& ltp> After all, prime numbers are hard to distinguish.
& ltp> So I'll just say my opinion.
& ltp> options 1
& ltp> First of all, eleven teams will draw lots to match, and one team will have a bye in the first round.
Five teams entered the loser group in the first round of<p>. The losing group doesn't play. & lt/p & gt；
& ltp> In the second round, three teams entered the loser group. At this time, there are eight losers in the loser group. This makes it easier to distribute the games of the loser group. & lt/p & gt；
There are still three teams left in the winner group of<p>, which is rather troublesome. I suggest that we have a round robin first, and then decide the champion. The first place entered the finals. The second is the third. & lt/p & gt；
& ltp> If there are three teams, there may be one win and one loss. It is suggested to judge according to the net head count, which will also increase the appreciation of the game. & lt/p & gt；
& ltp> The loser group consists of 8 teams, 8 into 4, 4 into 2. The winner enters the final and the loser competes for the third place. & lt/p & gt；
& ltp> option 2
& ltp> The winner group will proceed as above, and the loser group will proceed as usual. & lt/p & gt；
& ltp> In the first round of loser group, five teams, one by one, decided three teams. & lt/p & gt；
In the second round of<p>, we selected three teams from the three winners to play against and decided on three teams. & lt/p & gt；
& ltp> These three teams are also cyclical. Decide the second group. Refer to the above. & lt/p & gt；
& ltp> I sent a schematic diagram, and I wonder if you can understand it. & lt/p & gt；
& ltp> The winner group is the same, but the loser group has changed. & lt/p & gt；
& ltp & gt& lt/p & gt；