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 a[N][N];
  int c[N][N];
  int get_ans(int n, int m, int x, int y) {
    if (n == 1 && m == 1) return 0;
    int ans = 0;
    int tx = x, ty = y;
    while (tx < n && ty < m) tx++, ty++;
    c[tx][ty] = (1 - a[tx][ty]);
    for (int i = 1;; i++) {
      if (!tx || !ty) break;
      tx--, ty--;
      if (!tx || !ty) break;
      if (a[tx][ty])
        c[tx][ty] = 0;
      else
        c[tx][ty] = (1 - a[tx][ty]) + c[tx + 1][ty + 1];
    }
    tx++, ty++;
    int ttx = tx, tty = ty + 1;
    if (ty == m) {
      ttx = tx - 1, tty = m;
    }
    while (ttx < n && tty < m) ttx++, tty++;
    c[ttx][tty] = 1 - a[ttx][tty];
    for (int i = 1;; i++) {
      if (!ttx || !tty) break;
      ttx--, tty--;
      if (!ttx || !tty) break;
      if (a[ttx][tty])
        c[ttx][tty] = 0;
      else
        c[ttx][tty] = 1 - a[ttx][tty] + c[ttx + 1][tty + 1];
    }
    for (int i = 0; tx + i <= n && ty + i < m; i++) {
      if (c[tx + i][ty + i] == 0) continue;
      int l = c[tx + i][ty + i];
      int r = c[tx + i][ty + i + 1];
      ans += min(l, r);
      if (l > r)
        ans += r;
      else
        ans += l - 1;
    }
    ttx++, tty++;
    for (int i = 0; ttx + i < n && tty + i <= m; i++) {
      if (c[ttx + i][tty + i] == 0) continue;
      int u = c[ttx + i][tty + i];
      int d = c[ttx + i + 1][tty + i];
      ans += min(u, d);
      if (u > d)
        ans += d;
      else
        ans += u - 1;
    }
    ttx = tx + 1, tty = ty;
    if (tx == n) {
      ttx = tx;
      tty = ty - 1;
    }
    while (ttx < n && tty < m) ttx++, tty++;
    c[ttx][tty] = 1 - a[ttx][tty];
    for (int i = 1;; i++) {
      if (!ttx || !tty) break;
      ttx--, tty--;
      if (!ttx || !tty) break;
      if (a[ttx][tty])
        c[ttx][tty] = 0;
      else
        c[ttx][tty] = 1 - a[ttx][tty] + c[ttx + 1][tty + 1];
    }
    for (int i = 0; tx + i < n && ty + i <= m; i++) {
      if (c[tx + i][ty + i] == 0) continue;
      int u = c[tx + i][ty + i];
      int d = c[tx + i + 1][ty + i];
      ans += min(u, d);
      if (u > d)
        ans += d;
      else
        ans += u - 1;
    }
    ttx++, tty++;
    for (int i = 0; ttx + i <= n && tty + i < m; i++) {
      if (c[ttx + i][tty + i] == 0) continue;
      int l = c[ttx + i][tty + i];
      int r = c[ttx + i][tty + i + 1];
      ans += min(l, r);
      if (l > r)
        ans += r;
      else
        ans += l - 1;
    }
    return ans;
  }
  void solve() {
    int n, m, r;
    scanf("%d%d%d", &n, &m, &r);
    long long ans = n * m;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j < m; j++) {
        int l = min(n - i + 1, m - j + 1);
        int r = min(n - i + 1, m - j);
        ans += r;
        ans += l - 1;
      }
    }
    for (int i = 1; i < n; i++) {
      for (int j = 1; j <= m; j++) {
        int u = min(n - i + 1, m - j + 1);
        int d = min(n - i, m - j + 1);
        ans += d;
        ans += u - 1;
      }
    }
    while (r--) {
      int x, y;
      scanf("%d%d", &x, &y);
      ans -= get_ans(n, m, x, y);
      a[x][y] ^= 1;
      if (a[x][y])
        ans--;
      else
        ans++;
      ans += get_ans(n, m, x, y);
      printf("%lld\n", ans);
    }
  }
  int main() {
    int t = 1;
    while (t--) solve();
    return 0;
  }
  