CodeForces 550A Two Substrings 简单题_codeforrcestwo substrings-优快云博客

本文链接：https://blog.youkuaiyun.com/canghai_wangzi/article/details/52972186

本文介绍了一个简单的编程问题：如何判断一个给定的字符串中是否包含两个非重叠的子串“AB”和“BA”。通过示例说明了算法实现，并提供了具体的C++代码。

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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.

   Examples 
 
     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".

拿一个简单题来当做我的第一篇博客吧，就这样，代码都是自己写的，所有有的可能不是最佳的(๑•̀ㅂ•́)و✧

#include <bits/stdc++.h>

using namespace std;

int main()
{
    string a;
    int len,p,q,k;
    while(cin>>a)
    {
        len=a.size();
        k=p=q=0;
        for(int i=0;i<len;i++)
        {
            if(a[i]=='A')
            {
                if(i+2<len&&a[i+1]=='B'&&a[i+2]=='A')
                {
                    k++;
                    i+=2;
                }
                else if(i+1<len&&a[i+1]=='B')
                {
                    p=1;
                    i++;
                }
            }
            if(a[i]=='B')
            {
                if(i+2<len&&a[i+1]=='A'&&a[i+2]=='B')
                {
                    k++;
                    i+=2;
                }
                else if(i+1<len&&a[i+1]=='A')
                {
                    q=1;
                    i++;
                }
            }
        }
        if(p+q+k>=2)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}