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;
  bool st[100005];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      memset(st, false, sizeof st);
      int n;
      cin >> n;
      string str1;
      string str2;
      cin >> str1;
      cin >> str2;
      int ans = 0;
      for (int i = 0; i < str1.size(); i++) {
        if (str1[i] != str2[i]) {
          ans += 2;
          st[i] = true;
        } else {
          if (!st[i] && str1[i] == '0') {
            if (i - 1 >= 0 && !st[i - 1] && str1[i - 1] == '1' &&
                str2[i - 1] == '1') {
              ans += 2;
              st[i - 1] = true;
              st[i] = true;
            } else if (i + 1 < n && !st[i + 1] && str1[i + 1] == '1' &&
                       str2[i + 1] == '1') {
              ans += 2;
              st[i + 1] = true;
              st[i] = true;
            } else if (!st[i]) {
              ans += 1;
              st[i] = true;
            }
          } else if (!st[i]) {
            if (i - 1 >= 0 && !st[i - 1] && str1[i - 1] == '0' &&
                str2[i - 1] == '0') {
              ans += 2;
              st[i - 1] = true;
              st[i] = true;
            } else if (i + 1 < n && !st[i + 1] && str1[i + 1] == '0' &&
                       str2[i + 1] == '0') {
              ans += 2;
              st[i + 1] = true;
              st[i] = true;
            }
          }
        }
      }
      cout << ans << endl;
    }
    return 0;
  }
  