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題目：

There are two binary stringsaandbof lengthnn(a binary string is a string consisting of symbols0and1). In an operation, you select a prefix ofa, and simultaneously invert the bits in the prefix (0changes to1and1changes to0) and reverse the order of the bits in the prefix.

For example, ifa=001011and you select the prefix of length

Your task is to transform the stringaintobin at most2noperations. It can be proved that it is always possible.

思路：我們可以每次只處理一位二進位制，讓a的第一位和b的最後一位不同，然後反轉a未匹配的串，每次匹配一個b的後面的字元，這樣最多操作也是2n。

 1 #include <iostream>
 2 #include <cstdio>
 3
 
 #include <algorithm>
 4 #include <queue>
 5 #include <string>
 6 #include <vector>
 7 #include <cmath>
 8  
 9 using namespace std;
10  
11 #define ll long long
12 #define pb push_back
13 #define fi first
14 #define se second
15  
16 const int N = 2e5 + 10;
17 ll a[N], dp[N], sum; 
 
18 
19 void solve()
20 {      
21     int T;
22     cin >> T;
23     while(T--){
24         int n;
25         string a, b;
26         cin >> n >> a >> b;
27 
28         int k, al, ar, br, cnt, now;
29         vector<int > ans;
30         now = k = al = cnt = 0;
31         ar = br = n - 1;
32         while(1){
33             //printf("a[now] = %c  b[br] = %c\n", a[now], b[br]);
34             if(cnt & 1){
35                 if((a[now] - '0' + cnt) % 2 == b[br] - '0'){
36                     k += 2, br--, ar--, now = al;
37                     ans.pb(1);
38                     ans.pb(n - cnt);
39                 }else{
40                     k += 1, br--, ar--, now = al;
41                     ans.pb(n - cnt);
42                 }
43                 cnt++;
44             }else{
45                 if((a[now] - '0' + cnt) % 2 == b[br] - '0'){
46                     k += 2, br--, al++, now = ar;
47                     ans.pb(1);
48                     ans.pb(n - cnt);
49                 }else{
50                     k += 1, br--, al++, now = ar;
51                     ans.pb(n - cnt);
52                 }
53                 cnt++;
54             }
55             //cout << "k = " << k << endl;
56             //cout << " cnt = " << cnt << endl;
57             //cout << "br = " << br << endl;
58             if(br < 0) break;
59         }
60 
61         //cout << "ans = " << k << " ";
62         cout << k << endl;
63         for(auto x : ans) cout << x << " ";
64         cout << endl;
65     }
66 }
67  
68 int main()
69 {
70     ios::sync_with_stdio(false);
71     cin.tie(0);
72     cout.tie(0); 
73     solve();
74  
75     return 0;
76 }