CF 1381 A2 Prefix Flip (Hard Version)（思维 + 暴力）

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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。

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <string>
#include <vector>
#include <cmath>
 
using namespace std;
 
#define ll long long
#define pb push_back
#define fi first
#define se second
 
const int N = 2e5 + 10;
ll a[N], dp[N], sum; 

void solve()
{      
    int T;
    cin >> T;
    while(T--){
        int n;
        string a, b;
        cin >> n >> a >> b;

        int k, al, ar, br, cnt, now;
        vector<int > ans;
        now = k = al = cnt = 0;
        ar = br = n - 1;
        while(1){
            //printf("a[now] = %c  b[br] = %c
", a[now], b[br]);
            if(cnt & 1){
                if((a[now] - '0' + cnt) % 2 == b[br] - '0'){
                    k += 2, br--, ar--, now = al;
                    ans.pb(1);
                    ans.pb(n - cnt);
                }else{
                    k += 1, br--, ar--, now = al;
                    ans.pb(n - cnt);
                }
                cnt++;
            }else{
                if((a[now] - '0' + cnt) % 2 == b[br] - '0'){
                    k += 2, br--, al++, now = ar;
                    ans.pb(1);
                    ans.pb(n - cnt);
                }else{
                    k += 1, br--, al++, now = ar;
                    ans.pb(n - cnt);
                }
                cnt++;
            }
            //cout << "k = " << k << endl;
            //cout << " cnt = " << cnt << endl;
            //cout << "br = " << br << endl;
            if(br < 0) break;
        }

        //cout << "ans = " << k << " ";
        cout << k << endl;
        for(auto x : ans) cout << x << " ";
        cout << endl;
    }
}
 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0); 
    solve();
 
    return 0;
}