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 }

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原文地址：https://www.cnblogs.com/Kiven5197/p/5720419.html