temp = amount * 100 + .1;

if you have access to e-mail program and if you know how to use them, you can save your program in a disk and use your e-mail program to write an e-mail to your TA. You can either "attatch" the file to the e-mail, OR you can also copy N paste to the e-mail itself. Either way is fine.

If you DON'T have access to e-mail program, you can use your cse5a e-mail account and telnet to send an e-mail to your TA. Here is how:

Save part one of your hw#7 as H:\hw7p1.c
save part 2 of your hw#7 as H:\hw7p2.c
Save part 3 of yoru hw#7 as H:\hw7p3.c
Save part 4 of your hw#7 as H:\hw7p4.c
You can use what ever name you want to use, but make sure that you have in H:\ directory not H:\pc\bc45\bin\ something.. And remember the file names you saved. Please ask one of the TAs if you need help

Now the saving program part is done. If you look at the left bottom corner of the computer screen, you should be able to see a button saying START. Click that START button. Then it will show a menu looking somewhat like this:

'--------------------------`
|   ACS Account Utilities  |
|--------------------------|
|   Program Files          |
|   Documents              |
|   Find File              |
|   Settings               |
|   Yadda Yadda            |
|   Run ..                 |
|--------------------------|
|   Shutdown               |
 -----------------------------------
| START|   stuff
 -----------------------------------		
	

Please excuse my ascii pictuer ;) I hope it's somewhat close to reality. hehe. Anyways. See that blue RUN button? Click that button. Then it will open a small window looking like this:

'-------------------------------`
| Run                        _ x|
|-------------------------------|
|   Type yadda yadda ...yadda.. |
|   ..yadda yadda... ... you    |
|        --------------------   |
|OPEN : |                  |V|  |
|        --------------------   |
|              OK CANCEL BROWSE |
 ----------------------------------
| START|   stuff
 ----------------------------------

See that blue box next to OPEN ? Type telnet ieng9.ucsd.edu. And click OK button. So, it should look like this right before you click OK:

'-------------------------------`
| Run                        _ x|
|-------------------------------|
|   Type yadda yadda ...yadda.. |
|   ..yadda yadda... ... you    |
|        --------------------   |
|OPEN : |telnet ieng9.uc...|V|  |
|        --------------------   |
|              OK CANCEL BROWSE |
 ----------------------------------
| START|   stuff
 ----------------------------------

Then click the OK button in red....

Okey. THEN it will open a window that says

'----------------------------------------`
| Telnet ieng9.ucsd.edu               _ x|
|----------------------------------------|
| Connect Edit Terminal Help             |
|----------------------------------------|
|                                        |
|UNIX(r) System V Release 4.0 (ieng9...  |
|login:                                  |
|                                        |

Type your cse5a login name ( like cs5aaa, cs5adc, or somehting like that). Then it will ask for password

|
|UNIX(r) System V Release 4.0 (ieng9...  |
|login: cs5a6                            |
|Password:                               |
|                                        |

Don't type cs5a6.(that's me) Then type your password. If you type login name & passwd correctnly, you should get school notices saying useless stuff. Just press < SPACE BAR > until you get

|  ieng9.ucsd.edu %                      |
|                                        |

Yer half way though... okey.? we'll take a break here.

UNDER CONSTRUCTION!
MORE TO COME YET!

Then how do we find out the day other than January 1st? The assignment states that January 1st of 1984 was Sunday right? Then count the day between 1/1/1984 and the given date. For example, the given date is 1/11/1984, the days between 1/1/1984 and 1/8/1984 is 7 days. Notice that Sun, Mon, Tue,.. Sat repeats every 7 days. Thus 7 days after Sunday should be Sunday again. let's say it was 1/11/1984. Then there are 10 days between 1/1/1984 and 1/11/1984. Then it should be Wednesday. Notice that 10 is 7 + 3. how about 1/4/94? 4 - 1 = 3 = 7 * 0 + 3. 1/18/84? 18 - 1 = 17 = 7 * 2 + 3. right? All the other days could be represented the same way

if the day is Sunday, days between = 7 * ( some number ) + 0
Monday, days between = 7 * ( some number ) + 1
Tuesday, days between = 7 * ( some number ) + 2
Wednesday, days between = 7 * ( some number ) + 3
Thursday, days between = 7 * ( some number ) + 4
Friday, days between = 7 * ( some number ) + 5
Saturday, days between = 7 * ( some number ) + 6

So, for example, if you want to find the day of 6/12/1998, the mathematical procedure will look like this:

year 1984, January.
SUN MON TUE WED THU FRI SAT
 1   2   3   4   5   6   7
 8   ....

