$$ x(t) = \cos(\pi t) $$

The particle executes simple harmonic motion (SHM) with amplitude $$ A = 1 \text{ cm} $$ and period $$ T = 2 \text{ s} $$. In one complete cycle (2 s), the motion is as follows:

It starts at $$ x = 1 \text{ cm} $$.

It moves to $$ x = -1 \text{ cm} $$ (covering a distance of $$ 2 \text{ cm} $$).

It returns to $$ x = 1 \text{ cm} $$ (covering another $$ 2 \text{ cm} $$).

Thus, the total distance traveled in one complete cycle is:

$$ D_{\text{cycle}} = 2 + 2 = 4 \text{ cm}. $$

In 12.5 s, the number of cycles is:

$$ \frac{12.5}{2} = 6.25 \text{ cycles}. $$

For the 6 complete cycles (12 s), the distance traveled is:

$$ D_{\text{complete}} = 6 \times 4 = 24 \text{ cm}. $$

For the remaining 0.5 s, determine the displacement. At $$ t = 12 \text{ s} $$, the particle is at:

$$ x(12) = \cos(12\pi) = 1 \text{ cm}. $$

After an extra 0.5 s (i.e., at $$ t = 12.5 \text{ s} $$), the position becomes:

$$ x(12.5) = \cos(12.5\pi) = \cos\left(12\pi + \frac{\pi}{2}\right) = \cos\frac{\pi}{2} = 0 \text{ cm}. $$

The distance covered in this interval is:

$$ D_{\text{extra}} = |1 - 0| = 1 \text{ cm}. $$

So, the total distance traveled is:

$$ D = D_{\text{complete}} + D_{\text{extra}} = 24 + 1 = 25 \text{ cm}. $$

The net displacement, which is the difference between the final and initial positions, is calculated as follows:

Initial position at $$ t = 0 $$:

$$ x(0) = \cos 0 = 1 \text{ cm}, $$

Final position at $$ t = 12.5 \text{ s} $$:

$$ x(12.5) = 0 \text{ cm}. $$

Thus, the displacement is:

$$ d = |0 - 1| = 1 \text{ cm}. $$

The ratio of the total distance traveled to the displacement is:

$$ \frac{D}{d} = \frac{25}{1} = 25. $$

Therefore, the correct answer is $$25$$.