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 m, n, q;
  long long f[N][N][2];
  int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m >> q;
    memset(f, 0, sizeof f);
    long long cnt = n * m;
    long long total = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        f[i][j][0] = 1;
        f[i][j][1] = 1;
      }
    }
    for (int i = n - 1; i >= 0; i--) {
      for (int j = m - 1; j >= 0; j--) {
        f[i][j][0] += f[i][j + 1][1];
        f[i][j][1] += f[i + 1][j][0];
        total += (f[i][j][0] + f[i][j][1]);
      }
    }
    for (int i = 0; i < q; i++) {
      int x, y;
      cin >> x >> y;
      x--;
      y--;
      int change0, change1;
      if (f[x][y][0] > 0) {
        change0 = -f[x][y][1];
        change1 = -f[x][y][0];
        f[x][y][0] = f[x][y][1] = 0;
        total += (change0 + change1);
        cnt--;
      } else {
        change0 = f[x][y][1] = f[x + 1][y][0] + 1;
        change1 = f[x][y][0] = f[x][y + 1][1] + 1;
        total += (change0 + change1);
        cnt++;
      }
      int t = 1;
      int xx = x;
      int yy = y;
      bool failed = false;
      while (!failed) {
        if (t & 1) {
          x--;
          if (x < 0 || y < 0 || f[x][y][0] == 0)
            failed = true;
          else {
            f[x][y][1] += change1;
            total += change1;
          }
        } else {
          y--;
          if (x < 0 || y < 0 || f[x][y][0] == 0)
            failed = true;
          else {
            f[x][y][0] += change1;
            total += change1;
          }
        }
        t++;
      }
      x = xx;
      y = yy;
      t = 1;
      failed = false;
      while (!failed) {
        if (t & 1) {
          y--;
          if (x < 0 || y < 0 or f[x][y][0] == 0)
            failed = true;
          else {
            f[x][y][0] += change0;
            total += change0;
          }
        } else {
          x--;
          if (x < 0 || y < 0 or f[x][y][0] == 0)
            failed = true;
          else {
            f[x][y][1] += change0;
            total += change0;
          }
        }
        t++;
      }
      cout << total - cnt << endl;
    }
    return 0;
  }
  