CF Two Substrings

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




找到每组AB和BA的起点坐标，保存起来，然后用最远的一个BA的坐标减去最近的一个AB，或者最远的一个AB减去一个最近的BA,如果两者的差大于等于2，那么就YES.

#include <iostream>
#include <cstdio>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <climits>
using    namespace    std;

int    main(void)
{
    char    s[100005];
    int    ab[100005],ba[100005];
    int    p_ab = 0,p_ba = 0;
    
    scanf("%s",s);
    for(int i = 0;s[i];i ++)
    {
        if(s[i] == 'A' && s[i + 1] == 'B')
        {
            ab[p_ab] = i;
            p_ab ++;
        }
        else    if(s[i] == 'B' && s[i + 1] == 'A')
        {
            ba[p_ba] = i;
            p_ba ++;
        }
    }
    if(!p_ab || !p_ba)
    {
        puts("NO");
        return    0;
    }
    sort(ab,ab + p_ab);
    sort(ba,ba + p_ba);
    if(abs(ab[0] - ba[p_ba - 1]) >= 2 || abs(ba[0] - ab[p_ab - 1]) >= 2)
        puts("YES");
    else
        puts("NO");

    return    0;
}