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;
  pair<long long, long long> *matrix;
  bool *isOpen;
  int w, h, q;
  long long sum;
  void print() {
    for (int y = 0; y < h; y++) {
      for (int x = 0; x < w; x++) {
        if (isOpen[y * w + x]) {
          cout << "[" << matrix[y * w + x].first << ":"
               << matrix[y * w + x].second << "] ";
        } else {
          cout << "[!:!] ";
        }
      }
      cout << endl;
    }
    cout << endl;
    cout << endl;
  }
  void changeOpen(int x, int y) {
    int mult = isOpen[y * w + x] ? -1 : 1;
    sum += mult * (matrix[y * w + x].first + matrix[y * w + x].second + 1);
    isOpen[y * w + x] = (isOpen[y * w + x] + 1) % 2;
    bool R = true, D = true;
    for (int i = 1; i <= min(x, y); i++) {
      if (!isOpen[(y - i + 1) * w + (x - i)]) D = false;
      if (!isOpen[(y - i) * w + (x - i + 1)]) R = false;
      if (D) {
        matrix[(y - i) * w + (x - i)].first +=
            mult * (matrix[y * w + x].first + 1);
      }
      if (R) {
        matrix[(y - i) * w + (x - i)].second +=
            mult * (matrix[y * w + x].second + 1);
      }
      if (!isOpen[(y - i) * w + (x - i)] || (!D && !R)) break;
      if (D) sum += mult * (matrix[y * w + x].first + 1);
      if (R) sum += mult * (matrix[y * w + x].second + 1);
    }
    for (int i = 1; i <= min(x, y + 1); i++) {
      matrix[(y - i + 1) * w + (x - i)].second +=
          mult * (matrix[y * w + x].first + 1);
      if (x - i >= 0 && !isOpen[(y - i + 1) * w + (x - i)]) break;
      sum += mult * (matrix[y * w + x].first + 1);
      if (x - i >= 0 && y - i >= 0 && !isOpen[(y - i) * w + (x - i)]) break;
    }
    for (int i = 1; i <= min(x + 1, y); i++) {
      matrix[(y - i) * w + (x - i + 1)].first +=
          mult * (matrix[y * w + x].second + 1);
      if (y - i >= 0 && !isOpen[(y - i) * w + (x - i + 1)]) break;
      sum += mult * (matrix[y * w + x].second + 1);
      if (x - i >= 0 && y - i >= 0 && !isOpen[(y - i) * w + (x - i)]) break;
    }
  }
  int main() {
    cin >> w >> h >> q;
    sum = 0;
    matrix = new pair<long long, long long>[w * h];
    isOpen = new bool[w * h];
    int k1 = 0, k2 = 0;
    if (h > w) {
      k1 = (h - w);
      k2 = 0;
    } else if (w > h) {
      k1 = 0;
      k2 = (w - h);
    }
    for (int y = h - 1; y >= 0; y--) {
      for (int x = w - 1; x >= 0; x--) {
        int D = min(w - x - 1, h - y - 1);
        matrix[y * w + x].first = D * 2 + (y - k1 < x - k2 ? 1 : 0);
        matrix[y * w + x].second = D * 2 + (y - k1 > x - k2 ? 1 : 0);
        isOpen[y * w + x] = true;
        sum += matrix[y * w + x].first + matrix[y * w + x].second + 1;
      }
    }
    while (q--) {
      int x, y;
      cin >> x >> y;
      changeOpen(x - 1, y - 1);
      cout << sum << endl;
    }
    return 0;
  }
  