Shortest Statement Ever

The condition $p\&q=0$ means every bit can belong to at most one of the two numbers. So for each bit we may choose exactly one of

$$(p_i,q_i)\in\{(0,0),(1,0),(0,1)\}.$$

The trap is assuming bits are independent for the distance. They are not, because $|x-p|$ depends on the highest bit where $x$ and $p$ differ. So don't do local “fix bad common bits” hacks; carries and comparisons will clown on you.

Instead, remove the absolute values by trying all four order cases:

$$p \lessgtr x,\qquad q \lessgtr y.$$

For a fixed case, the objective becomes linear. Let

$$s_p = \begin{cases} +1,& p\ge x,\\ -1,& p\le x, \end{cases} \qquad s_q = \begin{cases} +1,& q\ge y,\\ -1,& q\le y. \end{cases}$$

Then inside that case,

$$|x-p|+|y-q|=s_p(p-x)+s_q(q-y),$$

where the terms involving $x,y$ are constants. So we only need to minimize

$$s_p p+s_q q$$

subject to:

• $p\&q=0$,
• $p\ge x$ or $p\le x$ depending on the case,
• $q\ge y$ or $q\le y$ depending on the case.

Now this is a standard bit DP.

Process bits from high to low. We need enough bits to allow answers up to below $2^{31}$, so bits $30,29,\dots,0$ are enough because $x,y<2^{30}$ and the statement guarantees a valid optimal answer within this range.

For each number, keep its comparison with the target prefix:

$$0 = \text{equal so far},\quad 1 = \text{already smaller},\quad 2 = \text{already larger}.$$

A DP state is

$$dp[i][c_p][c_q],$$

meaning after deciding all bits above $i$, the current prefix of $p$ compares to $x$ as $c_p$, and the current prefix of $q$ compares to $y$ as $c_q$. The value stored is the minimum signed contribution already paid.

When choosing bit $i$, try:

$$(p_i,q_i)\in\{(0,0),(1,0),(0,1)\}.$$

The contribution is

$$(s_p p_i+s_q q_i)2^i.$$

The comparison state updates normally:

• if it was already smaller/larger, it stays that way;
• if it was equal, comparing the chosen bit with the target bit decides whether it remains equal, becomes smaller, or becomes larger.

After all bits are chosen, only states satisfying the selected order case are allowed. For example:

• for $p\le x$, final comparison must be equal or smaller;
• for $p\ge x$, final comparison must be equal or larger.

Among valid final states, take the minimum DP value for that case, reconstruct $(p,q)$, and compute the real distance

$$|x-p|+|y-q|.$$

Finally, try all four cases and output the pair with smallest real distance.

Why this works:

Every feasible pair $(p,q)$ has exactly one relationship to each target: $p\le x$ or $p\ge x$ both include equality, and same for $q$. So the pair appears in at least one of the four cases. In any fixed case, the DP enumerates every bit assignment satisfying $p\&q=0$ and the required comparisons. Since the absolute-value objective is linear inside that case, minimizing the signed bit-sum is exactly minimizing the original distance among pairs in that case. Taking the best result over all four cases therefore gives a global optimum.

Edge cases are clean:

• $x=0$ or $y=0$ works naturally.
• $x=y=0$ returns $(0,0)$.
• If both numbers share many bits, the DP may either decrease one side or increase the other; it chooses whichever is actually cheaper.
• Bit $30$ is important: for values near $2^{30}-1$, the closest valid number may be $2^{30}$.

There are only $31$ bits, $3\cdot3$ comparison states, $3$ transitions per state, and $4$ sign cases. So each test case is tiny:

$$O(4\cdot31\cdot9\cdot3)=O(1)$$

time, and $O(31\cdot9)$ memory, or just constant memory plus parent pointers for reconstruction.