제출 #689802

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
689802		KiriLL1ca	Red-blue table (IZhO19_stones)	C++17	0 / 100	143 ms	1224 KiB

stones

#include <bits/stdc++.h>
#define sz(x) (int)((x).size())
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define fr first
#define sc second
#define pw(x) (1ll << x)
#define bcnt(x) (__builtin_popcountll(x))

using namespace std;

typedef long long ll;
typedef pair <int, int> pii;

template <typename T> inline bool umax (T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template <typename T> inline bool umin (T &a, const T &b) { if (a > b) { a = b; return 1; } return 0; }

pair <int, vector <int>> bst;
int N, M;
inline int cost (vector <int> a) {
    int res = 0;
    for (int i = 0; i < N; ++i) {
        int s = 0;
        for (int j = 0; j < M; ++j) s += ((a[i] >> j & 1) ? +1 : -1);
        res += s > 0;
    }
    for (int i = 0; i < M; ++i) {
        int s = 0;
        for (int j = 0; j < N; ++j) s += ((a[j] >> i & 1) ? +1 : -1);
        res += s < 0;
    }
    return res;
}

inline void brute (int n, int m, vector <int> a) {
    if (!~n) return void(umax(bst, {cost(a), a}));
    for (int j = 0; j < pw(m); ++j) {
        a.pb(j); brute(n - 1, m, a);
        a.pop_back();
    }
}

inline void solve () {
    cin >> N >> M;
    if (N <= 4 && M <= 4) {
        bst.fr = -1;
        brute(N, M, vector <int> ());
        cout << bst.fr << endl;
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < M; ++j) {
                cout << ((bst.sc[i] >> j & 1) ? '+' : '-');
            }
            cout << endl;
        }
    }
    if (N == M) {
        cout << N + 1 << endl;
        cout << string(N, '+') << endl;
        for (int i = 1; i < N; ++i) {
            cout << '-' << string(N - 1, '+') << endl;
        }
    }
}

signed main ()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    #ifdef LOCAL
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif // LOCAL

    int t = 1; cin >> t;
    while (t--) solve();
    return 0;
}

• Subtask 1: 0 / 17
• Subtask 2: 0 / 10
• Subtask 3: 0 / 16
• Subtask 4: 0 / 11
• Subtask 5: 0 / 15
• Subtask 6: 0 / 31