Nikita and string

最新推荐文章于 2025-02-22 22:00:00 发布

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博客围绕一个仅含字母a和b的字符串展开，探讨能否将其按规则分割为三部分，即第一和第三部分仅含a，第二部分仅含b，且各部分可为空。目标是通过移除部分字符（不改变顺序）得到最长的符合要求的字符串，并给出了输入输出示例及代码。

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

Input

bab

Output

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：

给出一个只有a,b且长度最长为5000的字符串。

能不能将这个字符串分成三部分，第一和第三部分只有a，第二部分由b组成，两部分皆可为空。

问你能不能分成这样的形式，如果能那么最长的长度是多少。

代码：

#include<stdio.h> 
#include<string.h>
#include<algorithm>
using namespace std;

char a[5001];
int b[5001], c[5001];

int main()
{
    while(~scanf("%s", a))
	{
	    memset(b, 0, sizeof b);
        memset(c, 0, sizeof c);
        
        int l = strlen(a); //长度 
        int sum = 0;
        
        for(int i = 0; i < l; i++)  //正着 
		{
            if(a[i] == 'b')  //第i个b前有几个a 
			    b[i] = sum;
            else 
			    sum++;
        }
 
        sum = 0;     //清空 
        for(int i = l-1; i >= 0; i--) //倒着 
		{
            if(a[i] == 'b') 
			   c[i] = sum;     
            else 
			   sum++;
        }

        int maxx = -0x3f3f3f3f;
        
        for(int i = 0; i < l; i++)
		{
            int s = 0;
            for(int j = i; j < l; j++)
			{
                if(a[j] == 'a') 
				   continue;
                s++;
                if(b[i]+s+c[j] > maxx)
				{
                    maxx = b[i]+s+c[j];
                }
            }
        }
        if(maxx == -0x3f3f3f3f) 
		    maxx = l;
        printf("%d\n", maxx);
    }
    return 0;
}