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;
  using LL = long long;
  const int MAXN = 1010;
  int n, m;
  bool z[MAXN][MAXN];
  LL ans = 0;
  int GetUp(int a, int b, int dir) {
    int ans = 0;
    while (1 <= a && a <= n && 1 <= b && b <= m && !z[a][b]) {
      ++ans;
      if (dir == 1)
        --a;
      else
        --b;
      dir ^= 1;
    }
    return ans;
  }
  int GetDown(int a, int b, int dir) {
    int ans = 0;
    while (1 <= a && a <= n && 1 <= b && b <= m && !z[a][b]) {
      ++ans;
      if (dir == 1)
        ++a;
      else
        ++b;
      dir ^= 1;
    }
    return ans;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int q;
    cin >> n >> m >> q;
    for (int i = 1; i <= n; ++i) {
      int L = GetDown(i, 1, 1);
      ans += L * (L + 1) / 2;
    }
    for (int i = 1; i <= m; ++i) {
      int L = GetDown(1, i, 0);
      ans += L * (L + 1) / 2;
    }
    ans -= n * m;
    while (q--) {
      int a, b;
      cin >> a >> b;
      if (z[a][b]) {
        int L = GetUp(a, b - 1, 1);
        int R = GetDown(a + 1, b, 0);
        ans -= L * (L + 1) / 2;
        ans -= R * (R + 1) / 2;
        ans += (L + R + 2) * (L + R + 1) / 2;
        L = GetUp(a - 1, b, 0);
        R = GetDown(a, b + 1, 1);
        ans -= L * (L + 1) / 2;
        ans -= R * (R + 1) / 2;
        ans += (L + R + 2) * (L + R + 1) / 2;
        --ans;
      } else {
        int L = GetUp(a, b - 1, 1);
        int R = GetDown(a + 1, b, 0);
        ans += L * (L + 1) / 2;
        ans += R * (R + 1) / 2;
        ans -= (L + R + 2) * (L + R + 1) / 2;
        L = GetUp(a - 1, b, 0);
        R = GetDown(a, b + 1, 1);
        ans += L * (L + 1) / 2;
        ans += R * (R + 1) / 2;
        ans -= (L + R + 2) * (L + R + 1) / 2;
        ++ans;
      }
      z[a][b] = !z[a][b];
      cout << ans << "\n";
    }
    return 0;
  }
  