General notes

When treating strings in a programming language, in many languages the first character is the \(0\)-th character, so be sure to check the language you use to implement.

Solution 1. Try all \(A, B, C\), and \(D\)

First, we introduce a naive solution. An efficient solution is explained in Solution 2.
There are only \(3025(=55 \times 55)\) tuples of integers \((A,B,C,D)\) such that

• \(1 \le A \le B \le 10\);
• \(1 \le C \le D \le 10\).

By actually generating from \(S_1\) through \(S_{10}\) and comparing with the input, you can get AC (accepted) for this problem with about \(302500\) operations (multiplied by a constant factor).

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  vector<string> s(10);
  for(int i=0;i<10;i++){cin >> s[i];}

  for(int a=1;a<=10;a++){
    for(int b=a;b<=10;b++){
      for(int c=1;c<=10;c++){
        for(int d=c;d<=10;d++){
          vector<string> t(10);
          for(int i=1;i<=10;i++){
            for(int j=1;j<=10;j++){
              if(a<=i && i<=b && c<=j && j<=d){
                t[i-1].push_back('#');
              }
              else{
                t[i-1].push_back('.');
              }
            }
          }
          if(s==t){
            cout << a << " " << b << "\n" << c << " " << d << "\n";
          }
        }
      }
    }
  }
  return 0;
}

Solution 2. An efficient solution

The generated string contains exactly one rectangle consisting of #s. Thus, one can determine \(A, B, C\), and \(D\) as follows:

• \(A=\) (The topmost row containing #);
• \(B=\) (The bottommost row containing #);
• \(C=\) (The leftmost column containing #);
• \(D=\) (The rightmost column containing #).

#include<bits/stdc++.h>

using namespace std;

int main(){
  vector<string> s(10);
  for(int i=0;i<10;i++){cin >> s[i];}

  int a=1e9,b=-1e9;
  int c=1e9,d=-1e9;

  for(int i=0;i<10;i++){
    for(int j=0;j<10;j++){
      if(s[i][j]=='#'){
        a=min(a,i+1);
        b=max(b,i+1);
        c=min(c,j+1);
        d=max(d,j+1);
      }
    }
  }
  
  cout << a << " " << b << "\n";
  cout << c << " " << d << "\n";
  return 0;
}