Counting the positive integer solutions of an equation
(Remark: The question mentions "positive number of solutions", but from your code I assume that you meant "number of non-negative integer solutions".)

Coding style:


#include &lt;cstring&gt; is not needed here.
Don't use namespace std;, see for example Why is “using namespace std;” considered bad practice?.
long c; is only used in main(), so there is no need at all to declare it as
a global variable.
Always use braces { ... } for the body of for- and if-statements, even
if it has only one statement. It helps to avoid errors if additional statements 
are added later.
Move the computation of the number of solutions to a separate function.
This keeps the main function small and is convenient if you add test cases.
I prefer a different spacing in for- and if-statements, but that may be a 
matter of taste.
Variable names: If the equation is given as \$ x + y + xy = n \$, then why
not use the same names in your program? And x is a quite non-descriptive
name of a number of solutions. 
It may not be immediately obvious why 2 is added to the number of solutions,
so an explaining comment would be appropriate here.


Then your code would look like this:

#include &lt;iostream&gt;

// Number of non-negative integer solutions to
//    x + y + x * y == n   .
long numberOfSolutions(long n) {

    if (n == 0) {
        return 1;
    }

    // Count all positive solutions:
    long count = 0;
    for (long x = 1; x &lt;= n / 2; x++) {
        for (long y = 1; y &lt;= n; y++) {
            if (x + y + x * y == n) {
                count++;
                break;
            }
        }
    }
    // Add 2 for the solutions x=0,y=n and x=0,y=n:
    return count + 2;
}

int main()
{
    long n;

    std::cin &gt;&gt; n;
    std::cout &lt;&lt; numberOfSolutions(n) &lt;&lt; std::endl;
}


Performance:

Your code tries all possible combinations of x, y to find solutions
and therefore runs in \$ O(n^2) \$. A small improvement could be to use the
symmetry of the problem, i.e. enumerate only pairs with x &lt;= y:

for (long x = 1; x &lt;= n / 2; x++) {
    for (long y = x; y &lt;= n; y++) {
        if (x + y + x * y == n) {
            // Counts a 2 solutions if x != y:
            count = count + 1 + (x != y);
            break;
        }
    }
}


This cuts the total number of iterations down by a factor of 2, but it is still
\$ O(n^2) \$.

Better algorithm:

If you write your equation as
$$
  n + 1 = x + y + xy + 1 = (x + 1)(y + 1)
$$
then it becomes obvious that it can be solved by determining the
number of divisors of \$ n+1 \$. This known as the Divisor function and can be computed 
efficiently.

Even the simplest implementation

long numberOfDivisors(long n) {
    long count = 0;

    for (long j = 1; j &lt;= n; j++) {
        if (n % j == 0) {
            count++;
        }
    }
    return count;
}


runs in \$ O(n) \$ instead of \$ O(n^2) \$. More sophisticated methods use
the prime factorization  of \$ n \$ .