Submission #1296113

#TimeUsernameProblemLanguageResultExecution timeMemory
1296113MunkhErdeneRed-blue table (IZhO19_stones)C++17
100 / 100
28 ms33840 KiB
#include <bits/stdc++.h>
using namespace std;
#define ull unsigned long long
#define lll __int128
#define ll long long
const ll mod = 1e9 + 7;
const ll mod1 = 998244353;
const ll naim = 1e9;
const ll max_bit = 60;
const ull tom = ULLONG_MAX;
const ll MAXN = 100005;
const ll LOG = 20;
const ll NAIM = 1e18;
const ll N = 2e6 + 5;
int main() {
    #define pb push_back
    #define ff first
    #define ss second
    #define _ << " " <<
    #define yes cout<<"YES\n"
    #define no cout<<"NO\n"
    #define all(x) x.begin(),x.end()
    #define rall(x) x.rbegin(),x.rend()
    #define BlueCrowner ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    #define FOR(i, a, b) for (ll i = (a); i < (b); i++)
    #define FORD(i, a, b) for (ll i = (a); i >= (b); i--)
    // ---------- GCD ----------
    auto gcd = [&](ll a, ll b) {
        while (b) {
            a %= b;
            swap(a, b);
        }
        return a;
    };
    // ---------- LCM ----------
    auto lcm = [&](ll a, ll b) {
        return a / gcd(a, b) * b;
    };
    // ---------- Modular Exponentiation ----------
    function<ll(ll, ll, ll)> modpow = [&](ll a, ll b, ll m) {
        ll c = 1;
        a %= m;
        while (b > 0) {
            if (b & 1) c = c * a % m;
            a = a * a % m;
            b >>= 1;
        }
        return c;
    };
    // ---------- Modular Inverse (Fermat’s Little Theorem) ----------
    function<ll(ll, ll)> modinv = [&](ll a, ll m) {
        return modpow(a, m - 2, m);
    };
    // ---------- Factorials and Inverse Factorials ----------
    vector<ll> fact(N), invfact(N);
    auto pre_fact = [&](ll n = N-1, ll m = mod) {
        fact[0] = 1;
        for (ll i = 1; i <= n; i++) fact[i] = fact[i-1] * i % m;
        invfact[n] = modinv(fact[n], m);
        for (ll i = n; i > 0; i--) invfact[i-1] = invfact[i] * i % m;
    };
    // ---------- nCr ----------
    auto nCr = [&](ll n, ll r, ll m = mod) {
        if (r < 0 || r > n) return 0LL;
        return fact[n] * invfact[r] % m * invfact[n-r] % m;
    };
    // ---------- Sieve of Eratosthenes ----------
    vector<ll> primes;
    vector<bool> is_prime(N);
    auto sieve = [&](ll n = N-1) {
        fill(is_prime.begin(), is_prime.begin() + n + 1, true);
        is_prime[0] = is_prime[1] = false;
        for (ll i = 2; i * i <= n; i++) {
            if (is_prime[i]) {
                for (ll j = i * i; j <= n; j += i)
                    is_prime[j] = false;
            }
        }
        for (ll i = 2; i <= n; i++)
            if (is_prime[i]) primes.pb(i);
    };
    function<void()> solve = [&]() {
        ll n, m; cin >> n >> m;
        vector<vector<char>> ans(n, vector<char>(m, '-'));
        ll sum = 0;
        if(n >= m){
            sum = n;
            ll cur = 0;
            ll x = (n - 1) / 2;
            vector<ll> cnt(n, 0);
            FOR(i, 0, m){
                FOR(j, 0, x){
                    ans[(cur + j) % n][i] = '+';
                    cnt[(cur + j) % n]++;
                }
                cur = (cur + x) % n;
            }
            FOR(i, 0, n){
                ll j = 0;
                while(cnt[i] < m - (m - 1) / 2){
                    if(ans[i][j] == '-'){
                        ans[i][j] = '+';
                        cnt[i]++;
                    }
                    j++;
                }
            }
            FOR(i, 0, m){
                ll cnt = 0;
                FOR(j, 0, n){
                    cnt += (ans[j][i] == '-');
                }
                if(cnt >= n - (n - 1) / 2) sum++;
            }
        }
        else{
            sum = m;
            ll cur = 0;
            ll x = (m - 1) / 2;
            vector<ll> cnt(m, 0);
            FOR(i, 0, n){
                FOR(j, 0, x){
                    ans[i][(cur + j) % m] = '+';
                    cnt[(cur + j) % m]++;
                }
                cur = (cur + x) % m;
            }
            FOR(j, 0, m){
                ll i = 0;
                while(cnt[j] < n - (n - 1) / 2){
                    if(ans[i][j] == '-'){
                        ans[i][j] = '+';
                        cnt[j]++;
                    }
                    i++;
                }
            }
            FOR(i, 0, n){
                ll cnt = 0;
                FOR(j, 0, m){
                    cnt += (ans[i][j] == '-');
                }
                if(cnt >= m - (m - 1) / 2) sum++;
            }
            FOR(i, 0, n){
                FOR(j, 0, m){
                    ans[i][j] = (ans[i][j] == '+') ? '-' : '+'; 
                }
            }
        }
        cout << sum << '\n';
        FOR(i, 0, n){
            FOR(j, 0, m){
                cout << ans[i][j];
            }
            cout << "\n";
        }
    };
    BlueCrowner;
    int t = 1;
    cin >> t;
    while (t--) {
        solve();
    }
}
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