lightoj 1122 - Digit Count(dp)

Time Limit: 2 second(s)

Memory Limit: 32 MB

Given a set of digits S, and an integer n, you have to find how many n-digit integers are there, which contain digits that belong to S and the difference between any two adjacent digits is not more than two.

Input

Input starts with an integer T (≤ 300), denoting the number of test cases.

Each case contains two integers, m (1 ≤ m < 10) and n (1 ≤ n ≤ 10). The next line will contain m integers (from 1 to 9) separated by spaces. These integers form the set S as described above. These integers will be distinct and given in ascending order.

Output

For each case, print the case number and the number of valid n-digit integers in a single line.

Sample Input	Output for Sample Input
3 3 2 1 3 6 3 2 1 2 3 3 3 1 4 6	Case 1: 5 Case 2: 9 Case 3: 9

Note

For the first case the valid integers are

11

13

31

33

66

题意：给你一些数字1~9不重复，这些数可以无限使用，问你可以构成多少种给n位数，并且满足相邻的两位数差值不超过2。

思路：dp[i][j]表示第i位为j的方案数，最后累加dp[n][j]就可以的了，这个dp比较简单。