1998 - 1984 = 14
but leap years in between
1984, thus 14+1
1988, thus 14+1+1
1992, thus 14+1+1+1
1996, thus 14+1+1+1+1 = 18

since 18 = 7 * 2 +4 this tells you that
1/1/1998 is Thursday.

year 1998, Januray.
SUN MON TUE WED THU FRI SAT
                 1   2   3
 4   5   6   7   8 .... 

the day you want to find is June, thus

18 + Jan + Feb + Mar + Apr + May 
 = 18 + 31 + 28 + 31 + 30 + 31 = 169

since 169 = 24 * 7 + 1,
6/1/1998 is Monday.

year 1998, June.
SUN MON TUE WED THU FRI SAT
     1   2   3   4   5   6
 7   8   9   ....

Since the date you wanted is 6/12/1998,
169 + 12 - 1 = 180 ( you need to subtarct 1
          because the first day a month is 1
          not 0 )
Since 180 = 25 * 7 + 5
6/12/1998 should be Friday.
SUN MON TUE WED THU FRI SAT
     1   2   3   4   5   6
 7   8   9  10  11  12  13
which is correct. !

Another thing. if you use "%f" to print out the sum, it will look like 12.0000000 or something like that. To make it look like "$20.00" (only two digits after "." ) use "%.2f" instead of "%f". So,

    printf ("change = $%f\n", 12.0);
    printf ("change = $%.2\f\n", 12.0);

    change = $12.000000
    change = $12.00

    int i;
    float f;

    i = 4.5;
    f = 4.5;
    

Then how do i print float numbers to screen? just like int except you need to use %f for float, instead of %d which is for int. For Example:

    int i;
    float f;

    i = 4.5;
    f = 4.5;

    printf (" i = %d\n", i);
    printf (" f = %f\n", f);
	

    i = 4
    f = 4.5000000
	

Now that we know what float is, what does (float) a do again? Let's think of the following code:

    int a = 4, b = 5;
    float f;
	
    f = a/b +1;
	
	printf ("%d / %d + 1 = %f\n", a, b, f);
	

    4 / 5 + 1 = 1.8
	

    4 / 5 + 1 = 1.0000000
	

To answer this, you need to know how compilers evaluate those expressions. When the compiler encounters this expression, it will read the expression as (int) / (int ) + (int) and the compiler will return int vaule for calculation between ints. In other words, int * int gives your int even though the result is decimal number. And int / int gives you int enen though the result might be decimal. Thus,

    3 / 6 = 0
    1 / 2 = 0
    5 / 2 = 2

    3.0 / 6.0 = 0.500000
    3 / 6.0 = 0.50000
    3.0 / 6 = 0.50000
BUT 3 / 6 = 0.00000

    (float) 3 / (float) 6 = 0.5000
    3 / (float) 6 = 0.500000
    (float) 3 / 6 = 0.500000
    3 / 6 = 0.50000

    int a = 11, b= 5, c= 3;
    float f;

    f = ((float)a ) / ((float) b ) + c;
    printf("f = %f\n", f);

what will happen if we get rid of (float)? Since the compiler will read it as (int) 11 / (int) 5 + 3 , the result will be 2 + 3, thus 5 will be printed to screen.

I hope this *short* explanation on float helps you guys understanding.. and if you have any questions, e-mail us! =)

main() {
    ....
    ....
    while ( condition1 ) {
        ....
        ....
        while ( condition2 ) {
            ....
            ....
        }
        ....
        ....
    }
}

The inner while loop will be used to get the sum of all the factors except the number itself. The inner while loop will go through the numbers from 1 to the original_number/2 and get the sum of the factors. ( you only need to go up to original_number/2 because original_number/2 is the greatest possible factor of original_number)

And after the inner while loop (the second while loop), you will need to check if the sum of all the factors is equal to the original_number. If they are the same, print out the original_number to screen and add one to the counter ( so the the outer loop can stop running if the counter becomes 4).

