New Year Permutations
Yeah, we failed to make up a New Year legend for this problem.A permutation of length $$$n$$$ is an array of $$$n$$$ integers such that every integer from $$$1$$$ to $$$n$$$ appears in it exactly once. An element $$$y$$$ of permutation $$$p$$$ is reachable from element $$$x$$$ if $$$x = y$$$, or $$$p_x = y$$$, or $$$p_{p_x} = y$$$, and so on. The decomposition of a permutation $$$p$$$ is defined as follows: firstly, we have a permutation $$$p$$$, all elements of which are not marked, and an empty list $$$l$$$. Then we do the following: while there is at least one not marked element in $$$p$$$, we find the leftmost such element, list all elements that are reachable from it in the order they appear in $$$p$$$, mark all of these elements, then cyclically shift the list of those elements so that the maximum appears at the first position, and add this list as an element of $$$l$$$. After all elements are marked, $$$l$$$ is the result of this decomposition.For example, if we want to build a decomposition of $$$p = [5, 4, 2, 3, 1, 7, 8, 6]$$$, we do the following:  initially $$$p = [5, 4, 2, 3, 1, 7, 8, 6]$$$ (bold elements are marked), $$$l = []$$$;  the leftmost unmarked element is $$$5$$$; $$$5$$$ and $$$1$$$ are reachable from it, so the list we want to shift is $$$[5, 1]$$$; there is no need to shift it, since maximum is already the first element;  $$$p = [\textbf{5}, 4, 2, 3, \textbf{1}, 7, 8, 6]$$$, $$$l = [[5, 1]]$$$;  the leftmost unmarked element is $$$4$$$, the list of reachable elements is $$$[4, 2, 3]$$$; the maximum is already the first element, so there's no need to shift it;  $$$p = [\textbf{5}, \textbf{4}, \textbf{2}, \textbf{3}, \textbf{1}, 7, 8, 6]$$$, $$$l = [[5, 1], [4, 2, 3]]$$$;  the leftmost unmarked element is $$$7$$$, the list of reachable elements is $$$[7, 8, 6]$$$; we have to shift it, so it becomes $$$[8, 6, 7]$$$;  $$$p = [\textbf{5}, \textbf{4}, \textbf{2}, \textbf{3}, \textbf{1}, \textbf{7}, \textbf{8}, \textbf{6}]$$$, $$$l = [[5, 1], [4, 2, 3], [8, 6, 7]]$$$;  all elements are marked, so $$$[[5, 1], [4, 2, 3], [8, 6, 7]]$$$ is the result. The New Year transformation of a permutation is defined as follows: we build the decomposition of this permutation; then we sort all lists in decomposition in ascending order of the first elements (we don't swap the elements in these lists, only the lists themselves); then we concatenate the lists into one list which becomes a new permutation. For example, the New Year transformation of $$$p = [5, 4, 2, 3, 1, 7, 8, 6]$$$ is built as follows:  the decomposition is $$$[[5, 1], [4, 2, 3], [8, 6, 7]]$$$;  after sorting the decomposition, it becomes $$$[[4, 2, 3], [5, 1], [8, 6, 7]]$$$;  $$$[4, 2, 3, 5, 1, 8, 6, 7]$$$ is the result of the transformation. We call a permutation good if the result of its transformation is the same as the permutation itself. For example, $$$[4, 3, 1, 2, 8, 5, 6, 7]$$$ is a good permutation; and $$$[5, 4, 2, 3, 1, 7, 8, 6]$$$ is bad, since the result of transformation is $$$[4, 2, 3, 5, 1, 8, 6, 7]$$$.Your task is the following: given $$$n$$$ and $$$k$$$, find the $$$k$$$-th (lexicographically) good permutation of length $$$n$$$.
The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.Then the test cases follow. Each test case is represented by one line containing two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 50$$$, $$$1 \le k \le 10^{18}$$$).
For each test case, print the answer to it as follows: if the number of good permutations of length $$$n$$$ is less than $$$k$$$, print one integer $$$-1$$$; otherwise, print the $$$k$$$-th good permutation on $$$n$$$ elements (in lexicographical order).
from math import factorial as fact
N = 55
c = [1]
for i in range(N):
    c.append(fact(i))
dp = [0] * N
dp[0] = 1
for i in range(1, N):
    for j in range(i):
        dp[i] += dp[j] * c[i - j - 1]

