Codeforces Round #306 (Div. 2) A

A. Two Substrings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input

ABA

output

NO

input

BACFAB

output

YES

input

AXBYBXA

output

NO

Note

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：给你一串字符串，若该字符串里含有 'AB' 和 'BA' 两个字串，并且这两个字串不重叠，如 ABA 即不可。（可有多个'AB' 和 'BA'）

一开始从头到尾扫了一遍，错了，特例如：ABAXXXAB，所以倒着在扫一遍就好咯，笨办法，嘻嘻(*^__^*) 

代码如下：

#include<cstdio>#include<cstring>const int N = 100005;char a[N];int main(){while(~scanf("%s", a)){int len = strlen(a);int ans = 0, tmp = 0;for(int i = 0; i < len ; i++){if(a[i] == 'B' && a[i+1] == 'A' && ans == 0){ans++;i++;}else if(a[i] == 'A' && a[i+1] == 'B' && tmp == 0){tmp++;i++;}}if(ans >= 1 && tmp >= 1) {printf("YES\n");continue;}tmp = 0, ans = 0;for(int i = len-1; i >= 0; i--){if(a[i] == 'B' && a[i-1] == 'A' && ans == 0){ans++;i--;}else if(a[i] == 'A' && a[i-1] == 'B' && tmp == 0){tmp++;i--;}}if(ans >= 1 && tmp >= 1) printf("YES\n");else printf("NO\n");}return 0;}