N4. Формальный исполнитель

#include <iostream>
#include <cmath>
#include <vector>
#include <set>
#include <string>
#include <algorithm>
using namespace std;
using ll = long long;
void cv(vector <int> v){
    for (auto x: v) cout<<x<<' ';
    cout<<'\n';
}
vector <int> dx = {-1, -1, 0, 1, 1, 1, 0, -1};
vector <int> dy = {0, 1, 1, 1, 0, -1, -1, -1};
vector <vector <int> > dsk(8, vector <int> (8,0));
vector <pair <int, int> > srt = {{0,2}, {1,3}, {2,1}, {2,2,}, {2,3} };

bool gd(int i, int j){
    return i >= 0 && i < 8 && j>=0 && j < 8;
}

void al(){
    vector <vector <int> > hlf(8, vector <int> (8, -9));
    for (int I = 0; I < 8 ;++I){
        for (int J = 0; J < 8;++J){
            int bl = 0;
            for (int q = 0;q<8;++q){
                int X = I + dx[q];
                int Y = J + dy[q];
                if (gd(X,Y) && dsk[X][Y] == 1){
                    bl++;
                }
            }
            if (dsk[I][J] == 0){
                    if (bl==3){
                        hlf[I][J] = 1;
                    }
                    else hlf[I][J] = 0;

            }else {
                if (bl == 2 || bl == 3){
                    hlf[I][J] = 1;
                }else hlf[I][J] = 0;
            }
        }
    }
    dsk = hlf;
}

int main()
{
    /*ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);*/
    for (auto c: srt) dsk[c.first][c.second] = 1;
    for (int i = 0; i < 8;++i){
            for (int j=0;j<8;++j){
                cout<<dsk[i][j];
            }cout<<'\n';
        }cout<<"\n\n";

    for (int i = 0;i<3;++i){
        al();
        for (int i = 0; i < 8;++i){
            for (int j=0;j<8;++j){
                cout<<dsk[i][j];
            }cout<<'\n';
        }cout<<"\n\n";
    }
    vector <pair<int,int> > ans;
    for (int i = 0; i < 8;++i){
        for (int j=0;j<8;++j){
            if (dsk[i][j]){
                ans.push_back({i,j});
            }
        }
    }
    string sh = "ABCDEFGH";
    for (auto f: ans) cout<<sh[f.second]<<f.first+1<<'\n';

}