description: You are given a matrix, consisting of n rows and m columns. The rows are numbered top to bottom, the columns are numbered left to right.
  
  Each cell of the matrix can be either free or locked.
  
  Let's call a path in the matrix a staircase if it: 
  
    * starts and ends in the free cell; 
    * visits only free cells; 
    * has one of the two following structures: 
      1. the second cell is 1 to the right from the first one, the third cell is 1 to the bottom from the second one, the fourth cell is 1 to the right from the third one, and so on; 
      2. the second cell is 1 to the bottom from the first one, the third cell is 1 to the right from the second one, the fourth cell is 1 to the bottom from the third one, and so on. 
  
  
  
  In particular, a path, consisting of a single cell, is considered to be a staircase.
  
  Here are some examples of staircases:
  
  <image>
  
  Initially all the cells of the matrix are free.
  
  You have to process q queries, each of them flips the state of a single cell. So, if a cell is currently free, it makes it locked, and if a cell is currently locked, it makes it free.
  
  Print the number of different staircases after each query. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.
  
  Input
  
  The first line contains three integers n, m and q (1 ≤ n, m ≤ 1000; 1 ≤ q ≤ 10^4) — the sizes of the matrix and the number of queries.
  
  Each of the next q lines contains two integers x and y (1 ≤ x ≤ n; 1 ≤ y ≤ m) — the description of each query.
  
  Output
  
  Print q integers — the i-th value should be equal to the number of different staircases after i queries. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.
  
  Examples
  
  Input
  
  
  2 2 8
  1 1
  1 1
  1 1
  2 2
  1 1
  1 2
  2 1
  1 1
  
  
  Output
  
  
  5
  10
  5
  2
  5
  3
  1
  0
  
  
  Input
  
  
  3 4 10
  1 4
  1 2
  2 3
  1 2
  2 3
  3 2
  1 3
  3 4
  1 3
  3 1
  
  
  Output
  
  
  49
  35
  24
  29
  49
  39
  31
  23
  29
  27
  
  
  Input
  
  
  1000 1000 2
  239 634
  239 634
  
  
  Output
  
  
  1332632508
  1333333000
solution: #include <bits/stdc++.h>
  using namespace std;
  int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    long long T, a, b, c, d, n, m, q;
    std::cin >> n >> m >> q;
    std::vector<vector<bool>> state(n, vector<bool>(m, false));
    long long poss = 2 * m - 1;
    std::vector<int> dp(m, 2);
    dp[0] = 1;
    std::vector<int> newDp(m);
    for (int i = 1; i < n; i++) {
      newDp[0] = 1;
      poss += 1;
      for (int j = 1; j < m; j++) {
        newDp[j] = dp[j - 1] + 2;
        poss += dp[j - 1] * 2 + 2;
      }
      poss += dp[m - 1];
      swap(dp, newDp);
    }
    for (int i = 0; i < q; i++) {
      std::cin >> a >> b;
      a--;
      b--;
      long long dir = -1;
      if (state[a][b]) {
        dir = 1;
      }
      state[a][b] = !state[a][b];
      valarray<long long> start[] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
      long long results[] = {0, 0, 0, 0};
      for (int j = 0; j < 4; j++) {
        long long temp[2] = {a, b};
        std::valarray<long long> cur(temp, 2);
        int index = j;
        cur += start[index];
        index = 3 - index;
        while (cur[0] >= 0 && cur[1] >= 0 && cur[0] < n && cur[1] < m &&
               state[cur[0]][cur[1]] == false) {
          results[j]++;
          cur += start[index];
          index = 3 - index;
        }
      }
      poss += results[0] * results[2] * dir;
      poss += results[1] * results[3] * dir;
      poss += results[0] * dir;
      poss += results[1] * dir;
      poss += results[2] * dir;
      poss += results[3] * dir;
      poss += dir;
      cout << poss << endl;
    }
  }
  