Shouldn't gravity simulations solve for acceleration instead of force? I'm learning computer programming right now, and one of the exercises I've seen quite a few people do is simulating gravity in space: i.e. planets orbiting a sun. All of the simulations of this kind that I've seen simply apply Newton's law of universal gravitation to all of the celestial bodies: 
The programs I've seen use F as the acceleration of the two celestial bodies in question, but isn't F force? For these programs to be accurate, wouldn't MA have to be substituted in for F? Which would make the equation for the acceleration of one celestial body toward another be . This feels correct to me, since if the first equation was used for the acceleration of two celestial bodies toward each other, then they would both accelerate toward each other at the same rate, even if their masses were drastically different, but I want to be sure because I've seen several implementation of gravity on a celestial scale that use the first equation.
EDIT: Some of the code I'm using, in Javascript (I apologize in advance, this is my first try), CBodies is an array with all of the Celestial Bodies in it, all Celestial Bodies have position and velocity, which are both vectors:
for (var i = 0; i < CBodies.length;i++) {
    var totalAcc = new PVector(0,0);
    for (var j = 0; j < CBodies.length; j++) {
        if (j !== i) {
            var accForce = grav * ((CBodies[i].mass * CBodies[j].mass) / sq(PVector.dist(CBodies[i].pos, CBodies[j].pos)));
            var tempVec = new PVector(CBodies[j].pos.x - CBodies[i].pos.x, CBodies[j].pos.y - CBodies[i].pos.y);
            var theta = tempVec.heading();
            var accX = accForce * cos(theta);
            var accY = accForce * sin(theta);
            var acc = new PVector(accX,accY);
            
            totalAcc.add(acc);
        }
    }
    totalAcc.limit(5);
    totalAcc.mult(time);
    CBodies[i].vel.add(totalAcc);
}

 A: Both options are possible in order to update your simulation variables. It depends whether you want to bookkeep only $(x, \dot{x})$ or also $(x,\dot{x}, \ddot{x})$ and what level of accuracy you want for your physics to enforce.
Computing all forces $\vec F_{ij}$ separately and then assigning them to an acceleration $\vec a_{i}$ is more computationally intensive, but it allows to construct algorithms that enforce symmetry $\vec F_{ij}=-\vec F_{ji}$ and inertiality of the frame $\vec a_{center-of-mass}=\vec0$.
A: Your concern is actually valid for forces other than gravity.
The obvious example: the Coulomb force (the electrostatic force)
Protons and positrons have the same charge: 1 unit of positive charge, but placed in the same electrostatic field the positron will undergo larger acceleration because the positron has far less mass than the proton.

In the case of gravity: as far as we know gravitational mass and inertial mass are the same.
We are compelled to grant that gravitational mass and inertial mass are the same because we see everywhere: all objects, when subjected to gravitational interaction, undergo the same acceleration.
The case of a two body system:
$m_1$ mass of the primary
$m_2$ mass of the secondary
$$ \begin{array}{rcl} 
F & = & G \frac{m_1 m_2}{r^2}   \\
F & = & m_2 a  \tag{1}
\end{array} $$
Combining:
$$ m_2 a = G \frac{m_1 m_2}{r^2} \tag{2} $$
At this point: the mass of the secondary is on both the left and side and right hand side, so it can be dropped:
$$ a = G \frac{m_1}{r^2} \tag{3} $$

Actually, in the case of the solar system what is known with accuracy is the value of the product $G m_1$. This product has the dimensions of acceleration.
We can for example use the letter '$g$' for the amount of gravitational acceleration that the Sun causes at 1 unit of distance.
For gravitational acceleration at any distance $r$ to the center of gravitational attraction:
$$ a = \frac{g}{r^2} \tag{4} $$
So:
Yeah, the universal convention is to express gravitational effect directly in the form of gravitational acceleration.
This comes about because of the equivalence of gravitational and inertial mass.
.
