Eating Game

Research check: the official Codeforces Round 1084 editorial code for 2200A and an independent write-up on cnblogs both reduce the problem to counting occurrences of the maximum value. The supplied statement matches that logic.

Let

$M = \max_{1 \le i \le n} a_i.$

Fix some starting player $s$. The game can be viewed as repeated full laps around the table in the cyclic order

$s, s+1, \dots, n, 1, 2, \dots, s-1.$

During the $k$-th lap, player $i$ gets their $k$-th turn. They eat on that turn if and only if they still have at least one dish left, which is exactly when

$a_i \ge k.$

So the last lap where any eating can happen is lap $M$. On that lap, the only players who can still eat are exactly the players with $a_i = M$. Every player with fewer dishes is already empty. That kills the tempting but wrong idea of using total dishes modulo $n$; empty players still take turns and skip, so that shortcut faceplants.

Now we know the winner must be a maximum player. We also need to show every maximum player can win.

Take any player $w$ with $a_w = M$. Choose the starting player to be the player immediately after $w$ in the circle, i.e.

$s = (w \bmod n) + 1.$

In this cyclic order, player $w$ appears last. Therefore, on the final lap $M$, all other maximum players eat their final dish before $w$, all non-maximum players skip, and then $w$ eats the last remaining dish. So $w$ can be the winner.

Thus the answer is simply the number of indices $i$ such that

$a_i = M.$

Edge cases are automatic: if $n=1$, the answer is $1$; if all players have the same number of dishes, any player can be made last by choosing the right start, so the answer is $n$.

Complexity per test case is $O(n)$ time and $O(1)$ extra memory.