题目：
There are two binary strings a and b of length nn (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001
Your task is to transform the string a into b in at most 2n operations. It can be proved that it is always possible.
思路：我们可以每次只处理一位二进制，让a的第一位和b的最后一位不同，然后反转a未匹配的串，每次匹配一个b的后面的字符，这样最多操作也是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 ", 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 }