description: A binary string is a string that consists of characters 0 and 1. A bi-table is a table that has exactly two rows of equal length, each being a binary string.
  
  Let \operatorname{MEX} of a bi-table be the smallest digit among 0, 1, or 2 that does not occur in the bi-table. For example, \operatorname{MEX} for \begin{bmatrix} 0011\\\ 1010 \end{bmatrix} is 2, because 0 and 1 occur in the bi-table at least once. \operatorname{MEX} for \begin{bmatrix} 111\\\ 111 \end{bmatrix} is 0, because 0 and 2 do not occur in the bi-table, and 0 < 2.
  
  You are given a bi-table with n columns. You should cut it into any number of bi-tables (each consisting of consecutive columns) so that each column is in exactly one bi-table. It is possible to cut the bi-table into a single bi-table — the whole bi-table.
  
  What is the maximal sum of \operatorname{MEX} of all resulting bi-tables can be?
  
  Input
  
  The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
  
  The first line of the description of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of columns in the bi-table.
  
  Each of the next two lines contains a binary string of length n — the rows of the bi-table.
  
  It's guaranteed that the sum of n over all test cases does not exceed 10^5.
  
  Output
  
  For each test case print a single integer — the maximal sum of \operatorname{MEX} of all bi-tables that it is possible to get by cutting the given bi-table optimally.
  
  Example
  
  Input
  
  
  4
  7
  0101000
  1101100
  5
  01100
  10101
  2
  01
  01
  6
  000000
  111111
  
  
  Output
  
  
  8
  8
  2
  12
  
  Note
  
  In the first test case you can cut the bi-table as follows:
  
    * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
    * \begin{bmatrix} 10\\\ 10 \end{bmatrix}, its \operatorname{MEX} is 2.
    * \begin{bmatrix} 1\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 0.
    * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
    * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.
    * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.
  
  
  
  The sum of \operatorname{MEX} is 8.
solution: #include <bits/stdc++.h>
  using namespace std;
  const long long inf = 1e18;
  long long t;
  long long mex(vector<long long> &v) {
    sort(v.begin(), v.end());
    long long ans = 0;
    for (long long x : v)
      if (x == ans) ans++;
    return ans;
  }
  long long main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      string s1, s2;
      cin >> s1 >> s2;
      vector<long long> dp(n);
      function<long long(long long, long long)> get_mex = [&](long long i,
                                                              long long delta) {
        vector<long long> v;
        for (long long j = 0; j < delta; j++) {
          v.push_back(s1[i - j] - '0');
          v.push_back(s2[i - j] - '0');
        }
        return mex(v);
      };
      for (long long i = 0; i < n; i++) {
        for (long long prv = i - 1; prv >= i - 2; prv--) {
          long long delta = i - prv;
          if (prv == -1) {
            dp[i] = max(dp[i], get_mex(i, delta));
          } else if (prv < -1) {
            continue;
          } else {
            dp[i] = max(dp[i], dp[prv] + get_mex(i, delta));
          }
        }
      }
      cout << dp[n - 1] << "\n";
    }
  }
  