https://codeforc.es/contest/1697/problem/C

因为规则中，两种字符串变换都与‘b’有关，所以我们根据b的位置来进行考虑；

先去掉所有的'b'，如果两字符串不相等就“NO”；

否则通过‘b'在a,b串中的位置，如果posa>posb,那么他们之间如果出现'a'就说明不可能

如果posb<posa，那么他们之间如果出现'c'说明不可能

 1 # include<iostream>
 2 # include<bits/stdc++.h>
 3 using namespace std;
 4 const int N = 3e5 + 10;
 5 #define int long long
 6 # define endl "\n"
 7 int suma[N], sumc[N];
 8 char a[N], b[N];
 9 char newa[N], newb[N];
10 int posa[N], posb[N];
11 void solve() {
12     int n;
13     cin >> n;
14     cin >> a + 1 >> b + 1;
15     int cnt1 = 0, cnt2 = 0;
16     for (int i = 1; i <= n; ++i) {
17         if (a[i] == 'a') suma[i] = suma[i - 1] + 1;
18         else suma[i] = suma[i - 1];
19         if (a[i] == 'c') sumc[i] = sumc[i - 1] + 1;
20         else sumc[i] = sumc[i - 1];
21         if (a[i] != 'b') newa[++cnt1] = a[i];
22         if (b[i] != 'b') newb[++cnt2] = b[i];
23     }
24 
25     if (cnt1 != cnt2) {
26         cout << "NO" << endl;
27         return;
28     }
29     for (int i = 1; i <= cnt1; ++i) {
30         if (newa[i] != newb[i]) {
31             cout << "NO" << endl;
32             return;
33         }
34     }
35     cnt1 = 0, cnt2  = 0;
36     for (int i = 1; i <= n; ++i) {
37         if (a[i] == 'b') posa[++cnt1] = i;
38         if (b[i] == 'b') posb[++cnt2] = i;
39     }
40     if (cnt1 != cnt2) {
41         cout << "NO" << endl;
42         return;
43     }
44     for (int i = 1; i <= cnt1; ++i) {
45         int x = posa[i], y = posb[i];
46         if (x < y) {
47             if (suma[y] - suma[x - 1] != 0) /*前缀和判断a的位置*/
　　　　　　　　　{
48                 cout << "NO" << endl;
49                 return;
50             }
51         }
52         if (x > y) {
53             if (sumc[x] - sumc[y - 1] != 0) {
54                 cout << "NO" << endl;
55                 return;
56             }
57         }
58 
59     }
60     cout << "YES" << endl;
61 }
62 int tt;
63 signed main() {
64     ios::sync_with_stdio(false);
65     cin.tie(0);
66     cout.tie(0);
67 
68     cin >> tt;
69 
70     while (tt--)solve();
71     return 0;
72 }
73 /*
74 
75 */

主要是通过前缀和来判断两个pos之间是否存在'a'或者'c'的操作