Codeforces Round #354 (Div. 2) C Vasya and String

Codeforces Round #354 (Div. 2) C Vasya and String

time limit per test
1 second

memory limit per test
256 megabytes

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 a substring (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 and k (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 than k 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".

题目在这：点击打开链接

ac渣渣没用什么算法

思路：

当k大于 “a”的个数，并大于“b”的个数时，为n
当k=0时 为‘a’,'b'个数中教大的；
其他情况：统计连续‘a’，连续“b”的个数存在两个数组中（连续为a时，连续b的个数为0，其他情况类似）；分别求出两个数组中连续k+1个数相加的最大值；其中较大的一个即答案

代码：

#include<stdio.h>
#include<string.h>

char str[100005];
int num[100005];
int num2[100005];
int main()
{
int n,k;
while(~scanf("%d %d",&n,&k))
{
getchar();
int cont=0;
int cont2=0;
int maxa=0;
int maxb=0;
int l=0;
int cont4=0;
int g=0;
for(int i=0;i<n;i++)
{
scanf("%c",&str[i]);
if(str[i]=='a')
cont++;
if(str[i]=='a')
cont2++;
else if(str[i]=='b')
{
if(maxa<cont2)
maxa=cont2;
num[l++]=cont2;
cont2=0;
}
if(str[i]=='b')
cont4++;
else if(str[i]=='a')
{
if(maxb<cont4)
maxb=cont4;
num2[g++]=cont4;
cont4=0;
}
}
if(maxa<cont2)
maxa=cont2;
num[l++]=cont2;
if(maxb<cont4)
maxb=cont4;
num2[g++]=cont4;
if(k>=cont||k>=(strlen(str)-cont))
{
printf("%d\n",n);
}
else
{
if(k==0)
{
printf("%d\n",(maxa>maxb? maxa:maxb));
}
else
{
int max1=0;
int max2=0;
for(int i=0;i<l-k;i++)
{
int cont3=num[i];
for(int j=1;j<=k;j++)
{
cont3+=num[i+j];
cont3++;
}
if(max1<cont3)
max1=cont3;
}

for(int i=0;i<g-k;i++)
{
int cont3=num2[i];
for(int j=1;j<=k;j++)
{
cont3+=num2[i+j];
cont3++;
}
if(max2<cont3)
max2=cont3;
}
printf("%d\n",max1>max2? max1:max2);
}
}
}
return 0;
}