Devu and his Class

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Read problems statements in Mandarin Chinese and Russian.
Devu is a class teacher of a class of n students. One day, in the morning prayer of the school, all the students of his class were standing in a line. You are given information of their arrangement by a string s. The string s consists of only letters 'B' and 'G', where 'B' represents a boy and 'G' represents a girl.
Devu wants intergender interaction among his class should to be maximum. So he does not like seeing two or more boys/girls standing nearby (i.e. continuous) in the line. e.g. he does not like the arrangements BBG and GBB, but he likes BG, GBG etc.
Now by seeing the initial arrangement s of students, Devu may get furious and now he wants to change this arrangement into a likable arrangement. For achieving that, he can swap positions of any two students (not necessary continuous). Let the cost of swapping people from position i with position j (i ≠ j) be c(i, j). You are provided an integer variable type, then the cost of the the swap will be defined by c(i, j) = j − i^{type}.
Please help Devu in finding minimum cost of swaps needed to convert the current arrangement into a likable one.
Input
The first line of input contains an integer T, denoting the number of test cases. Then T test cases are follow.
The first line of each test case contains an integer type, denoting the type of the cost function. Then the next line contains string s of length n, denoting the initial arrangement s of students.
Note that the integer n is not given explicitly in input.
Output
For each test case, print a single line containing the answer of the test case, that is, the minimum cost to convert the current arrangement into a likable one. If it is not possible to convert the current arrangement into a likable one, then print 1 instead of the minimum cost.
Constraints and Subtasks
Subtask 1: 25 points
 1 ≤ T ≤ 10^{5}
 1 ≤ n ≤ 10^{5}
 type = 0
 Sum of n over all the test cases in one test file does not exceed 10^{6}.
Subtask 2: 25 points
 1 ≤ T ≤ 10^{5}
 1 ≤ n ≤ 10^{5}
 type = 1
 Sum of n over all the test cases in one test file does not exceed 10^{6}.
Subtask 3: 25 points
 1 ≤ T ≤ 10^{5}
 1 ≤ n ≤ 10^{5}
 type = 2
 Sum of n over all the test cases in one test file does not exceed 10^{6}.
Subtask 4: 25 points
 1 ≤ T ≤ 10^{2}
 1 ≤ n ≤ 10^{3}
 type can be 0, 1 or 2, that is type ∈ {0, 1, 2}.
Example
Input: 8 0 BB 0 BG 0 BBGG 1 BGG 1 BGGB 1 BBBGG 2 BBGG 2 BGB Output: 1 0 1 1 1 3 1 0
Explanation
Note type of the first 3 test cases is 0. So c(i, j) = 1. Hence we just have to count minimum number of swaps needed.
Example case 1. There is no way to make sure that both the boys does not stand nearby. So answer is 1.
Example case 2. Arrangement is already valid. No swap is needed. So answer is 0.
Example case 3. Swap boy at position 1 with girl at position 2. After swap the arrangement will be BGBG which is a valid arrangement. So answer is 1.
Now type of the next 3 test cases is 1. So c(i, j) = j − i, that is, the absolute value of the difference between i and j.
Example case 4. Swap boy at position 0 with girl at position 1. After swap the arrangement will be GBG which is a valid arrangement. So answer is 1  0 = 1.
Example case 5. Swap boy at position 0 with girl at position 1. After swap the arrangement will be GBGB which is a valid arrangement. So answer is 1  0 = 1.
Example case 6. Swap boy at position 1 with girl at position 4. After swap the arrangement will be BGBGB which is a valid arrangement. So answer is 4  1 = 3.
Then type of the last 2 test cases is 2. So c(i, j) = (j − i)^{2}
Example case 7. Swap boy at position 1 with girl at position 2. After swap the arrangement will be BGBG which is a valid arrangement. So answer is (2  1)^{2} = 1.
Example case 8. Arrangement is already valid. No swap is needed. So answer is 0.
Author:  admin2 
Tester:  laycurse 
Editorial  http://discuss.codechef.com/problems/DEVCLASS 
Tags  adhoc, admin2, easy, greedy, march15 
Date Added:  23122014 
Time Limit:  2 sec 
Source Limit:  50000 Bytes 
Languages:  C, CPP14, JAVA, PYTH, PYTH 3.6, CS2, PAS fpc, PAS gpc, RUBY, PHP, GO, NODEJS, HASK, SCALA, D, PERL, FORT, WSPC, ADA, CAML, ICK, BF, ASM, CLPS, PRLG, ICON, SCM qobi, PIKE, ST, NICE, LUA, BASH, NEM, LISP sbcl, LISP clisp, SCM guile, JS, ERL, TCL, PERL6, TEXT, PYP3, CLOJ, FS 
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