Codeforces Round #306 (Div. 2)——A——Two Substrings

2444 단어 codeforces
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".
지능이 눌리다.hack 당했어, 거꾸로 판단만 하면 돼.
#include<cstdio>

#include<algorithm>

#include<cstring>

using namespace std;

int main()

{

    char a[100110];

    while(~scanf("%s",&a)){

        int n = strlen(a);

        int t1 ,t2 ;

        t1 = t2 = 0;

        int i = 0;

        while(i < n){

            if(a[i] == 'A' && a[i+1] == 'B'&&t1 == 0){

            t1++;

            i++;}

            else if(a[i] == 'B' && a[i+1] == 'A'&&t2 == 0){

            t2++;

            i++;

            }

            i++;

        }

        if(t1 > 0 && t2 > 0){

            printf("YES
"); continue; } i = n-1; t1 = t2 = 0; while(i >= 0){ if(a[i] == 'A' && a[i+1] == 'B'&& t1 == 0){ t1++; i--; } else if(a[i] == 'B' && a[i+1] == 'A' && t2 == 0){ t2++; i--; } i--; } if(t1 > 0 && t2 > 0) printf("YES
"); else printf("NO
"); } return 0; }

  
 

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