zoukankan      html  css  js  c++  java
  • codeforces 877b

    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.

    思路:记录每个位置后面有多少个a,有多少个b      存在arr[5005][2]里面  arr[i][0]表示i后面包括i有arr[i][0]个a    arr[i][1]表示i后面包括i有arr[i][1]个b

    遍历每个段位

    for(i =0 ; i < strlen(str);++i)

    {

      if(str[j] == 'b')

      {

        for(j = i; j < strlen(str);++j)

        {

           if(str[j] == 'b')

             maxn = max(maxn,  arr[0][0] - arr[i][0]    +    arr[i][2] - arr[j][2] + 1      + arr[j][0]);

        }

      }

    }

    注意全是a的情况

    丑陋的代码

    #include <iostream>

    #include <cstdio>

    #include <cstring>

    #include <string>

    using namespace std;

    int main()

    {

    //    char str[5010];

        string str;

        int arr[5010][2];

        int i,j,k,a,b,a1,b1;

    //    memset(str,0,sizeof(str));

        int maxn = 0;

        a1 = b1 = a = b = 0;

    //    scanf("%s",str);

        cin >> str;

        memset(arr, 0, sizeof(arr));

        for(i = 0; i < str.length(); ++i)

        {

            if(str[i] == 'a')

                a++;

            else

                b++;

        }

        if(b == 0)

        {

            printf("%d ",(int)str.length());

            return 0;

        }

        a1 = b1 = 0;

        for(i = 0; i < str.length(); ++i)

        {

            if(str[i] == 'a')

            {

                arr[i][0] = a - a1;

                arr[i][1] = b - b1;

                a1++;

            }

            else

            {

                arr[i][0] = a - a1;

                arr[i][1] =  b - b1;

                b1++;

            }

        }

        a1 = b1 = 0;

        for(i = 0; i < str.length(); ++i)

        {

            if(str[i] == 'b')

            {

                for(j = i; j < str.length(); ++j)

                {

                    if(str[j] == 'b')

                        maxn = max(maxn,a - arr[i][0] + arr[i][1] - arr[j][1] + arr[j][0]+1);

                }

            }

        }

        printf("%d ",maxn);

    }

  • 相关阅读:
    如何解决MathPage.wll或MathType.dll文件找不到问题
    2-构建模型
    R语言 ur.df函数(2)
    平稳过程趋势项变点的 CUSUV 检验 秦瑞兵,郭娟
    时间序列的弱相依与强相依
    Cent OS|使用httpd发布网页
    C++练习 | 基于栈的中缀算术表达式求值(double类型
    C++练习 | 不使用头插法逆转单链表
    C++练习 | 单链表的创建与输出(结构体格式)
    C++练习 | 最长公共字符串(DP)
  • 原文地址:https://www.cnblogs.com/mltang/p/7764136.html
Copyright © 2011-2022 走看看