2 types of query:

1. add value to x y
2. output the sum of $ value*(|a – x|+|b-y|)$ for every pair(a, b, v).

In the start you are given $ n$ (number of querys), $ m$ (maximal dimension for square matrix).

In the next $ n$ lines you are given either:

1. 1 x y v, you should add value v to x y coordinate, or
2. 2 x y calculate the $ distance * value$ between every $ a$ and $ b$ that have some values (distance is being calculated: $ |a- x| + |b – y|$ ). So the formula for one spot in matrix would be $ v * (|a – x| + |b – y|)$ for a given $ a$ and $ b$ .

My code works, but it is too slow.

Constraints:
$ n <= 250’000$
$ m <= 10^9$
$ X_i$ and $ Y_i <= m$
$ V_i$ <= $ 10^4$

time limit: 2.5s

My code:

for (int i = 0; i < n; ++i){      int el;     cin >> el;      if (el == 1){         int a, b, c;         cin >> a >> b >> c;          x.push_back(a);         y.push_back(b);         v.push_back(c);     }      if (el == 2){         int a, b;         cin >> a >> b;          ll sol = 0;          for (int j = 0; j < x.size(); ++j){             sol += v[j] * (abs(a - x[j]) + abs(b - y[j]));         }         cout << sol << "\n";      }  }