time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a strip of paper s that is n cells long. Each cell is either black or white. In an operation you can take any k consecutive cells and make them all white.
Find the minimum number of operations needed to remove all black cells.
Input
The first line contains a single integer t (1≤t≤1000) — the number of test cases.
The first line of each test case contains two integers n and k (1≤k≤n≤2⋅105) — the length of the paper and the integer used in the operation.
The second line of each test case contains a string s of length n consisting of characters B (representing a black cell) or W (representing a white cell).
The sum of n over all test cases does not exceed 2⋅105.
Output
For each test case, output a single integer — the minimum number of operations needed to remove all black cells.
Example
input
Copy
8
6 3
WBWWWB
7 3
WWBWBWW
5 4
BWBWB
5 5
BBBBB
8 2
BWBWBBBB
10 2
WBBWBBWBBW
4 1
BBBB
3 2
WWW
output
Copy
2 1 2 1 4 3 4 0
Note
In the first test case you can perform the following operations:WBWWWB→WWWWWB→WWWWWW
In the second test case you can perform the following operations:WWBWBWW→WWWWWWW
In the third test case you can perform the following operations:BWBWB→BWWWW→WWWWW文章来源:https://www.toymoban.com/news/detail-719985.html
解题说明:此题是一道模拟题,每次只能修改连续K个位置,遍历查找其中的B字母位置,直接将其与后面K-1个位置全变成W即可。文章来源地址https://www.toymoban.com/news/detail-719985.html
#include <stdio.h>
int main()
{
int t, n, k, i, j, b;
scanf("%d", &t);
for (i = 0; i < t; i++)
{
scanf("%d %d", &n, &k);
char a[200005];
scanf("%s", a);
b = 0;
for (j = 0; j < n; j++)
{
if (a[j] == 'W')
{
continue;
}
else
{
j += (k - 1);
b += 1;
}
}
printf("%d\n", b);
}
return 0;
}
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