CF550A Two Substrings

原题链接

• 题意：求字符串中是否存在一个 \(AB\) 和 \(BA\)，他们不重叠。
• 题解：直接把 \(BAB\) 和 \(ABA\) 记录数量的同时，转化为别的字符，然后统计 \(AB\) 和 \(BA\) 然后判断答案。
• 代码：

#include <iostream>
#include <cstring>
using namespace std;
typedef long long ll;
const int N = 5e5 + 9;
const ll mod = 1e9 + 7;
bool vis[N];
char s[N];
int main() {
    int ab_cnt = 0;
    int ba_cnt = 0;
    int aba_cnt = 0;
    cin >> (s + 1);
    int n = strlen(s + 1);
    for (int i = 1; i <n-1; i ++) {
        if (s[i] == ‘B‘ && s[i + 1] == ‘A‘ && s[i + 2] == ‘B‘) {
            aba_cnt++;
            s[i] = s[i + 1] = s[i+2] = ‘?‘;
        }
        if (s[i] == ‘A‘ && s[i + 1] == ‘B‘ &&s[i + 2] == ‘A‘) {
            aba_cnt++;
            s[i] = s[i + 1] = s[i + 2] = ‘?‘;
        }
    }
    for (int i = 1; i < n; i ++) {
        if (s[i] == ‘A‘ && s[ i + 1] == ‘B‘) {
            ab_cnt++;
        } 
        if (s[i] == ‘B‘ && s[i + 1] == ‘A‘)ba_cnt++;
    }
    cout << s << endl;
    if ( (ab_cnt >= 1&&ba_cnt >= 1) || (ab_cnt >=1 && aba_cnt >= 1 )|| (ba_cnt >= 1 && aba_cnt >= 1)||(aba_cnt > 1)){
        cout << "YES\n";
    } else cout << "NO\n";
}

CF550A Two Substrings

原文：https://www.cnblogs.com/Xiao-yan/p/14734579.html