Halloween candy codecoach!

Question

#include <stdio.h>
#include<math.h>
int main() {
int houses;
scanf("%d", &houses);
float b=(200.0/houses); printf("%.0f",ceil(b));

//your code goes here

return 0;
}
Is there any other ways to do it?

Answer 1

#include <stdio.h>

int main() {
int houses;
scanf("%d", &houses);

int perc = (int)(((2.0/(double)houses)*100.0) + 0.99);
printf("%d", perc);

return 0;
}

Answer 2

Hi Dan Delgado
If 200 were to be used, the result would be the integer part of the division 200/houses, as both terms are integer. But using 200.0 the compiler automatically cast the division as a floating point operation.

Answer 3

Oh that makes sense, thanks for the explaining, now i get it. :)

Answer 4

why did you put 200.0 instead of 200?

Answer 5

I thought alike and I still don't understand why it's wrong.

Follows below my code:

#include <stdio.h>

int main() {
int houses;
scanf("%d", &houses);

//your code goes here
float prob = 200 / houses;

printf("%.0f",prob);
return 0;
}

Since I declared 'prob' as float and used '%.0f' on printf() shouldn't it round prob to the nearest whole number?