Still confused???

okey. then, here is a little sample program that might give you an idea about what to do in your program. This program will print out the factors of numbers 2 through 10.

The basic structure of this sample program is same as above.. two while loops (inner loop and outer loop). The outer loop will make sure that the inner part will run until the original_num becomes 10 (make sure that inner stætements will be executed when original_num is 10). The inner loop will find all the factors of given original_num. First think about what you need to do in each loop.

void main()
{
    /* initize variables..*/
    ....
    ....
    /* this loop will run until original number
       is 10.. */
    while ( condition1 ) {

        /* now we need to find factors. */
        /* this inner loop will go through
           2 to orignial_num/2 and check 
           if the number is a factor or not.*/
        while ( condition2 {
	
            /* find if the number is factor 
               of original number and 
               if factor, print the value */
            /* increment the number to test
               the number */
        }
		
        /* increment original_number */
    }
		
}

/* 
 * FileName   : factors.c
 * Name       : Ji Lee
 * Login      : cs5a6
 * Descrpition: This program will print out
 *              all the factors of numbers
 *              from 2 to 10.
 */
#include < stdio.h >
void main()
{
    /* initializing variables. */
    int original_num, factor;
    original_num = 2;   /* starts from 2 not 1 */

    /* outer loop.. iterates the original_num
       from 2 to 10 */
    while ( original_num <= 10 ) {

        printf ("factors of %d are = ", original_num);
        
        /* finding all the factors of 
           orignianl_num */
        factor = 1;
        while ( factor <= original_num /2 ) {

            /* if the remainder is 0, it's
               factor of the number..*/
            if ( original_num % factor == 0 ) {
                printf (" %d ", factor);
            }
            factor++;
        }

        printf ("\n");
        original_num++;
		
    }

    return;
}

factors of 2 are =  1 
factors of 3 are =  1 
factors of 4 are =  1  2 
factors of 5 are =  1 
factors of 6 are =  1  2  3 
factors of 7 are =  1 
factors of 8 are =  1  2  4 
factors of 9 are =  1  3 
factors of 10 are =  1  2  5 

printf ("2 * a %% - b + c + 1\n");

or

printf ("2 * a \045\045 -b + c +1\n");

Second, you can change your tutor to whoever you like and whoever fits your schedule the best. Just make sure that you e-mail both of the tutors (the one you are currently assignmed to and the new one ) because they have to exchange your records & stuff. However, the change is not guaranteed if the tutor you want has too many students already.

a * +-++b                      , then
a *   +   -  ++b               , then
3 * ( + ( - ( 5 ) ) )          , then
3 * ( + ( -5 ) )               , then
3 * ( -5 )                     , then
= -15.

#include < stdio.h >
void main()
{
    int a,b,c;
    a = 3; b = 4; c = 5;
    printf("a++ * a++ = %d\n", a++ * a++);
    printf("a++ + a++ + a++ + ++a = %d\n", 
		a++ + a++ + a++ + ++a);
}

gcc compiler from ieng9.ucsd.edu -------
a++ * a++ = 12
a++ + a++ + a++ + ++a = 27

cc compiler from ieng9.ucsd.edu -------- 
a++ * a++ = 9
a++ + a++ + a++ + ++a = 24

gcc compiler for pc --------------------
a++ * a++ = 9
a++ + a++ + a++ + ++a = 21

/*
 * FileName   : Sample.c 
 * Name       : Ji Lee 
 * Login Name : cs5a6 
 * Assignment : #1 
 * Description: This programs reads two numbers 
 *              from user, adds five other 
 *              numbers to sum and prints out 
 *              the result to the screen.
 */
#include < stdio.h >
void main()
{
	int a, b, c, d, e, f, g, sum;
	/* initializing the variables */
	scanf ("%d%d", &a, &b);
	c = 3;
	d = 4;
	e = 5;
	f = 6;
	g = 7;

	/* computing the sum */
	sum = a + b + c + d + e + f + g;

	/* output the result */
	printf ("the sum is = %d\n", sum);
}