Note: The main parts of the statements of the problems "Reversort" and "Reversort Engineering" are identical, except for the last paragraph. The problems can otherwise be solved independently.
Reversort is an algorithm to sort a list of distinct integers in increasing order. The algorithm is based on the "Reverse" operation. Each application of this operation reverses the order of some contiguous part of the list.
The pseudocode of the algorithm is the following:
Reversort(L):
for i := 1 to length(L) - 1
j := position with the minimum value in L between i and length(L), inclusive
Reverse(L[i..j])
After iterations, the positions
of the list
contain the
smallest elements of
, in increasing order.
During the
iteration, the process reverses the sublist going
from the
position to the current position of the
minimum
element. That makes the
minimum element end up in the
position.
For example, for a list with 4 elements, the algorithm would perform 3 iterations.
Here is how it would process :
,
,
,
The most expensive part of executing the algorithm on our architecture is the Reverse operation.
Therefore, our measure for the cost of each iteration is simply the length of the sublist passed
to Reverse, that is, the value . The cost of the whole algorithm is the sum of the
costs of each iteration.
In the example above, the iterations cost 3, 1, and 2, in that order, for a total of 6.
Given the initial list, compute the cost of executing Reversort on it.
Input Specification
The first line of the input gives the number of test cases, .
test cases follow.
Each test case consists of 2 lines. The first line contains a single integer
, representing
the number of elements in the input list. The second line contains
distinct integers
, representing the
elements of the input list
, in order.
Output Specification
For each test case, output one line containing Case #x: y
, where is the test case number (starting from 1) and
is the total cost of executing Reversort on the list given as input.
Limits
Time limit: 10 seconds.
Memory limit: 1 GB.
Test Set 1
.
.
, for all
.
, for all
.
Sample Input
3
4
4 2 1 3
2
1 2
7
7 6 5 4 3 2 1
Sample Output
Case #1: 6
Case #2: 1
Case #3: 12
Sample Case #1 is described in the statment above.
In Sample Case #2, there is a single iteration, in which Reverse is applied to a sublist of size 1. Therefore, the total cost is 1.
In Sample Case #3, the first iteration reverses the full list, for a cost of 7. After that, the list is already sorted, but there are 5 more iterations, each of which contributes a cost of 1.
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