permutation-equation

Sequence Equation ⬀

Given a sequence of n integers, p(1), p(2) where each element is distinct and satisfies 1 ≤ p(x) ≤ n. For each x where 1 ≤ x ≤ n, that is x increments from 1 to n, find any integer y such that p(p(y)) ≡ x and keep a history of the values of y in a return array.

Example

p = [5, 2, 1, 3, 4]

Each value of x between 1 and 5, the length of the sequence, is analyzed as follows:

1. x = 1 ≡ p[3], p[4] = 3, so p[p[4]] = 1
2. x = 2 ≡ p[2], p[2] = 3, so p[p[2]] = 2
3. x = 3 ≡ p[3], p[5] = 3, so p[p[5]] = 3
4. x = 4 ≡ p[5], p[1] = 3, so p[p[1]] = 4
5. x = 5 ≡ p[1], p[3] = 3, so p[p[3]] = 5

The values for y are [4, 2, 5, 1, 3].

Function Description

Complete the permutationEquation function in the editor below.

permutationEquation has the following parameter(s):

• int p[n]: an array of integers

Returns

• int[n]: the values of y for all x in the arithmetic sequence 1 to n

Input Format

The first line contains an integer n, the number of elements in the sequence.

The second line contains n space-separated integers p[i] where 1 ≤ i ≤ n.

Constraints

• 1 ≤ n ≤ 50
• 1 ≤ p[i] ≤ 50, where 1 ≤ i ≤ n.
• Each element in the sequence is distinct.

Sample Input 0

3
2 3 1

Sample Output 0

2
3
1

Explanation 0

Given the values of p(1) = 2, p(2) = 3, and p(3) = 1, we calculate and print the following values for each x from 1 to n:

1. x = 1 ≡ p(3) = p(p(2)) = p(p(y)), so we print the value of y = 2 on a new line.
2. x = 2 ≡ p(1) = p(p(3)) = p(p(y)), so we print the value of y = 3 on a new line.
3. x = 3 ≡ p(2) = p(p(1)) = p(p(y)), so we print the value of y = 1 on a new line.

Sample Input 1

5
4 3 5 1 2

Sample Output 1

1
3
5
4
2