8 人解决，18 人已尝试。
12 份提交通过，共有 61 份提交。
7.4 EMB 奖励。
单点时限: 2.0 sec
内存限制: 256 MB
Bruce Force has had an interesting idea how to encode strings. The following is the description of how the encoding is done:
Let x1,x2,…,xn be the sequence of characters of the string to be encoded.
Choose an integer m and n pairwise distinct numbers p1,p2,…,pn from the set {1, 2, …, n} (a permutation of the numbers 1 to n).
Repeat the following step m times.
For 1 ≤ i ≤ n set yi to xpi, and then for 1 ≤ i ≤ n replace xi by yi.
For example, when we want to encode the string “hello”, and we choose the value m = 3 and the permutation 2, 3, 1, 5, 4, the data would be encoded in 3 steps: “hello” -> “elhol” -> “lhelo” -> “helol”.
Bruce gives you the encoded strings, and the numbers m and p1, …, pn used to encode these strings. He claims that because he used huge numbers m for encoding, you will need a lot of time to decode the strings. Can you disprove this claim by quickly decoding the strings?
The input contains several test cases. Each test case starts with a line containing two numbers n and m (1 ≤ n ≤ 80, 1 ≤ m ≤ 10^9). The following line consists of n pairwise different numbers p1,…,pn (1 ≤ pi ≤ n). The third line of each test case consists of exactly n characters, and represent the encoded string. The last test case is followed by a line containing two zeros.
For each test case, print one line with the decoded string.
5 3 2 3 1 5 4 helol 16 804289384 13 10 2 7 8 1 16 12 15 6 5 14 3 4 11 9 scssoet tcaede n 8 12 5 3 4 2 1 8 6 7 encoded? 0 0
hello second test case encoded?
8 人解决，18 人已尝试。
12 份提交通过，共有 61 份提交。
7.4 EMB 奖励。