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 = 1e3 + 5;
  int n, m, q, ans, f[N][N][2], g[N][N];
  bool st[N][N];
  int main() {
    scanf("%d %d %d", &n, &m, &q);
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        f[i][j][0] = (j == 1) ? 0 : (f[i][j - 1][1] + 1);
        f[i][j][1] = (i == 1) ? 0 : (f[i - 1][j][0] + 1);
        g[i][j] = f[i][j][0] + f[i][j][1] + 1;
      }
    }
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        ans += g[i][j];
      }
    }
    for (int i = 0; i <= n; ++i) {
      st[i][0] = st[i][m + 1] = 1;
    }
    for (int i = 0; i <= m; ++i) {
      st[0][i] = st[n + 1][i] = 1;
    }
    while (q--) {
      int x, y, sign;
      scanf("%d %d", &x, &y);
      sign = st[x][y] ? 1 : -1;
      int A = min(x - 1, y - 1);
      int B = min(n - x, m - y);
      int p = 0, q = 0, s = 0, t = 0;
      for (int i = 1; i <= A + 1; ++i) {
        if (st[x - i][y - i + 1]) break;
        ++p;
        if (st[x - i][y - i]) break;
        ++p;
      }
      for (int i = 1; i <= A + 1; ++i) {
        if (st[x - i + 1][y - i]) break;
        ++q;
        if (st[x - i][y - i]) break;
        ++q;
      }
      for (int i = 1; i <= B + 1; ++i) {
        if (st[x + i - 1][y + i]) break;
        ++s;
        if (st[x + i][y + i]) break;
        ++s;
      }
      for (int i = 1; i <= B + 1; ++i) {
        if (st[x + i][y + i - 1]) break;
        ++t;
        if (st[x + i][y + i]) break;
        ++t;
      }
      ans += (p * s + q * t + s + p + q + t + 1) * sign;
      st[x][y] ^= 1;
      printf("%d\n", ans);
    }
    return 0;
  }
  