The idea of this problem was originally proposed by Keyence.

This problem can be solved with a for statement. If you didn’t manage to solve it, let’s understand the solution of this problem to grasp the for statement.

A for statement is a statement to describe a loop structure, that is, an algorithm of repeating a specific process until a specific condition is satisfied.
Especially, the statement for(int i = 0; i < N; i++) (or for i in range(N) in Python) is the commonly used one. With this statement, one can apply the same process for all \(i=0, 1, \dots, N-1\), so we can use it to perform some operation on an array of length \(N\). For example, in ABC272A, you are asked to “find the sum of \(A_0, A_1, \dots\), and \(A_{N-1}\),” which can be solved with a for statement as follows:

// A : array of type int of length N
int s = 0;
for(int i = 0; i < N; i++){
  s += A[i];
}

A bit advanced use of the for statement is to nest for statements to describe an algorithm with nested loops, like a double or triple loop.
In this problem, you are asked to find the number of integers \(i\) such that \(C_{i,j}\) is #; this can be solved by iterating a double loop over the variables \(i\) and \(j\), which can be described by combining for statements in the form for(int i = 0; i < N; i++) in a nested manner as follows:

// C : two-dimensional array of type char of size H x W
vector<int> ans(W);
for (int i = 0; i < H; i++) {
  for (int j = 0; j < W; j++) {
    if (C[i][j] == '#') ans[j]++;
  }
}

The complexity is \(\mathrm{O}(N^2)\), which is fast enough. Therefore, the problem has been solved. The idea of nested for statement can be found in the past Problems B several times. If you struggled with this problem, try solving the following past problems as well. Past Problem 1 Past problem 2

• Sample code (C++)

#include <iostream>
#include <string>
#include <vector>
using namespace std;
int main() {
  int H, W;
  cin >> H >> W;
  vector<string> C(H);
  for (int i = 0; i < H; i++) cin >> C[i];
  vector<int> ans(W);
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      if (C[i][j] == '#') ans[j]++;
    }
  }
  for (int j = 0; j < W; j++) cout << ans[j] << " \n"[j == W - 1];
}