2018年省赛热身赛第7场 L - One-Dimensional Maze

L - One-Dimensional Maze

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

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Description

BaoBao is trapped in a one-dimensional maze consisting of \(n\) grids arranged in a row! The grids are numbered from 1 to \(n\) from left to right, and the \(i\)-th grid is marked with a character \(s_i\), where \(s_i\) is either 'L' or 'R'.

Starting from the \(m\)-th grid, BaoBao will repeatedly take the following steps until he escapes the maze:

• If BaoBao is in the 1st grid or the \(n\)-th grid, then BaoBao is considered to arrive at the exit and thus can escape successfully.
• Otherwise, let BaoBao be in the \(t\)-th grid. If \(s_t = \text{'L'}\), BaoBao will move to the \((t-1)\)-th grid; If \(s_t = \text{'R'}\), Baobao will move to the \((t+1)\)-th grid.

Before taking the above steps, BaoBao can change the characters in some grids to help himself escape. Concretely speaking, for the \(i\)-th grid, BaoBao can change \(s_i\) from 'L' to 'R', or from 'R' to 'L'.

But changing characters in grids is a tiring job. Your task is to help BaoBao calculate the minimum number of grids he has to change to escape the maze.

Input

There are multiple test cases. The first line of the input contains an integer \(T\), indicating the number of test cases. For each test case:

The first line contains two integers \(n\) and \(m\) (\(3 \le n \le 10^5\), \(1 < m < n\)), indicating the number of grids in the maze, and the index of the starting grid.

The second line contains a string \(s\) (\(|s| = n\)) consisting of characters 'L' and 'R'. The \(i\)-th character of \(s\) indicates the character in the \(i\)-th grid.

It is guaranteed that the sum of \(n\) over all test cases will not exceed \(10^6\).

Output

For each test case output one line containing one integer, indicating the minimum number of grids BaoBao has to change to escape the maze.

Sample Input

3
3 2
LRL
10 4
RRRRRRRLLR
7 4
RLLRLLR

Sample Output

0
2
1

Hint

For the first sample test case, BaoBao doesn't have to change any character and can escape from the 3rd grid. So the answer is 0.

For the second sample test case, BaoBao can change \(s_8\) to 'R' and \(s_9\) to 'R' and escape from the 10th grid. So the answer is 2.

For the third sample test case, BaoBao can change \(s_4\) to 'L' and escape from the 1st grid. So the answer is 1.

题意：就是给你n和m，起始点在第m，如果是'L'，就往左走一步，如果是'R'，就往右走一步，这题可以改左右，走到第一个或者最后一个就是走到出口，问从第m的位置走到出口最少要change几次。

题解：这题就是在m这个点，左边肯定是往左走，右边往右走，所以只要统计m的左边有多少个不是L，右边统计有多少个不是R，然后选出最小的就ok啦。

AC：

#include
using namespace std;
char s[100010];

int main(void)
{
	int T;scanf("%d",&T);
	while(T--)
	{
		int n,m;
		scanf("%d%d",&n,&m);getchar();
		scanf("%s",s+1);
		int l1=0,r1=0,l2=0,r2=0;
		for(int i=m;i>1;i--)
		{
			if(s[i]!='L') r1++;
		}
		for(int i=m;i