java help for beginners

Compute the remainder of seconds entered by the user by the number of seconds
in an hour (SECONDS_X_MINUTE).


import java.util.Scanner;

public class Time {

    public static final int SECONDS_X_HOUR = 3600;
    public static final int SECONDS_X_MINUTE = 60;
    public static final int AM_PERIOD = 43199;
    public static void main(String[] args) {
   
        int timeInSeconds;
        int timeInHours;
        int remainderHours;
        int timeInMinutes;
        int standardTimeInHours;
       
        Scanner keyboard = new Scanner(System.in);
        System.out.println ("Please Enter Time in Seconds :\t");
        timeInSeconds = keyboard.nextInt();
        timeInHours = timeInSeconds/SECONDS_X_HOUR;   
        remainderHours = timeInHours&SECONDS_X_HOUR;
        timeInMinutes = remainderHours * SECONDS_X_MINUTE;
       
       
        System.out.println (timeInSeconds + " " + timeInHours + " " + remainderHours + " " +timeInMinutes);
        System.out.println ("Please enter the time in seconds: " + timeInSeconds);
        System.out.println ("Standard Format: " + (int)timeInHours + "h. " + timeInMinutes + "m. " + timeInSeconds + "s.");
        System.out.println ("AM/PM Format: " + timeInHours + "h. " + timeInMinutes + "m. " + timeInSeconds + "s.");


    }

}

basically i am trying to figure out how to get the remainder from the timeInHours calculation

I think I just threw up a little

I think I just threw up a lot

ok its an int so something like 26/3 is going to come up as 8 i realize but if you could me out that would be great

JAVA HAS A FUCKING MODULUS OPERATOR FOR FUCKS SAKE USE THAT

oh fuck yeah i meant % instead of & but im still not getting the remainder like i want

DON'T HELP HIM!!!

god. thanks.

>>8 I wasn't going to help him till you did that, now i will

remainderHours = timeInHours%SECONDS_X_HOUR;
You're taking modulus of timeInHours, so the answer you get will always be the same as timeInHours

Now if you take timeInSeconds%SECONDS_X_HOUR you will get the number of seconds remaining.

i.e. 4.30 am =  16200s,
     16200/3600 = 4h
     16200%3600 = 1800s

/prog/ - /pr/

THIS THREAD IS NOW ABOUT HOW JAVA SUCKS BALLS!!!!!!!!

>>10
thanks for not being a prick

>>13
no thanks for not reading SICP

1. Divide the seconds entered by the user by the number of seconds in an hour
(SECONDS_X_HOUR), the result will be the hours in standard time.

5. The number of hours in AM/PM format is the remainder of dividing the hours in
standard time computed in (1) by 12.

again i think this might be a logic error on my part.

is this this what the problem is asking? if so my output is messed up.

int AmPmTimeInHours = (timeInSeconds%SECONDS_X_HOUR)/12;

i figured it out
int AmPmTimeInHours = timeInHours%12; its that right?

Do your own homework.  Nothing you are being asked to do is hard, the majority of your errors are because you aren't even thinking about what you want to do, you are just writing statements.  This isn't a "We'll write your programs" board.

will someone write a program doing the pythageron law for me plz?

>>18
I will, along with haxing your anus.

>>18
Fuck. IHBT

We have been trolled constantly.

>>18
int main(int argc,char**argv){
 int a,b,c;
 if(argc<2)return 1;
 a=atoi(argv[1])
 b=atoi(argv[2])
 c=sqrt(a*a+b*b);
 printf("%d",c);
}

pyth a b = sqrt $ a**2 + b**2

>>22
int

Horrible!

>>24
Not really -- int arithmetic is faster.

>>23
pyth a b = sqrt . foldl (+) 0 $ map (**2) [a, b]

>>26
I do not use folds because I am GvR.

I do not use folds because I am "GRUNNER"

>>23,26
pyth a b = sqrt . sum $ map (**2) [a, b]
or you could use a better language:
: pyth ( a b -- c ) [ sq ] bi@ + sqrt ;

and then there's this bullshit:
sub pyth($$){sqrt sum map{$_**2}@_}

>>27
GvR would probably do something like this:
(a*a+b*b)oh god how did i get here im not good with computer

>>25
But you can't sqrt an int.

>>30
But you can sqrt a fish

>>29
sub pyth($$){sqrt sum map{$_**2}@_}

Prototypes considered harmful.

>>30
int isqrt(n){
 /* calculates integer square root.    *
  * if n is less than zero, calculates *
  * integer square root divided by i.  */
  int r = 0, i = 1;
  if(0 > n) n = -n;
  for(i = i << ((sizeof(int) * CHAR_BIT - (0 > ~i ? 2 : 1)) / 2) * 2; i; i >>=2)
    if(n >= (i | r)){
      n -= i | r;
      r = r >> 1 | i;
    }else r >>= 1;
  return r;
}

>>25,33
int i = sqrt(2);
float f = sqrt(2);
assert( (float) i == f);

>>34
int i = 2;
assert(sqrt(i) == sqrt(2));

>>34-35
Actually, i = sqrt(-1)

>>36
no, i = sqrt(-1)

41 Name: Anonymous 2013-01-19 23:14

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