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 n, m, a[1005][1005], f[1005][1005], ans;
  int calc(int x, int y, int dx, int dy, int type) {
    int cnt = 0;
    if (type == 0) {
      while (x - dx >= 1 && y - dy >= 1) {
        x -= dx, y -= dy;
        if (!a[x][y]) break;
        ++cnt;
        if (dx == 1)
          dy = 1, dx = 0;
        else
          dy = 0, dx = 1;
      }
    } else {
      while (x + dx <= n && y + dy <= m) {
        x += dx, y += dy;
        if (!a[x][y]) break;
        ++cnt;
        if (dx == 1)
          dy = 1, dx = 0;
        else
          dy = 0, dx = 1;
      }
    }
    return cnt;
  }
  signed main() {
    int q;
    cin >> n >> m >> q;
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j) a[i][j] = 1;
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j)
        ans += 1 + calc(i, j, 0, 1, 1) + calc(i, j, 1, 0, 1);
    while (q--) {
      int i, j;
      cin >> i >> j;
      int sum = (1 + calc(i, j, 0, 1, 1)) * (1 + calc(i, j, 1, 0, 0)) +
                (1 + calc(i, j, 1, 0, 1)) * (1 + calc(i, j, 0, 1, 0)) - 1;
      if (a[i][j])
        ans -= sum;
      else
        ans += sum;
      a[i][j] ^= 1;
      cout << ans << "\n";
    }
  }
  