6 人解决，32 人已尝试。
11 份提交通过，共有 101 份提交。
8.7 EMB 奖励。
单点时限: 2.0 sec
内存限制: 256 MB
In this problem, you have to find the last three digits before the decimal point for the number (3 + √5)^n.
For example:
when n = 5, (3 + √5)^5 = 3935.73982… The answer is 935.
For n = 2, (3 + √5)^2 = 27.4164079… The answer is 027.
The first line of input gives the number of cases, T. T test cases follow, each on a separate line. Each test case contains one positive integer n.
1 <= T <= 10
2 <= n <= 2000000000
For each input case, you should output:
Case #X: Y
where X is the number of the test case and Y is the last three integer digits of the number (3 + √5)^n. In case that number has fewer than three integer digits, add leading zeros so that your output contains exactly three digits.
2 5 2
Case #1: 935 Case #2: 027
6 人解决，32 人已尝试。
11 份提交通过，共有 101 份提交。
8.7 EMB 奖励。