There is a empty sack on the floor which area of density is s(the sack's) and air can't go through it's surface.
Sack's circumference is L and length l(so when it's filled with air it's like a sausage, also l>L)
Inside the sack is overpressure( p. Air density is meager and free-fall acceleration is g.
Define the length of contact(c) between sack and the floor.
Answer is:
The pressure at the floor P = p + sg, hence sLg=(p+sg)c and on and on(I can handle the rest)
But I don't understand the meaning of the P = p + sg. I thought it like how does the overpressure inside "the box" affect to the floor. And then there is that devil sg.
Could somebody explain that P part, please?

RKJP, I think I understand your question. At first I thought you meant that P was the pressure exerted on the floor. In that case, the answer doesn't make any sense at all. Then I realized that P is simply the net pressure inside the sack when it is in contact with the floor.

The overpressure within the sack is p. That means that if the sack was just floating in the air, any point on its interior surface would be experiencing an outward pressure p. In the case at hand, however, there is an additional force acting on part of the sack. That additional force comes from the floor, and it's equal and opposite to the weight of the sack, mg. This upward force from the floor causes the sack to collapse somewhat, forming a flat spot on the floor, thereby decreasing its volume and correspondingly increasing the interior pressure until an equilibrium is reached. Since the new equilibrium pressure, P, is exactly counteracting the weight of the sack, P must be equal to the original pressure, p, plus the additional pressure due to the weight of the sack. That additional pressure is the weight, mg, divided by the surface area, A. So P = p + mg/A = p + sg.

Aaah... Now I understand...
Thanks