C arithmetic, modulus - c

Question

So I have this problem I'm working on where I must determine the cost for time spent at a go kart track; they charge by the hour and if you spend 30 minutes or more for a specific hour, you are charged the full amount. Note: I am not looking for someone to just answer my hw for me, much like I don't think anyone here would do so.
I will now post my source code:
#include <math.h>
int main(){
float price;
int minutes;
int hours;
int total_cost;
printf("How much does this go-kart location charge per hour?\n");
scanf("%f", &price);
printf("How many minutes can you spend there?\n");
scanf("%d", &minutes);
hours= minutes/60;
total_cost=hours * price;
printf("Your total will be $%d" ,&total_cost);
return 0;
}
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My problem is that
. I'm not sure how to accurately calculate someones time spent, and thus the cost will be wrong. If someones spends 1.5 hours at this place, they are billed for 2 hours.And likewise if someone spends just 30 min. they are billed for 1 hour in full.
Part of me believes I am to use the modulus operator in order to round up the remainder but I'm not sure how to do so.60 minutes in an hour and I spend 30 minutes at this track, 60 % 30 would be the time spent. But this returns a remainder of 30, and spending 30 minutes would mean I pay full price for 1 hour.
Again I'm not looking for someone to answer my hw for me, it would serve me no purpose , just some general help on how to calculate a number that must always be rounded up from the half of the hour and onwards.
Thanks to anyone and everyone who helps, or just stops by.
Update; SO I added an if else statement to my problem, and it seems to be working. I'll post the updated source code below.
................................................................................
This appears to work based on some trial runs i've done, I'll have to edit the output for 2 decimal place(%.2f, I think, to make it look nicer, thanks to all of you for your help, I'm new to programming in C, and it confuses me deeply, but It's important to me to learn this, not just for some grade, so thanks for being so helpful.

Answer 1

If you can use floating-point types, the simplest thing to do is just round to the nearest integer:
int billable_hours(int minutes) {
return (int)round(minutes / 60.);
}
it's worth reading up on the rounding modes to verify this will do what you need.
There's a trick to avoid worrying about rounding modes, which is to add a bias so that simply rounding down does what you need:
int billable_hours(int minutes) {
return (int)floor(0.5 + minutes / 60.);
}
But that this works the same if we stick with integers (integer division discards the fractional part, effectively rounding towards zero):
int billable_hours(int minutes) {
return (minutes + 30) / 60;
}
because any fractional hour >= 30 minutes, we force up to the next whole hour before discarding the fractional part.
Notes:
these will behave differently with negative inputs. Not (hopefully) relevant here, but worth being aware of
be aware of type promotions: the literal 60. is a double, while 60 would be an integer. When you write int / double, it is promoted to double / double instead, so you don't have to cast. The division in the third function is still integer division, though.
by the simplest thing, I mean the simplest thing to type and/or read. Computationally the third is by far the simplest.

Answer 2

Try this,
int extra;
hours= minutes/60;
extra = minutes % 60;
if(extra >= 30)
hours += 1;
total_cost=hours * price;
printf("Your total will be $%d" ,total_cost);
hours = minutes/60 gives the number of full hours, since hours is an integer variable. hours%60 gives you the extra minutes spent. And there is no need for an & before total_cost in the printf statement.

Answer 3

You are rounding times to the nearest integer. When you do integer (modulo N) arithmetic, one way to round things to the nearest integer is to add N/2 to the original number of minutes (this is equivalent to the int(x + 0.5) problem to arithmetic (module 1)), so when you integer divide by N, you'll get results rounded to the nearest integer. Try hours_to_charge = (minutes_run + 30) / 60; and you'll get the desired results.
Example
So, if you have, for example 37 minutes and want to know how many hours to charge, just add 30 min to that (getting 67) and then divide by 60 (getting one hour)
But, be carefull that for times of under 30 minutes you are going to charge 0 to the cart runners.

Answer 4

Just cast minutes to a float, and than round the number using ceil() and floor(). A very common way of doing this is defining a macro called ROUND to accomplish this:
#define ROUND(x) ((x - floor(x) >= 0.5) ? ceil(x) : floor(x))
hours = ROUND((float)minutes/(float)60);
This will turn your hours into billable hours, and the rest of your code will work as coded.