Codeforces Round #306 (Div. 2) A

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".

題意：給你一串字符串，若該字符串裡含有 'AB' 和 'BA' 兩個字串，並且這兩個字串不重疊，如 ABA 即不可。（可有多個'AB' 和 'BA'）

一開始從頭到尾掃了一遍，錯了，特例如：ABAXXXAB，所以倒著在掃一遍就好咯，笨辦法，嘻嘻(*^__^*)

代碼如下：

 

#include
#include
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;
}