7. A snail initially located at point $(0,1)$, every day it crawls from point $(x, y)$ to point $(2 y, x+y)$, at the end of the first day, it crawls to point $(2,1)$. Then on the 2017th day, it is located at $\qquad$ .
7. $\left(\frac{2}{3} \times 2^{2016}-\frac{2}{3}, \frac{1}{3} \times 2^{2017}+\frac{1}{3}\right)$.

Suppose the snail is at point $\left(a_{n}, b_{n}\right)(n \in \mathbf{N})$ on the $n$-th day.
Then $a_{0}=0, b_{0}=1, a_{n+1}=2 b_{n}, b_{n+1}=a_{n}+b_{n}$.
Thus, $b_{n+2}=2 b_{n}+b_{n+1}$
$$
\begin{aligned}
\Rightarrow & b_{n+2}+b_{n+1}=2\left(b_{n+1}+b_{n}\right) \text { or } \\
& b_{n+2}-2 b_{n+1}=-\left(b_{n+1}-2 b_{n}\right) .
\end{aligned}
$$

Therefore, $\left\{b_{n+1}+b_{n}\right\}$ is a geometric sequence with the first term $b_{1}+b_{0}$ and common ratio 2, and $\left\{b_{n+1}-2 b_{n}\right\}$ is a geometric sequence with the first term $b_{1}-2 b_{0}$ and common ratio -1.
$$
\begin{array}{l}
\text { Hence } b_{n+1}+b_{n}=2^{n}, b_{n+1}-2 b_{n}=(-1)^{n} \\
\Rightarrow b_{n}=\frac{1}{3}\left(2^{n}-(-1)^{n}\right) \\
\Rightarrow a_{n}=\frac{2}{3}\left(2^{n-1}-(-1)^{n-1}\right) .
\end{array}
$$

Thus, the snail is at
$$
\left(\frac{2}{3} \times 2^{2016}-\frac{2}{3}, \frac{1}{3} \times 2^{2017}+\frac{1}{3}\right) 
$$
on the 2017th day.