BestCoder Round #59 (div.2)B.Reorder the Books

Reorder the Books

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 316 Accepted Submission(s): 221



Problem Description

dxy has a collection of a series of books called "The Stories of SDOI",There are n(n≤19) books
in this series.Every book has a number from 1 to n.

dxy puts these books in a book stack with the order of their numbers increasing from top to bottom. dxy takes great care of these books and no one is allowed to touch them.

One day Evensgn visited dxy's home, because dxy was dating with his girlfriend, dxy let Evensgn stay at home himself. Evensgn was curious about this series of books.So he took a look at them. He found out there was a story about "Little E&Little Q". While losing
himself in the story,he disrupted the order of the books.

Knowing that dxy would be back soon,Evensgn needed to get the books ordered again.But because the books were too heavy.The only thing Evensgn could do was to take out a book from the book stack and and put it at the stack top.

Give you the order of the disordered books.Could you calculate the minimum steps Evensgn would use to reorder the books? If you could solve the problem for him,he will give you a signed book "The Stories of SDOI 9: The Story of Little E" as a gift.



Input

There are several testcases.

There is an positive integer T(T≤30) in
the first line standing for the number of testcases.

For each testcase, there is an positive integer n in
the first line standing for the number of books in this series.

Followed n positive
integers separated by space standing for the order of the disordered books,the ith integer
stands for the ith book's
number(from top to bottom).

Hint:

For the first testcase:Moving in the order of book3,book2,book1 ,(4,1,2,3)→(3,4,1,2)→(2,3,4,1)→(1,2,3,4),and
this is the best way to reorder the books.

For the second testcase:It's already ordered so there is no operation needed.



Output

For each testcase,output one line for an integer standing for the minimum steps Evensgn would use to reorder the books.



Sample Input

2
4
4 1 2 3
5
1 2 3 4 5

Sample Output

3
0

Source

BestCoder Round #59 (div.1)


n<19肯定不能模拟，肯定有规律。
首先可以发现最大的数n是不用操作的（其他数操作好了，数"n"自然就在最后面了）。 于是我们先找到最大的数"n"的位置，从这个位置往前找，直到找到(n−1)。假如找到头也没找到(n-1)，那么数"(n-1)"需要操作，而一旦操作了(n-1)，根据前面结论，总共就需要(n−1)次操作了;假如找到了(n-1)，那么数"(n−1)"也不需要操作（和数"n"不需要操作一个道理）。 同理，我们接着从(n-1)的位置往前找(n−2),再从(n−2)的位置往前找(n−3)...假如数k找不到了，那么就至少需要k次操作。这种做法的复杂度是O(n)的