3 4 1 2 4 2 1 3 3 3 1 1 1
Takahashi For each $i = 1, 2, 3$, let $A'_i$ be the number of stones remaining in the $i$\-th pile. Now, we use the sequence $A' = (A'_1, A'_2, A'_3)$ to represent the numbers of stones remaining in the piles. Before the start of the game, we have $A' = (1, 2, 4)$. One possible progression of the game is as follows. * First, Takahashi removes $1$ stone from the $3$\-rd pile. Now, $A' = (1, 2, 3)$. * Next, Aoki removes $2$ stones from the $2$\-nd pile. Now, $A' = (1, 0, 3)$. * Then, Takahashi removes $2$ stones from the $3$\-rd pile. Now, $A' = (1, 0, 1)$. At this point, the $1$\-st and $3$\-rd piles still have $1$ stone each, but it is forbidden ー as the fourth forbidden move ー to remove exactly $1$ stone from a pile composed of exactly $1$ stone, so Aoki cannot play. Thus, Takahashi wins.
1 5 5 5 1 5 2 5 3 5 4 5 5
Aoki
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"content": "Takahashi and Aoki will play a game against each other using $N$ piles of stones.\nInitially, for each $i = 1, 2, \\ldots, N$, the $i$\\-th pile is composed of $A_i$ stones. The players alternately perfo...",
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