Official Analysis (C++ and Java)

Explanation

If the cow in position $i$ moves to position $a_i$ after one shuffle, then after that shuffle the cow sitting at $a_i$ must have come from $i$.

To undo a single shuffle, for each current position $j$ we find the unique $i$ satisfying $a_i = j$ and move the cow at $j$ back to $i$. Repeating this step three times recovers the original ordering before any shuffles.

Implementation

Time Complexity: $\mathcal{O}(N)$

Copy
SHUFFLE_NUM = 3

with open("shuffle.in") as read:
	n = int(read.readline())
	shuffle = list(map(int, read.readline().split()))
	ids = list(map(int, read.readline().split()))

for _ in range(SHUFFLE_NUM):
	past_order = [0] * n
	for i in range(n):