Total number of common factors for two numbers LARGE VALUES up to 10^12
long long count = 0, m, n;
    ...
    for (long long int i = 1; i &lt;= ((min(m, n)) / 2) + 1; i++) {


long long or long long int? I'm not a C++ programmer so I don't know whether style guides have a preference, but I'm quite certain that you should be consistent.




  Inputs are two values \$1 \le m , n \le 10^{12}\$. I don't know why my code is taking so long
  ...

   for (long long int i = 1; i &lt;= ((min(m, n)) / 2) + 1; i++) {



There you go. The main loop runs up to \$5 \times 10^{11}\$ times. As a rule of thumb, when brute-forcing a problem I work on the basis that \$10^9\$ is about as high as I want to go.

Solution: change your algorithm. This will require learning some mathematics, and I'm not going to write a detailed tutorial, but there are two things you need:


Basic number theory: unique prime factorisation, greatest common divisor, finding it efficiently;
Sane factorisation of a small number (and \$10^{12}\$ is very small in these terms)


If you wanted to do it properly then you could extend the number theory research and learn about Euler's totient function.