问题描述 |
---|
This is a very easy problem. ACMeow
loves GTX1920. Now he has m RMB, but no GTX1920s. In the next n days, the unit
price of GTX1920 in the ith day is Ci RMB. In other words, in the
ith day, he can buy one GTX1920 with Ci RMB, or sell one GTX1920 to
gain Ci RMB. He can buy or sell as many times as he wants in one
day, but make sure that he has enough money for buying or enough GTX1920 for
selling. Now
he wants to know, how many RMB can he get after the n days. Could you please
help him?
It’s
really easy, yeah? |
输入描述 |
First line contains an
integer T(1 ≤ T ≤ 20), represents there are T test cases.
For each test case:
first line contains two integers n(1 ≤ n ≤ 2000) and m(0 ≤ m ≤ 1000000000).
Following n integers in one line, the ith integer represents Ci(1 ≤
Ci ≤ 1000000000). |
输出描述 |
For each test case, output "Case #X:
Y" in a line (without quotes), where X is the case number starting from 1,
and Y is the maximum number of RMB he can get mod 1000000007. |
样例输入复制样例 |
2 3 1 1 2 3 4 1
1 2 1 2 |
样例输出 |
Case
#1: 3
Case
#2: 4 |
来源 |
2017年第八届福建省大学生程序设计竞赛正式赛J |