26.22 * There are 100 balls, and two people, A and B, take turns to take balls from the bag. Each time, they can take 1 to 5 balls. A goes first, and the one who takes the last ball wins. Who will be the final winner (assuming both are sufficiently smart)?
To obtain the last ball, before taking the last ball, when the opponent takes the ball, there must be no less than 6. And if there are more than 6, the opponent can take away some, making the number of balls still more than or equal to 6, so that the last one cannot be taken either. Therefore, the winner, before the last turn, must leave the opponent with exactly 6 balls. By reasoning backward, we can conclude that A will win. The strategy is: the first time, take 4 balls, and afterward, if B takes $k(1 \leqslant k \leqslant 5)$ balls, then A takes 6- $k$ balls. Thus, in the last round, there will definitely be 6 balls left, and no matter how many balls B takes, A can take the last ball.