def get_kth_cycle(n, k):
    if n == 1:
        return [1]
    ans = [-1] * n
    ans[0] = n - 1
    fin = [i for i in range(n)]
    fin[0] = n - 1
    init = [i for i in range(n)]
    init[n - 1] = 0
    used = [False] * n
    used[n - 1] = True
    for i in range(1, n - 1):
        j = 0
        cur = fact(n - i - 2)
        while True:
            while used[j] or (i < n - 1 and j == init[i]):
                j += 1
            if k > cur:
                k -= cur
                j += 1
            else:
                fin[init[i]] = fin[j]
                init[fin[j]] = init[i]
                ans[i] = j
                used[j] = True
                break
    ans[-1] = init[-1]
    return [x + 1 for x in ans]

def f(n, k):
    if n == 0:
        assert k == 1
        return []
    cl = 1
    while c[cl - 1] * dp[n - cl] < k:
        k -= c[cl - 1] * dp[n - cl]
        cl += 1
    rest = f(n - cl, (k - 1) % dp[n - cl] + 1)
    rest = [x + cl for x in rest]
    k = (k - 1) // dp[n - cl] + 1
    return get_kth_cycle(cl, k) + rest

def solve():
    (n, k) = map(int, input().split())
    if k > dp[n]:
        print(-1)
        return
    print(*f(n, k))

def main():
    t = int(input())
    while t > 0:
        t -= 1
        solve()
main()
from math import factorial as fact
N = 55
c = [1]
for i in range(N):
    c.append(fact(i))
dp = [0] * N
dp[0] = 1
for i in range(1, N):
    for j in range(i):
        dp[i] += dp[j] * c[i - j - 1]

def get_kth_cycle(n, k):
    if n == 1:
        return [1]
    ans = [-1] * n
    ans[0] = n - 1
    fin = [i for i in range(n)]
    fin[0] = n - 1
    init = [i for i in range(n)]
    init[n - 1] = 0
    used = [False] * n
    used[n - 1] = True
    for i in range(1, n - 1):
        j = 0
        cur = fact(n - i - 2)
        while True:
            while used[j] or (i < n - 1 and j == init[i]):
                j += 1
            if k > cur:
                k -= cur
                j += 1
            else:
                fin[init[i]] = fin[j]
                init[fin[j]] = init[i]
                ans[i] = j
                used[j] = True
                break
    ans[-1] = init[-1]
    return [x + 1 for x in ans]

def f(n, k):
    if n == 0:
        assert k == 1
        return []
    cl = 1
    while c[cl - 1] * dp[n - cl] < k:
        k -= c[cl - 1] * dp[n - cl]
        cl += 1
    rest = f(n - cl, (k - 1) % dp[n - cl] + 1)
    rest = [x + cl for x in rest]
    k = (k - 1) // dp[n - cl] + 1
    return get_kth_cycle(cl, k) + rest

def solve():
    (n, k) = map(int, input().split())
    if k > dp[n]:
        print(-1)
        return
    print(*f(n, k))

def main():
    t = int(input())
    while t > 0:
        t -= 1
        solve()
main()
from math import factorial as fact
N = 55
c = [1]
for i in range(N):
    c.append(fact(i))
dp = [0] * N
dp[0] = 1
for i in range(1, N):
    for j in range(i):
        dp[i] += dp[j] * c[i - j - 1]

def get_kth_cycle(n, k):
    if n == 1:
        return [1]
    ans = [-1] * n
    ans[0] = n - 1
    fin = [i for i in range(n)]
    fin[0] = n - 1
    init = [i for i in range(n)]
    init[n - 1] = 0
    used = [False] * n
    used[n - 1] = True
    for i in range(1, n - 1):
        j = 0
        cur = fact(n - i - 2)
        while True:
            while used[j] or (i < n - 1 and j == init[i]):
                j += 1
            if k > cur:
                k -= cur
                j += 1
            else:
                fin[init[i]] = fin[j]
                init[fin[j]] = init[i]
                ans[i] = j
                used[j] = True
                break
    ans[-1] = init[-1]
    return [x + 1 for x in ans]

def f(n, k):
    if n == 0:
        assert k == 1
        return []
    cl = 1
    while c[cl - 1] * dp[n - cl] < k:
        k -= c[cl - 1] * dp[n - cl]
        cl += 1
    rest = f(n - cl, (k - 1) % dp[n - cl] + 1)
    rest = [x + cl for x in rest]
    k = (k - 1) // dp[n - cl] + 1
    return get_kth_cycle(cl, k) + rest

def solve():
    (n, k) = map(int, input().split())
    if k > dp[n]:
        print(-1)
        return
    print(*f(n, k))

def main():
    t = int(input())
    while t > 0:
        t -= 1
        solve()
main()