I. Slot Machine

Key Observations

• Fix all wheels except one wheel $i$, and rotate only $i$ through a full cycle. For each position of wheel $i$, the reported number of distinct symbols is minimal exactly when the symbol shown by wheel $i$ is already present on at least one of the other wheels. So one full scan of a wheel gives a binary mask of which symbols are currently present elsewhere.

• First scan wheel $1$. From this mask we can pick one symbol $s$ that is absent from all other wheels, and one symbol $r$ that is present on some other wheel.

• Now fix wheel $1$ to show $s$ and scan another wheel $j$. Then fix wheel $1$ to show $r$ and scan $j$ again. The only difference between the two scans is the position where wheel $j$ itself shows symbol $s$. Therefore comparing the two masks reveals exactly how much wheel $j$ must be rotated to show $s$.

• Repeating this for every wheel gives a common target symbol, so the final phase is just rotating every wheel to its discovered offset.

Algorithm

1. Maintain the current offset of each wheel modulo $n$.

2. Define scan(i): rotate wheel $i$ through all $n$ positions, record the reported distinct-count value for each one, and mark the positions that attain the minimum value.

3. Scan wheel $1$ once. Choose one position labeled $0$ in the resulting mask and call its symbol $s$; choose one position labeled $1$ and call its symbol $r$.

4. Rotate wheel $1$ to $s$. For each wheel $j \ge 2$:

   • scan $j$ with wheel $1$ fixed at $s$;

   • rotate wheel $1$ to $r$ and scan $j$ again;

   • compare the two masks to find the unique position where wheel $j$ shows symbol $s$;

   • rotate wheel $1$ back to $s$.

   • Finally, rotate every wheel to the stored position that makes it show $s$. enumerate

     Correctness Proof

     We prove that the algorithm returns the correct answer.

     Lemma 1.

     For a scan of wheel $i$, a position is marked by the algorithm if and only if the symbol shown by wheel $i$ at that position is already shown by at least one other wheel.

     Proof.

     Fix the other $n-1$ wheels. If wheel $i$ is rotated to a symbol that is absent from all other wheels, the total number of distinct visible symbols increases by one compared to showing any already-present symbol. Therefore the minimum possible distinct-count value is attained exactly at the positions where wheel $i$ shows a symbol already present elsewhere. □

     Lemma 2.

     For every wheel $j \ge 2$, comparing the two scans described above identifies the unique offset at which wheel $j$ shows the chosen target symbol $s$.

     Proof.

     The two scans differ only in the symbol shown by wheel $1$: once it is $s$, once it is $r$. All other wheels are unchanged. By construction, $s$ is absent from the other wheels when wheel $1$ shows $s$, while $r$ is present on at least one other wheel when wheel $1$ shows $r$. Hence the membership test from Lemma 1 changes exactly at the position where wheel $j$ itself shows $s$. That position is therefore identified uniquely. □

     Theorem.

     The algorithm outputs the correct answer for every valid input.

     Proof.

     By Lemma 2, for every wheel we discover an offset that makes it show the same symbol $s$. After the final rotation phase, all wheels therefore display $s$, so the machine shows only one distinct symbol and the jackpot condition is met. □

     Complexity Analysis

     One scan uses exactly $n$ actions. We perform one scan for wheel $1$, two scans for every other wheel, and then $O(n)$ final rotations. Thus the number of actions is at most \[ 1 + n + 2(n-1)n + O(n) < 2n^2 + 3n \le 5150 \] for $n \le 50$, well below the limit of $10\,000$. The local memory usage is $O(n)$.

     Implementation Notes

     • The helper rotateTo always uses the shorter cyclic rotation.

     • The solution exits immediately when the judge reports $k=1$, as required by the interaction protocol.