This solution would not accept any floats - c

Question

Currently on one of the cs50x problem sets 'Cash', which is a simple 'ask for how much change is owed, then calculate how many coins are required' task, so not here asking for a solution but, I don't understand why this won't work.
While it does ask for an input, when I type in a float such as 5.96, it simply hangs. No returns, no errors whatsoever. I have to force it shut. The other thing is the while loop doing the same when set to 0, which is the intended way of doing things in order to get the exact number of coins.
I know how inefficient this code is and there are simpler ways of doing things. I just wish to understand the whys in order to avoid making the same mistakes moving on. Thanks.
#include <stdio.h>
#include <cs50.h>
#include <math.h>
int main(void)
{
// changes and other containers
int p = 1;
int n = 5;
int d = 10;
int q = 25;
int x = 0;
float c;
// get how much change is owed in float
do
{
c = get_float("Change owed: ");
}
while (c < 0);
// int conversion to avoid imprecision
int a = round(c * 100);
// 1 because 0 spits out an unknown error
while (a >= 1)
{
// if the converted amount is bigger than a quarter
if (a >= q)
{
// x = number of coins, a = amount left
x = a / q;
a = a % q;
}
else if (a >= d)
{
x = x + a / d;
a = a - a % d;
}
else if (a >= n)
{
x = x + a / n;
a = a - a % n;
}
else
{
x = x + a / p;
a = a - a % p;
}
}
printf("%i\n", x);
printf("%i\n", a);
}

Answer 1

Thanks to WhozCraig, I figured out that my logic was at fault.
#include <stdio.h>
#include <cs50.h>
#include <math.h>
int main(void)
{
// changes and other containers
int p = 1;
int n = 5;
int d = 10;
int q = 25;
int x = 0;
float c;
// get how much change is owed in float
do
{
c = get_float("Change owed: ");
}
while (c < 0);
// int conversion to avoid imprecision
int a = round(c * 100);
// 1 because 0 spits out an unknown error
while (a >= 1)
{
// if the converted amount is bigger than a quarter
if (a >= q)
{
// x = number of coins, a = amount left
x = a / q;
a = a % q;
}
else if (a >= d)
{
x = x + a / d;
a = a % d;
}
else if (a >= n)
{
x = x + a / n;
a = a % n;
}
else
{
x = x + a / p;
a = a % p;
}
}
printf("%i\n", x);
printf("%i\n", a);
}