354 div2 C：两指针滑动区间

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                                                                                                        C. Vasya and String

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotesbeauty
of the string as the maximum length of asubstring (consecutive subsequence) consisting of equal letters.

Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve?

Input
The first line of the input contains two integers n andk (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n) — the length of the string and the
maximum number of characters to change.

The second line contains the string, consisting of letters 'a' and 'b' only.

Output
Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more thank characters.

Examples

Input

4 2
abba

Output

4

Input

8 1
aabaabaa

Output

5

Note
In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".

In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".

               

题意：一串只有ab的字符串，数量n可改变其中K个字符，求改变后的最长相同子串。

思路：最长相同子串要不是a，要不是b，分别扫一遍。两个指针lr，r一直向右延伸直到不能延伸（改变次数k用完），记录最大值然后向右移动l，使l越过一个需要改变的字符（腾出位置才能改变r下一个字符）。ab两趟取最大。

总结：求最长子串，因此改变的字符（区间）肯定是连续的，所以想到滑动区间（尺取法？？），需要注意的是l指针舍去的时候（跳过需要改变的字符）

           还要注意的是学会字符串的读取...！！！！！

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
char a[100000 + 5];
int main()
{
int n,k,l = 1,suma = 0, sumb = 0;
scanf("%d%d",&n,&k);
scanf("%s",a + 1);
for(int i = 0,add = 0, num = 0; i <= n; i++)
{
if(a[i] == 'b')
{
if(add == k)
{
i--;
while(a[l] != 'b')
l++;
l++;
add--;
}
else
add++;
}
if(suma < i - l + 1)
suma = i - l + 1;
}
for(int i = 0,add = 0, num = 0, l = 1; i <= n; i++)
{
if(a[i] == 'a')
{
if(add == k)
{
i--;
while(a[l] != 'a')
l++;
l++;
add--;
}
else
add++;
}
if(sumb < i - l + 1)
sumb = i - l + 1;
}

printf("%d",max(suma,sumb));
return 0;
}