Fold and fold again

If $A$ coincides with $O$ after a fold along $y = -x + 2^{2018}$ , then $AO$ is perpendicular to $y = -x + 2^{2018}$ , and $A$ and $O$ are also equidistant to $y = -x + 2^{2018}$ . Since $AO$ is perpendicular to $y = -x + 2^{2018}$ it has a slope of $1$ , and since $AO$ passes through the origin it has a $y$ -intercept of $0$ , so $AO$ is on the line $y = x$ . Let $D$ be the intersection of $y = -x + 2^{2018}$ and $y = x$ . Then $D$ is $(2^{2017}, 2^{2017})$ , and since $A$ and $O$ are equidistant to $D$ on the line $y = x$ , $A$ is $(2^{2018}, 2^{2018})$ .

If $B$ coincides with $O$ after a fold along $y = 2^{2016}$ , then $BO$ is perpendicular to $y = 2^{2016}$ , and $B$ and $O$ are also equidistant to $y = 2^{2016}$ . Since $BO$ is perpendicular to $y = 2^{2016}$ it is a vertical line, and since $BO$ also passes through the origin it is on the line $x = 0$ . Let $E$ be the intersection of $y = 2^{2016}$ and $x = 0$ . Then $E$ is $(0, 2^{2016})$ , and since $B$ and $O$ are equidistant to $E$ on the line $x = 0$ , $B$ is $(0, 2^{2017})$ .

If $C$ coincides with $B$ after a fold along $y = -x + 2^{2018}$ , then $BC$ is perpendicular to $y = -x + 2^{2018}$ , and $B$ and $C$ are also equidistant to $y = -x + 2^{2018}$ . Since $BC$ is perpendicular to $y = -x + 2^{2018}$ it has a slope of $1$ , and since $B$ is $(0, 2^{2017})$ , $BC$ has a $y$ -intercept of $2^{2017}$ , so $BC$ is on the line $y = x + 2^{2017}$ . Let $F$ be the intersection of $y = -x + 2^{2018}$ and $y = x + 2^{2017}$ . Then $F$ is $(2^{2016}, 2^{2016} + 2^{2017})$ , and since $B$ and $C$ are equidistant to $F$ on the line $y = x + 2^{2017}$ , $C$ is $(2^{2017}, 2^{2018})$ .

Then $ABCO$ forms an isosceles trapezoid, and letting $G$ be $(0, 2^{2018})$ , the area of $ABCO$ is equivalent to the difference of the areas of the right isosceles triangles $\triangle AGO$ and $\triangle CGB$ . $\triangle AGO$ has legs of $2^{2018}$ , so its area is $\frac{1}{2}(2^{2018})^2 = 2^{4035}$ , and $\triangle CGB$ has legs of $2^{2017}$ , so its area is $\frac{1}{2}(2^{2017})^2 = 2^{4033}$ . This means the area of $ABCO$ is $2^{4035} - 2^{4033} = 4 \cdot 2^{4033} - 2^{4033} = 3 \cdot 2^{4033}$ , which means $a = 3$ , $b = 2$ , and $c = 4033$ , and $a + b + c = \boxed{4038}$ .