CodeForces 877 B Nikita and string

最新推荐文章于 2021-06-02 15:24:00 发布

swqeaaa

最新推荐文章于 2021-06-02 15:24:00 发布

阅读量269

点赞数

本文链接：https://blog.youkuaiyun.com/qq_32259423/article/details/81268233

版权

CodeForces 专栏收录该内容

122 篇文章

订阅专栏

本文介绍了一个算法问题：如何通过删除字符使字符串变得“美丽”。美丽字符串定义为可以分割成三部分的字符串，首尾部分只包含字母a，中间部分只包含字母b。文章探讨了如何最大化这种美丽字符串的长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

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.

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

char a[5505];

int main()
{
    cin>>a;
    int n=strlen(a);
    int dp[5505];
    dp[0]=1;
    int MAX=1;
    int anumber=0;
    int bnumber=0;
    int beforeb =0;
    if(a[0]=='b')
        bnumber++;
    else
        anumber++;
    for(int i=1; i<n; i++) {
        if(a[i]==a[i-1]) {
            dp[i]=dp[i-1]+1;
        } else if(a[i]=='a') {
            dp[i]=MAX+1;
        } else {
            dp[i]=max(bnumber+1,anumber+1);
            dp[i]=max(dp[i],beforeb+1);

        }
        if(a[i]=='a')
            anumber++;
        else {
            beforeb=max(dp[i],beforeb);
            bnumber++;
        }
        MAX=max(MAX,dp[i]);
    }
    cout<<MAX;
}