p229.ans1
======================
YES
100
010
000

=================
p229.ans2
======================
NO

=================
p229.cpp
======================
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;
vector<string> g;

void read() {
    cin >> n;
    g.assign(n, "");
    cin >> g;
}

void solve() {
    // We need to split Q into A and B such that B is the image of A under a
    // rotation by a multiple of 90 degrees followed by a translation. Since
    // N <= 20, |Q| <= 400 and we can afford to brute force over the
    // transformation. There are only four rotations to try, and for each one
    // we can fix the lex-smallest cell c0 of Q to lie in A. Then f(c0) must
    // be some other cell of Q, which leaves only |Q| candidate shift vectors
    // per rotation - O(|Q|) transformations in total. If c0 actually belongs
    // to B in the true partition, the inverse rotation handles that case, so
    // anchoring c0 in A loses nothing.
    //
    // Fix a transformation f = R^k + v. The requirement Q = A U f(A) induces
    // a functional graph on Q: draw a directed edge c -> f(c) whenever both
    // c and f(c) are in Q. Each vertex has in- and out-degree at most one,
    // so the components are simple paths and simple cycles. Along every
    // component the labels A/B must alternate, because c in A forces f(c)
    // in B and vice versa. A path endpoint with no out-edge is forced into
    // B, an endpoint with no in-edge into A, so the path needs even length.
    // Cycles also need even length, but f has order dividing 4 and so the
    // cycles have length 2 or 4 - the parity is automatic. The only test
    // worth running per transformation is "are all path components even?".
    //
    // If some f passes the test, we recover A by taking the cells coloured
    // first along each component and print it. If nothing works, the answer
    // is "NO".

    vector<pair<int, int>> cells;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            if(g[i][j] == '1') {
                cells.push_back({i, j});
            }
        }
    }

    int total = cells.size();
    if(total == 0) {
        cout << "YES" << '\n';
        for(int i = 0; i < n; i++) {
            cout << string(n, '0') << '\n';
        }
        return;
    }

    if(total % 2 != 0) {
        cout << "NO" << '\n';
        return;
    }

    auto rotate_k = [](int r, int c, int k) {
        for(int i = 0; i < k; i++) {
            int nr = c, nc = -r;
            r = nr;
            c = nc;
        }
        return make_pair(r, c);
    };

    auto in_q = [&](int r, int c) {
        return 0 <= r && r < n && 0 <= c && c < n && g[r][c] == '1';
    };

    auto [c0r, c0c] = cells[0];

    auto try_transform = [&](int k, int dr, int dc) -> bool {
        auto f = [&](int r, int c) {
            auto [nr, nc] = rotate_k(r, c, k);
            return make_pair(nr + dr, nc + dc);
        };
        auto f_inv = [&](int r, int c) {
            auto [nr, nc] = rotate_k(r - dr, c - dc, (4 - k) % 4);
            return make_pair(nr, nc);
        };

        vector<vector<int>> color(n, vector<int>(n, -1));

        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                if(g[i][j] != '1' || color[i][j] != -1) {
                    continue;
                }

                int sr = i, sc = j;
                while(true) {
                    auto [pr, pc] = f_inv(sr, sc);
                    if(!in_q(pr, pc) || (pr == i && pc == j)) {
                        break;
                    }
                    sr = pr;
                    sc = pc;
                }

                int col = 0, length = 0;
                int cr = sr, cc = sc;
                while(color[cr][cc] == -1) {
                    color[cr][cc] = col;
                    col ^= 1;
                    length++;
                    auto [nr, nc] = f(cr, cc);
                    if(!in_q(nr, nc)) {
                        break;
                    }
                    cr = nr;
                    cc = nc;
                }

                if(length % 2 != 0) {
                    return false;
                }
            }
        }

        cout << "YES" << '\n';
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                if(g[i][j] == '1' && color[i][j] == 0) {
                    cout << '1';
                } else {
                    cout << '0';
                }
            }
            cout << '\n';
        }
        return true;
    };

    for(int k = 0; k < 4; k++) {
        auto [rr, rc] = rotate_k(c0r, c0c, k);
        for(auto [tr, tc]: cells) {
            if(tr == c0r && tc == c0c) {
                continue;
            }
            int dr = tr - rr, dc = tc - rc;
            if(try_transform(k, dr, dc)) {
                return;
            }
        }
    }

    cout << "NO" << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}

=================
p229.in1
======================
3
100
011
010

=================
p229.in2
======================
3
100
101
010

=================
statement.txt
======================
229. Divide and conquer
time limit per test: 0.25 sec.
memory limit per test: 4096 KB
input: standard
output: standard



You are to write a program which will divide set Q of squares into two parts, in such a way that one of them can be presented as other after rotations on 90 degrees and shifts along some vector. Set Q is the subset of NxN square.

Input
The first line of the input contains the integer N (1 <= N <= 20). The following N lines contain description of the set Q (symbol '1' means that square belongs to Q and '0' otherwise).

Output
Write 'YES' in the first line of the output, if such division exists, and 'NO' otherwise. In the first case output one of the subsets of Q.

Sample test(s)

Input

Test #1
3
100
011
010

Test #2
3
100
101
010

Output

Test #1
YES
100
010
000

Test #2
NO
Author:	Andrew V. Lazarev
Resource:	
Date:	







=================
