different way to solve it and do it that way. If it helps I have learned a lot.

Lines 17 and 19 handle it. It's so simple once you see it.

You will have several of these moments of < facepalm > with C. If you don't, you're not learning anything new.

I don't think you're cheating... well, not unless you tell me you "borrowed" someone else's code... We call that "scoop and poop coding" mostly because that's what the end result usually is... poop ...

Not to worry... I figure you just got a little too deep into experimental code...

I've always found programming to be a deeply interresting challenge. Every new project brings it's own set of questions and problems and for me it's just about the best puzzle solving exercise on the planet. Being self-educated, I can attest to how much you learn and how well you remember when you're struggling for every character...

Keep going... finish your assignment... then I'll show you how I solved it...
And yes, I used my little 4 step plan, just like always. :D

Quote:

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Originally Posted by CommonTater

I don't think you're cheating... well, not unless you tell me you "borrowed" someone else's code... We call that "scoop and poop coding" mostly because that's what the end result usually is... poop ...

Not to worry... I figure you just got a little too deep into experimental code...

I've always found programming to be a deeply interresting challenge. Every new project brings it's own set of questions and problems and for me it's just about the best puzzle solving exercise on the planet. Being self-educated, I can attest to how much you learn and how well you remember when you're struggling for every character...

Keep going... finish your assignment... then I'll show you how I solved it...
And yes, I used my little 4 step plan, just like always. :D

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Thanks for that! I really enjoy the problem solving aspect of programming, and I usually try to figure it all out on my own. This time after working for several hours and getting a little too close to the due date I decided to seek help. You both have been a huge help and I greatly appreciate it. I really hope I can adapt your 4 step plan because I think it would really help and it would keep me from getting so frustrated.
I am still having issues figuring out the printing, but hopefully I can figure it out. I look forward to seeing how you solved it.

Ok... I have doors 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 as open. Is that what you have?

Yeah, the last thing you want to do is to start coding before you even know what you're writing. It's always a good idea to spend some time thinking it over before you start. Of course on complex programs, you write piles and piles of notes. The one thing that never happens in real world programming is that you hand a programmer an assignment and he heads on over and starts typing...

As I'm sure my friend Adak would agree... C is easy... Programming is hard.

Quote:

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Originally Posted by COG92

Ok... I have doors 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 as open. Is that what you have?

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Yep that's what I got ... so post your code then I'll post mine... Hopefully Adak is still on line and can add his too..

Quote:

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Originally Posted by CommonTater

Yeah, the last thing you want to do is to start coding before you even know what you're writing. It's always a good idea to spend some time thinking it over before you start. Of course on complex programs, you write piles and piles of notes. The one thing that never happens in real world programming is that you hand a programmer an assignment and he heads on over and starts typing...

As I'm sure my friend Adak would agree... C is easy... Programming is hard.

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That makes complete sense. And it is probably a good idea to get in that habit now, even if the codes are fairly small. I am starting to realize how C would be easy and programming would be hard. Lol.
Here is my code:

Code:

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#include <stdio.h>
 
 
int main(void)
{
    int i,s,d[101],n=1;
   
    for (i=0;i<=100;i++)
        d[i]=0;
   
    for(s=1;s<=100;s++)
    {
        for(n=1;(s*n)<=100;n++)
        { 
            if(d[s*n]==1)
                    d[s*n]=0;
            else
                    d[s*n]=1;
        }
       
    }
    for(i=0;i<=100;i++)
            if(d[i]==1)
            printf("%d\t",i);
    return (0);
}

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Looks pretty good...

Here's mine, but I tossed in a little extra...

Code:

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#include <stdio.h>

int main (void)
  { char doors[101] = {0}; // doors closed
    int stu, dor, tmp;    // for counters

    printf("100 students, 100 doors...\n");

    // calculate door status
    for (stu = 1; stu < 101; stu++)
      for (dor = 1; (dor * stu) < 101; dor++)
        doors[dor * stu]++;
 
    printf("The following doors would be open...\n");
    for (tmp = 1; tmp < 101; tmp++)
      if (doors[tmp] & 1)
        printf("%d\t",tmp);

    printf("\n\nEach door was entered [times] ...\n");
    for (tmp = 1; tmp < 101; tmp++)
      printf("%d [%d]\t",tmp,doors[tmp]);

  printf("\n\n");
  return 0; }

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That's what I got, also. I have not analyzed it to see if it's correct or not though. I believe it is, but that's as far as I can go to affirm the accuracy.

Yes, programming is a lot more than using the language of your choice. I have spent more than a week, working on a problem by hand, (an hour or two each day), before I saw the pattern I needed to program, and could put it into the small steps that the program could use.

Practice improves your problem solving skills a great deal, and programming is all about solving problems, imo. It's like any other language or sports skill. You want to be good at it, you practice, practice, and practice.

Quote:

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Originally Posted by Adak

Practice improves your problem solving skills a great deal, and programming is all about solving problems, imo.

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Yeppers... that's why they call it "coding a solution"...

Nice code CommonTater! I like the added part that mentions how many times the door was opened. Thank you both for all the help, tips, and advice! I really appreciate it! Now, I get to go work on the second problem that's due :) This time I will make sure to do the 4 step process! And...I will continue to practice! Hope you both have a good one!

No worries... It's been fun.

If you get stuck on the new one, just post up your code and we'll see what we can do...

Quote:

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Originally Posted by CommonTater

No worries... It's been fun.

If you get stuck on the new one, just post up your code and we'll see what we can do...

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Thanks! I plan on working on it for the next couple days so if I have issues I will certainly see if you guys can help!

You're welcome.