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;
  const int N = 1010;
  int n, m, Q;
  long long ans, cnt, f[N][N][2];
  bool vis[N][N];
  void update(int i, int j) {
    if (j < 1 || j > m) return;
    ans -= (f[i][j][0] + f[i][j][1]);
    if (vis[i][j])
      f[i][j][0] = f[i][j][1] = 0;
    else
      f[i][j][0] = f[i][j + 1][1] + 1, f[i][j][1] = f[i + 1][j][0] + 1;
    ans += (f[i][j][0] + f[i][j][1]);
  }
  int main() {
    scanf("%d%d%d", &n, &m, &Q);
    for (int i = n; i >= 1; i--)
      for (int j = m; j >= 1; j--) update(i, j), cnt++;
    while (Q--) {
      int x, y;
      scanf("%d%d", &x, &y);
      vis[x][y] ^= 1;
      if (vis[x][y] == 1)
        cnt--;
      else
        cnt++;
      for (int i = x, j = y - 1; i >= 1; i--, j--)
        update(i, j + 2), update(i, j + 1), update(i, j);
      cout << ans - cnt << "\n";
    }
    return 0;
  }
  