zoukankan      html  css  js  c++  java
  • A

    Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

    Description

    You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order).

    Input

    The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters.

    Output

    Print "YES" (without the quotes), if string s contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise.

    Sample Input

    Input
    ABA
    Output
    NO
    Input
    BACFAB
    Output
    YES
    Input
    AXBYBXA
    Output
    NO

    Hint

    In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".

    In the second sample test there are the following occurrences of the substrings: BACFAB.

    In the third sample test there is no substring "AB" nor substring "BA".

    题意:

    给定一字符串,求能否找出“AB”“BA”两不重叠字符串。

    啊啊啊,坑比的字符串题!!卡了三组数据TAT

    附AC代码:

     1 #include<iostream>
     2 #include<cstring>
     3 using namespace std;
     4 
     5 int main(){
     6     string s;
     7     int t=-1,v=-1,ans,temp,x,y,a,b,c,d;
     8     cin>>s;
     9     int len=s.size();
    10     ans=0;
    11     temp=0;
    12     x=0;
    13     y=0;
    14     for(int i=0;i<len;i++){
    15         if(s[i]=='A'&&s[i+1]=='B'){
    16             if(i!=t&&!ans){
    17                 ans++;
    18                 t=i+1;
    19                 a=i;
    20                 break;
    21             }
    22         }
    23     }
    24     for(int i=0;i<len;i++){
    25         if(s[i]=='B'&&s[i+1]=='A'){
    26             if(i!=t&&i+1!=a&&!temp){
    27                 temp++;
    28                 t=i+1;
    29                 break;
    30             }    
    31         }
    32     }
    33     for(int i=0;i<len;i++){
    34         if(s[i]=='B'&&s[i+1]=='A'){
    35             if(i!=v&&!x){
    36                 x++;
    37                 v=i+1;
    38                 b=i;
    39                 break;
    40             }    
    41         }
    42     }
    43     for(int i=0;i<len;i++){
    44         if(s[i]=='A'&&s[i+1]=='B'){
    45             if(i!=v&&i+1!=b&&!y){
    46                 y++;
    47                 v=i+1;
    48                 break;
    49             }
    50         }
    51     }
    52     if(ans&&temp){
    53         cout<<"YES"<<endl;
    54         return 0;
    55     }
    56     else if(x&&y){
    57         cout<<"YES"<<endl;
    58         return 0;
    59     }
    60     cout<<"NO"<<endl;
    61     return 0;
    62 } 
  • 相关阅读:
    记录ICallbackEventHandler 同时并发访问容易引发的问题
    IIS 属性
    Solaris10 安装
    VerifyRenderingInServerForm和EnableEventValidation引发的两个问题
    Solaris 上网配置
    动态数据类型转换
    RDLC 折线图
    codesmith复制中文乱码解决
    关于手机等品牌型号搜索与采集的中文分词分离
    说点包租公限制共享上网的破事
  • 原文地址:https://www.cnblogs.com/Kiven5197/p/5720419.html
Copyright © 2011-2022 走看看