Codeforces 877(442 Div.2) B. Nikita and string

B. Nikita and string

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

题意：

一个只有'a','b'组成的字符串可以分割为3个字符串且不能打乱之前的顺序，要保证第1个和第3个字符串全为'a'，第二个全为'b'(这3个字符串可为空，如"a"，"b"，""是满足条件的)。现在你可删除若干个字符，在满足条件下尽可能使得字符串长度最大，输出它的长度。

思路：

我们记dp[i][0]表示最长aaa......形式的子序列，dp[i][1]表示最长aaa......bbb......形式的子序列，dp[i][2]表示最长aaa......bbb......aaa......形式的子序列，最后三种情况取最大即可。

那么状态转移方程为：

if(s[i]=='a'){    dp[i][0]=dp[i-1][0]+1;    dp[i][1]=max(dp[i-1][1],dp[i][0]);    dp[i][2]=max(dp[i-1][2]+1,dp[i-1][1]+1);}else{    dp[i][0]=dp[i-1][0];    dp[i][1]=max(dp[i-1][1]+1,dp[i][0]);    dp[i][2]=dp[i-1][2];}

示例程序：

#include <bits/stdc++.h>#define LDQ 1000000007#define QAQ 0x3f3f3f3fusing namespace std;int dp[5001][3];int main(){    int len,i;    char s[5001];    scanf("%s",s);    len=strlen(s);    for(i=1;len>=i;i++)    {        if(s[i-1]=='a')        {            dp[i][0]=dp[i-1][0]+1;            dp[i][1]=max(dp[i-1][1],dp[i][0]);            dp[i][2]=max(dp[i-1][2]+1,dp[i-1][1]+1);        }        else        {            dp[i][0]=dp[i-1][0];            dp[i][1]=max(dp[i-1][1]+1,dp[i][0]);            dp[i][2]=dp[i-1][2];        }    }    printf("%d",max(dp[len][0],max(dp[len][1],dp[len][2])));    return 0;}