Codeforces Round #442 (Div. 2) B - Nikita and string

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本文介绍了一个编程挑战，目标是在不改变字符顺序的情况下，通过删除部分字符来形成最长的ABA形式的字符串（A代表若干个a，B代表若干个b）。文章提供了一种解决方案，包括计算a和b的前缀和，然后遍历所有可能的子串，找到满足条件的最长子串。

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.

题意给出一个只包含a，b 的字符串，可以删除任意数量的字符（不可调换位置），使得ABA的形式的字符串最长（A代表若干个a，B代表若干个b）

求出a，b的前缀和，遍历一遍i，j之间b的个数，找出最大长度

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#include <set>
using namespace std;

int main()
{
	char s[5005];
	int a[5005] = {0}, b[5005] = {0};
	scanf("%s", s + 1);
	int l = strlen(s + 1), A = 0, B = 0, sum = 0;
	for (int i = 1; i <= l; i ++) {
		b[i] = s[i] == 'b' ? b[i - 1] + 1 : b[i - 1];
		a[i] = s[i] == 'a' ? a[i - 1] + 1 : a[i - 1];
	}
	for (int i = 1; i <= l; i ++) {
		for (int j = i - 1; j <= l; j ++) {
			if (a[i - 1] + b[j] - b[i - 1] + a[l] - a[j] > sum) {
				sum = a[i - 1] + b[j] - b[i - 1] + a[l] - a[j];
			}
		}
	}
	printf("%d\n", sum);
}