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;
  void solve() {
    long long n, m, q;
    cin >> n >> m >> q;
    vector<vector<bool>> a(n, vector<bool>(m, 1));
    long long total = 0;
    vector<vector<pair<long long, long long>>> dp(
        n, vector<pair<long long, long long>>(m));
    for (long long i = 0; i < n; ++i) {
      for (long long j = 0; j < m; ++j) {
        if (i == 0) {
          if (j == 0)
            ;
          else
            dp[i][j].second = 1;
        } else if (j == 0)
          dp[i][j].first = 1;
        else
          dp[i][j].first = (dp[i - 1][j].second + 1),
          dp[i][j].second = (dp[i][j - 1].first + 1);
        total += (1 + dp[i][j].first + dp[i][j].second);
      }
    }
    while (q--) {
      long long r, c;
      cin >> r >> c;
      --r, --c;
      long long cur_c = c, cur_r = r;
      bool add = 0;
      if (!a[r][c]) add = 1;
      long long ts = 0, trn = 0, t1u = 0, t2u = 0, t1d = 0, t2d = 0;
      while (cur_c >= 0 && cur_r >= 0) {
        if (cur_r == r && cur_c == c) {
          t1u += 1;
          cur_r -= 1;
          trn++;
          continue;
        }
        if (a[cur_r][cur_c] == 0) break;
        if (trn == 0)
          cur_r -= 1;
        else
          cur_c -= 1;
        trn++;
        trn %= 2;
        t1u++;
      }
      cur_r = r, cur_c = c;
      trn = 0;
      while (cur_r >= 0 && cur_c >= 0) {
        if (cur_r == r && cur_c == c) {
          cur_c -= 1;
          trn++;
          continue;
        }
        if (a[cur_r][cur_c] == 0) break;
        if (trn == 0)
          cur_c -= 1;
        else
          cur_r -= 1;
        trn++;
        trn %= 2;
        t2u++;
      }
      cur_r = r, cur_c = c;
      trn = 0;
      while (cur_r < n && cur_c < m) {
        if (cur_r == r && cur_c == c) {
          cur_r += 1;
          trn++;
          continue;
        }
        if (a[cur_r][cur_c] == 0) break;
        if (trn == 0)
          cur_r += 1;
        else
          cur_c += 1;
        trn++;
        trn %= 2;
        t1d++;
      }
      cur_r = r, cur_c = c;
      trn = 0;
      while (cur_r < n && cur_c < m) {
        if (cur_r == r && cur_c == c) {
          cur_c += 1;
          trn++;
          continue;
        }
        if (a[cur_r][cur_c] == 0) break;
        ;
        if (trn == 0)
          cur_c += 1;
        else
          cur_r += 1;
        trn++;
        trn %= 2;
        t2d++;
      }
      ts = (t1u + t2u + t1d + t2d);
      ts += ((t1u - 1) * t2d + t2u * t1d);
      a[r][c] = (!a[r][c]);
      if (add)
        total += ts;
      else
        total -= ts;
      cout << total << '\n';
    }
  }
  int32_t main() {
    long long nt;
    nt = 1;
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    while (nt--) {
      solve();
    }
  }
  