codeforces877B（div2）Nikita and string 暴力枚举+前缀和

题目：

One day Nikita found the string containing letters “a” and “b” only.

Nikita thinks that string is beautiful if it can be cut into 3 strings (possibly empty) without changing the order of the letters, where the 1-st and the 3-rd one contain only letters “a” and the 2-nd contains only letters “b”.

Nikita wants to make the string beautiful by removing some (possibly none) of its characters, but without changing their order. What is the maximum length of the string he can get?

Input

The first line contains a non-empty string of length not greater than 5 000 containing only lowercase English letters “a” and “b”.

Output

Print a single integer — the maximum possible size of beautiful string Nikita can get.

Examples

input

abba

output

4

input

bab

output

2

Note

It the first sample the string is already beautiful.

In the second sample he needs to delete one of “b” to make it beautiful.

策略：暴力枚举第二个区间的边界，并事先处理好这个区间内的b的个数，这个区间左边a的个数，这个区间右边a的个数，那么枚举出来的区间内的b的个数+区间左边a的个数+区间右边a的个数的最大值就是答案。当时写的很挫。

#include<bits/stdc++.h>
using namespace std;
int a[5010],frl[5010],lv[5010],n=0,luc0[5010],luc1[5010],luc;
int main()
{
string s;
cin>>s;
char pre;
int cnt;
pre=s[0];
cnt=1;
for(int i=1;i<s.length();i++)
{
if(s[i]!=pre)
{
a
=cnt;
frl
=s[i-1]-97;
n++;
cnt=0;
}
cnt++;
pre=s[i];
}
if(cnt)
{
a
=cnt;
frl
=s[s.length()-1]-97;
n++;
}
luc=0;
for(int i=0;i<n;i++)
{
if(frl[i]==0)
{
luc+=a[i];
}
else
luc0[i]=luc;
}
luc=0;
for(int i=n-1;i>=0;i--)
{
if(frl[i]==0)
{
luc+=a[i];
}
else
luc1[i]=luc;
}
if(n==1)
{
cout<<a[0]<<endl;
return 0;
}
int maxx=0;
for(int i=0;i<n;i++)
{
if(frl[i]==0)
continue;
int sumb=0;
for(int j=i;j<n;j++)
{
if(frl[j]==0)
continue;
sumb+=a[j];
maxx=max(luc0[i]+luc1[j]+sumb,maxx);
}
}
cout<<maxx<<endl;
}
/*
abaaaaababbbbababb 14
*/