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For the second question, we only require the sequence of ants sorted by position number after t seconds to calculate the answer. Assuming that the ants are arranged in the order of the size of the initial position number at the beginning, position 0 (the middle of position 1 and position m) is a critical point, consider two cases to find the order of ants after t seconds:

• All ants did not collide within t seconds: When an ant walks from 1 to 0 to m, then this ant will become the last ant, and the second An ant will become the first one, and so on. Similarly, when an ant goes from 0 to 1 from m, this ant will become the first ant. Therefore, we only need to calculate how many times the ants have walked past 0 o’clock from right to left or left to right in t seconds, and the difference can get the order of the ants after t seconds.

• An ant collides within t seconds: If an ant collides, because the ant’s trajectory can be seen as passing through each other, we can still Ignoring the collision situation to calculate the number of times the ant passes through the 0 point from two directions, it is regarded as the above situation. Every time we pass 0 o’clock from right to left, the sequence of ants moves to the left once, and the same goes to the right.

For example, if the sequence of ant numbers is [1,3,2] at the beginning, in a certain second 2 passes from 0 to 1 from point m Point, the sequence becomes [2,1,3]. If 2 and 1 collide, 2 passes through 0 again in the reverse direction, and the arrangement order changes back to [1,3,2]. It seems that we need to judge that Ant 2 has passed 0 o’clock from different directions, but we can completely ignore the collision as Ant 2 passes 0 o’clock from the left and Ant 1 passes 0 o’clock from the right. It is also fully adapted to complex situations. .

At this time we know:

• There are ants in those positions after t seconds
• The ants are sorted by position after t seconds

We can directly map the positions of the ants one by one