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| using namespace std; | |
| int N; | |
| deque<int> A, B; | |
| bool is_metalike_now() { | |
| for (int i = 0; i < N; i++) { | |
| if (i+1 < (N+1)/2 && !(A[i] < B[i])) { | |
| return false; | |
| } | |
| if (i+1 > (N+1)/2 && !(A[i] > B[i])) { | |
| return false; | |
| } | |
| if (A[i] != B[N - i - 1]) { | |
| return false; | |
| } | |
| } | |
| return true; | |
| } | |
| // Simulate 2*N seconds. | |
| // Given a vector M of O(1) candidate midpoints (index of the middle elem if N | |
| // is odd, or right after the middle if N even), return the first time that: | |
| // - some m in M becomes the midpoint floor(N/2) (both even and odd N), and | |
| // - is_meta_like() holds true. | |
| int check(vector<int> M) { | |
| for (int t = 0; t <= 2*N; t++) { | |
| for (int m : M) { | |
| if (m == N/2 && is_metalike_now()) { | |
| return t; | |
| } | |
| } | |
| A.push_back(B[0]); | |
| B.push_back(A[0]); | |
| A.pop_front(); | |
| B.pop_front(); | |
| for (int i = 0; i < (int)M.size(); i++) { | |
| if (--M[i] < 0) { | |
| M[i] += N; | |
| } | |
| } | |
| } | |
| return -1; | |
| } | |
| int solve() { | |
| cin >> N; | |
| A.resize(N); | |
| B.resize(N); | |
| for (int i = 0; i < N; i++) { | |
| cin >> A[i]; | |
| } | |
| for (int i = 0; i < N; i++) { | |
| cin >> B[i]; | |
| } | |
| // Handle any equal pairs. | |
| vector<int> eq; | |
| for (int i = 0; i < N; i++) { | |
| if (A[i] == B[i]) { | |
| if (N % 2 == 0) { | |
| return -1; | |
| } | |
| eq.push_back(i); | |
| } | |
| } | |
| if (eq.size() > 1) { | |
| return -1; | |
| } | |
| if (eq.size() == 1) { | |
| return check({eq[0]}); | |
| } | |
| // From here on, guaranteed that A[i] != B[i]. | |
| // Check the number of times that A[i] < B[i] flips. | |
| vector<int> flipped; | |
| for (int i = 1; i < N; i++) { | |
| if ((A[i - 1] < B[i - 1]) != (A[i] < B[i])) { | |
| flipped.push_back(i); | |
| } | |
| } | |
| if (flipped.size() > 1) { | |
| return -1; | |
| } | |
| // Might as well consider elements 0 and N-1 to be midpoints. | |
| flipped.push_back(0); | |
| flipped.push_back(N-1); | |
| return check(flipped); | |
| } | |
| int main() { | |
| int T; | |
| cin >> T; | |
| for (int t = 1; t <= T; t++) { | |
| cout << "Case #" << t << ": " << solve() << endl; | |
| } | |
| return 0; | |
| } |