c++ quadratic equation not working properly for a code checker - c++

Question

I am submitting code to my university's tester, the code for me works thoroughly for the the test cases I provided.
The checker tells there are some test cases that can't pass, I am testing in visual studio, and that's exactly compatible with the online judge compiler.
The problem description:
input : Contains 3 integer numbers: a,b,c (|a,b,c|<=1000).
output; The number of the equation's roots, then all roots in an ascending order. If you cannot do this, output "-1".
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
int a, b, c, delta;
double x1, x2, x;
cin >> a >> b >> c;
delta = ((b*b) - (4 * a * c));
if (a == 0) {
if (b != 0) {
x = (-(c)*1.0 / (b));
cout << 1 << " " << x;
}
else
cout << "-1";
}
else
if (delta > 0)
{
x1 = ((-(b + sqrt(delta))*1.0) / (2 * a));
x2 = ((-(b - sqrt(delta))*1.0) / (2 * a));
if (x1 <= x2) {
cout << 2 << " " << x1 << " " << x2;
}
else
cout << 2 << " " << x2 << " " << x1;
}
else if (delta == 0)
{
x = (-(b)*1.0 / (2 * a));
cout << 1 << " " << x;
}
else
cout << "-1";
return 0;
}
I can't find any test case that fails my program, please if you can find some test cases, hints whatever I am just stuck at the problem. Thanks

Answer 1

the example output is described as follows; intput: 1 -8 15 output: 2 3 5
If the input was -1 8 -15, this program would output 2 5 3, while the two solutions are requested to be printed in ascending order. So you need to sort them before printing. Also I'd add a newline to every output statement.
-edit-
Without knowing the exact format required and the numbers in the failing tests, we can only speculate on how to modify the posted program. Some changes, though, could be beneficial.
I/O
There isn't any check of the input. On-line judges usually give well defined input, but you can always add some test.
cin >> a >> b >> c;
if (!cin)
cout << "-1\n";
I already mentioned the newline at the end of the line, another (unknown) requirement could be a particular precision of the outputted numbers (in fixed or scientific format).
In the comments I also pointed out the particular case of -0.0 or (-0), which while beeing a perfectly valid floating point value, could be for some reason rejected by the tester. Unlikely, but possible. So you could write the first case like this:
if (a == 0)
{
if (b != 0)
{
if ( c == 0 )
x = 0.0;
else
x = double(-c) / b;
cout << "1 " << x << '\n';
}
else
cout << "-1\n";
}
precision and corner cases
In the unlikely (given the ranges of the input values) case that you are running into precision issues, you could choose other formulas (see e.g. numerically stable algorithm for solving the quadratic equation).
else if (delta > 0)
{
if ( c == 0 )
{
x1 = 0.0;
x2 = double(-b) / a;
}
else
{
if ( b > 0 )
x1 = -0.5 * (b + std::sqrt(delta)) / a;
else
x1 = -0.5 * (b - std::sqrt(delta)) / a;
x2 = c / (a * x1);
}
if (x1 > x2)
cout << "2 " << x2 << " " << x1 << '\n';
else
cout << "2 " << x1 << " " << x2 << '\n';
}
The last part really boils down to how to consider two coincident solutions.
else if (delta == 0)
{
if ( b == 0 )
x = 0.0;
else
x = -0.5 * b / a;
cout << "1 " << x << '\n';
// Assuming they don't require ("all the roots")
// cout << "2 " << x << ' ' << x << '\n';
}

Answer 2

The problems would likely lie in the realm of int to double conversion.
To debug this, you need to separate out your operations into distinct lines the debugger can stop on. For instance the expression -(c*1.0) is better expressed as (double)-c. It at least makes it clearer and utilizes a negation (free) and cast (more precise [linguistically] than multiplication by 1.0).
So for instance, your code would become something like this :
#include <iostream>
#include <math.h>
using namespace std;
int mainC(int argc , char** argv)
{
(void)argc;
(void)argv;
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
int a, b, c, delta;
double x1, x2, x;
cin >> a >> b >> c;
delta = ((b*b) - (4 * a * c));
if (a == 0) {
if (b != 0) {
/// x = (-(c*1.0) / (b));
/// separate the operation and avoid multiplication and implicit casts
x = ((double)-c);
x /= (double)b;
cout << 1 << " " << x;
}
else
cout << "-1";
}
else
if (delta > 0)
{
/// x1 = ((-(b + sqrt(delta))*1.0) / (2 * a));
/// x2 = ((-(b - sqrt(delta))*1.0) / (2 * a));
/// separate the operation and avoid multiplication and implicit casts
// discriminant
const double discrim = sqrt(delta);
x = (double)-b;
x1 = x + discrim;
x2 = x - discrim;
x1 /= (double)(2*a);
x2 /= (double)(2*a);
cout << 2 << " " << x1 << " " << x2;
}
else if (delta == 0)
{
/// x = (-(b*1.0) / (2 * a));
/// separate the operation and avoid multiplication and implicit casts
x = (double)-b;
x /= (double)(2*a);
cout << 1 << " " << x;
}
else
cout << "-1";
return 0;
}

Answer 3

I think default cout precision does not math to your problem. Try
std::cout << std::setprecision(10);
or greater.
Also u can try use
std::cout << std::flush