SMPCIRC - Two Circles - c++

Question

I dont know where I have a mistake
It work for the example but the result shows me "bad answer"
THE PROBLEM:
https://www.spoj.com/problems/SMPCIRC/
SMPCIRC - Two Circles
plane-geometry #basics
Given two circles: O1 with the center o1 = (xo1, yo1) and a radius r1
and O2 with the center o2 = (xo2, yo2) and radius r2, please compute
if O1 is inside O2 or if O2 is inside O1.
Input description First t < 1000, the number of test cases. In each of
the following t lines, 6 integers: xo1 yo1 r1 xo2 yo2 r2. Where 0 ≤
xo1, yo1, xo2, yo2 ≤ 10000 and 0 < r1, r2 ≤ 10000.
Output description For each test case print one character: I, if O1 is
inside O2 (or if O2 is inside O1), E, if O1 is internally tangent to
O2 (or if O2 is internally tangent to O1), O, in other cases.
Example
Input: 2 103 104 5 100 100 10 103 104 10 100 100 10
Output: E O
MY CODE:
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int t;
cin>>t;
for(int i=0;i<t;i++)
{
double x1, y1, r1, x2, y2, r2, dl;
cin>>x1>>y1>>r1>>x2>>y2>>r2;
dl=sqrt(pow(x2-x1,2) + pow(y2-y1,2));
if(dl+r1==r2)
cout<<"E";
else if(dl+r2==r1)
cout<<"E";
else if (dl+r1<r2)
cout<<"I";
else if(dl+r2<r1)
cout<<"I";
else
cout<<"O";
}
return 0;
}

Answer 1

When using floating point types, a certain amount of numerical errors are to be expected. Sometimes this may be unacceptable for the kind of problem we are trying to solve.
The quote in OP's question, clearly state that the inputted variables are (emphasis mine)
6 integers: xo1 yo1 r1 xo2 yo2 r2. Where 0 ≤ xo1, yo1, xo2, yo2 ≤ 10000 and 0 < r1, r2 ≤ 10000.
Given the nature of the underlying mathematical problem, it's safe to use integer variables for the calculation too, applying just a bit of algebra:
#include <iostream>
constexpr long int square(long int x)
{
return x * x;
}
int main()
{
int t;
// Usually, the input is well defined in this kind of assignments
std::cin >> t;
for(int i=0; i < t; i++)
{
long int x1, y1, r1, x2, y2, r2;
std::cin >> x1 >> y1 >> r1 >> x2 >> y2 >> r2;
// Calculte the square of the distance
long int d2 = square(x2 - x1) + square(y2 - y1);
// Instead of d == r2 - r1, check if d^2 == (r2 - r1)^2
long int dr21 = square(r2 - r1);
if ( d2 > dr21 )
std::cout << "O\n";
else if ( d2 < dr21 )
std::cout << "I\n";
else // This is probably the least likely case
std::cout << "E\n";
}
return 0;
}

Answer 2

You shouldn't use == operator to compare two double values. Use
fabs(a - b) < eps instead of == operator. You can choose eps = 1e-7.
Also print a new line after each output.