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

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最新推荐文章于 2022-11-22 23:15:14 发布

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本文介绍了一道 CodeForces 上的题目 877B 的解题思路，该题要求将一个仅包含字母 'a' 和 'b' 的字符串通过删除部分字符的方式重构为 x*a+y*b+z*a 的形式，并求得重构后字符串的最大长度。文章给出了两种解决方案，一种是使用动态规划思想，另一种是采用简单的遍历优化方法。

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http://codeforces.com/problemset/problem/877/B

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

把一个字符串通过删去一些字符分成x*a+y*b+z*a的形式，问这个字符串最长几个字符

题解

用a[i]和b[i]记录i之前的a和b的数量，然后得到动态转移方程：
ans=max(ans,a[i]+(b[j]-b[i])+(a[l]-a[j]));

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
    char ch[5050];
    scanf("%s",ch);
    int l=strlen(ch);
    int a[5050];
    int b[5050];
    a[0]=b[0]=0;
    for(int i=0;i<l;i++)
    {
        a[i+1]=a[i]+(ch[i]=='a');
        b[i+1]=b[i]+(ch[i]=='b');
    }
    int ans=0;
    for(int i=0;i<=l;i++)
        for(int j=i;j<=l;j++)
            ans=max(ans,a[i]+(b[j]-b[i])+(a[l]-a[j]));
        cout << ans << endl;
        return 0;
}

简单的方法

#include<bits/stdc++.h>
using namespace std;
int main()
{
    string s;
    cin>>s;
    int a=0,aba=0,ab=0;
    int l=s.length();
    for(int i=0; i<l; i++)
    {
        if(s[i]=='a') aba++,a++;
        if(s[i]=='b') ab++;
        aba=max(aba, ab);
        ab=max(ab, a);
    }
    cout<<aba;
}