8.58 Two players, A and B, are playing a game on a strip of paper with $1 \times 100$ squares: there is 1 chess piece placed in the leftmost square. Each move can shift the piece 1 square to the right, 10 squares to the right, or 11 squares to the right. Players A and B take turns making moves, with A going first, until one player is unable to make a move according to the rules and loses. If both players play optimally, who will win?
[Solution] Number the squares from left to right as $1, 2, \cdots, 100$. For a square, if a player can definitely win from it, it is called a "winning square"; if a player will definitely lose from it, it is called a "losing square". Clearly, if a player can only move to a winning square from a certain square, then that square is a losing square; if a player can move to at least one losing square from a certain square, then that square is a winning square. It is easy to see that the 100th square is a losing square. Therefore, the 99th square is a winning square, the 98th square is a losing square, and so on, alternating between winning and losing squares, until the 91st square is a winning square. The 81st to 90th squares are all winning squares, because from each of these squares, a player can move 10 or 11 squares to enter one of the last 10 squares, which are losing squares. Similarly, it can be seen that among the 10 squares from 71 to 80, the even-numbered squares are losing squares and the odd-numbered squares are winning squares; and the 10 squares from 61 to 70 are all winning squares. By analogy, the 1st square is a winning square. Therefore, if player A goes first, he can definitely win.