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, a[N][N];
  long long tot;
  int main() {
    ios::sync_with_stdio(0);
    cin >> n >> m >> q;
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j) {
        int x = n - i + 1, y = m - j + 1;
        int tt = min(x, y);
        tot += tt * 4 - 2;
        if (x == y) tot--;
      }
    while (q--) {
      int x, y;
      cin >> x >> y;
      int flag;
      if (a[x][y] == 1)
        flag = -1;
      else
        flag = 1;
      a[x][y] ^= 1;
      int last;
      long long num1 = 0, num2 = 0;
      int tx = x, ty = y;
      last = 1;
      while (tx >= 1 && ty >= 1) {
        num1++;
        if (last == 1)
          tx--;
        else
          ty--;
        if (a[tx][ty] == 1) break;
        last ^= 1;
      }
      last = 0;
      tx = x, ty = y;
      while (tx <= n && ty <= m) {
        num2++;
        if (last == 1)
          tx++;
        else
          ty++;
        if (a[tx][ty] == 1) break;
        last ^= 1;
      }
      tot -= num1 * num2 * flag;
      num1 = 0, num2 = 0;
      tx = x, ty = y;
      last = 0;
      while (tx >= 1 && ty >= 1) {
        num1++;
        if (last == 1)
          tx--;
        else
          ty--;
        if (a[tx][ty] == 1) break;
        last ^= 1;
      }
      last = 1;
      tx = x, ty = y;
      while (tx <= n && ty <= m) {
        num2++;
        if (last == 1)
          tx++;
        else
          ty++;
        if (a[tx][ty] == 1) break;
        last ^= 1;
      }
      tot -= num1 * num2 * flag;
      tot += flag;
      cout << tot << endl;
    }
    return 0;
  }
  