Parkour Design

The clean observation is that all three moves have the same value of $\Delta x + \Delta y$:

• $(2,1)$ gives $2+1=3$,
• $(3,0)$ gives $3+0=3$,
• $(4,-1)$ gives $4-1=3$.

So after exactly $k$ moves, we must have

$$x+y=3k.$$

That gives the first hard requirement: $x+y$ must be divisible by $3$. If not, the answer is instantly NO; no amount of parkour cope fixes arithmetic.

Now assume

$$k=\frac{x+y}{3}.$$

This $k$ is the number of moves. For these $k$ moves, each move contributes one of the horizontal values $2$, $3$, or $4$. Therefore the total $x$ coordinate must be at least

$$2k$$

and at most

$$4k.$$

So we need

$$2k \le x \le 4k.$$

The only subtle point is proving that every integer $x$ in this range is actually reachable, not just the endpoints. Write

$$x = 2k + r,$$

where $0 \le r \le 2k$. Think of each move as starting from horizontal value $2$, then adding an extra amount:

• move $(2,1)$ adds extra $0$,
• move $(3,0)$ adds extra $1$,
• move $(4,-1)$ adds extra $2$.

With $k$ moves, we need the extras to sum to $r$, using $k$ numbers from $\{0,1,2\}$. Every value from $0$ to $2k$ is possible: use enough $2$'s, then maybe one $1$, and fill the rest with $0$'s. So there are no missing horizontal totals.

Once $x$ is valid, $y$ is automatically valid too because $x+y=3k$ fixes it:

$$y=3k-x.$$

So the complete condition is:

1. $x+y$ is divisible by $3$;
2. with $k=(x+y)/3$, we have $2k \le x \le 4k$.

This also handles negative $y$ naturally. For example, going downward means using more $(4,-1)$ moves, which increases horizontal distance faster, exactly reflected by the upper bound $x \le 4k$.

Complexity is $O(1)$ per test case, with $O(1)$ memory.

Research checked: Codeforces problem statement, official editorial thread, and a public C++ solution. The supplied statement remains canonical.