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() {
    int n, m, q;
    cin >> n >> m >> q;
    bool available[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        available[i][j] = true;
      }
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        for (int b = 0; b <= 1; b++) {
          int len = 1;
          while (true) {
            int x = i + (len + !b) / 2;
            int y = j + (len + b) / 2;
            if (x >= n || y >= m) {
              break;
            }
            len++;
          }
          ans += len - 1;
        }
      }
    }
    ans += n * m;
    while (q--) {
      int x, y;
      cin >> x >> y;
      x--, y--;
      vector<int> vec1;
      int new_ans1 = 1;
      for (int b : vector<int>({0, 1})) {
        int len = 1;
        while (true) {
          int x1 = x + (len + !b) / 2;
          int y1 = y + (len + b) / 2;
          if (x1 >= n || y1 >= m || !available[x1][y1]) {
            break;
          }
          len++;
        }
        new_ans1 += len;
        vec1.push_back(len);
      }
      vector<int> vec2;
      int new_ans2 = 1;
      for (int b = 0; b <= 1; b++) {
        int len = 1;
        while (true) {
          int x1 = x - (len + !b) / 2;
          int y1 = y - (len + b) / 2;
          if (x1 < 0 || y1 < 0 || !available[x1][y1]) {
            break;
          }
          len++;
        }
        new_ans2 += len;
        vec2.push_back(len);
      }
      int d = vec1[0] * vec2[1] + vec1[1] * vec2[0] - 1;
      if (available[x][y]) {
        ans -= d;
      } else {
        ans += d;
      }
      available[x][y] = !available[x][y];
      cout << ans << endl;
    }
  }
  