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：
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#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; }
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