====== n제곱 계산 ======

===== 풀이 =====
  * [[ps:problems:boj:12925]]와 동일한 문제. 풀이는 그쪽 참고

===== 코드 =====
<dkpr py>
"""Solution code for "BOJ 12925. Numbers".

- Problem link: https://www.acmicpc.net/problem/12925
- Solution link: http://www.teferi.net/ps/problems/boj/12925

Tags: [Math]
"""

from teflib import combinatorics

MOD = 1000
COEF = [6, -4]  # A(n)=6A(n-1)-4A(n-2)  ( A(n) = (3+sqrt(5))^n + (3-sqrt(5))^n )
SEED = [2, 6]  # A(0)=2, A(1)=6


def main():
    T = int(input())
    for i in range(T):
        n = int(input())

        answer = combinatorics.linear_homogeneous_recurrence(
            COEF, SEED, n, MOD) - 1
        print(f'Case #{i + 1}: {answer:>03}')


if __name__ == '__main__':
    main()
</dkpr>
  * Dependency: [[:ps:teflib:combinatorics#linear_homogeneous_recurrence|teflib.combinatorics.linear_homogeneous_recurrence]]

{{tag>BOJ ps:problems:boj:플래티넘_1}}