Submission #570719

#	Time	Username	Problem	Language	Result	Execution time	Memory
570719		Theo830	Gardening (RMI21_gardening)	C++17	5 / 100	27 ms	852 KiB

Main

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF = 1e9+7;
const ll MOD = 998244353;
typedef pair<ll,ll> ii;
#define iii pair<ii,ll>
#define ull unsigned ll
#define f(i,a,b) for(ll i = a;i < b;i++)
#define pb push_back
#define vll vector<ll>
#define F first
#define S second
#define all(x) (x).begin(), (x).end()
///I hope I will get uprating and don't make mistakes
///I will never stop programming
///sqrt(-1) Love C++
///Please don't hack me
///@TheofanisOrfanou Theo830
///Think different approaches (bs,dp,greedy,graphs,shortest paths,mst)
///Stay Calm
///Look for special cases
///Beware of overflow and array bounds
///Think the problem backwards
///Training
const ll dx[4] = {0,0,1,-1};
const ll dy[4] = {1,-1,0,0};
bool in(ll x,ll y,ll n,ll m){
    if(min(x,y) >= 0 && x < n && y < m){
        return true;
    }
    return false;
}
int main(void){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    ll t;
    cin>>t;
    while(t--){
        ll n,m,k;
        cin>>n>>m>>k;
        bool ok = 1;
        if(n % 2 == 1 || m % 2 == 1){
            ok = 0;
        }
        else if(k > (n / 2) * (m / 2)){
            ok = 0;
        }
        else if(k < max(n / 2,m / 2)){
            ok = 0;
        }
        else{
            ll num = (n * m) / 4;
            num -= k;
            ll X = sqrt(num) + 1;
            if(X * (X - 1) != num){
                ok = 0;
            }
        }
        if(!ok){
            cout<<"NO\n";
        }
        else{
            ll ans[n+5][m+5];
            ll N = n,M = m;
            ll lx = 0,rx = N - 1,ly = 0,ry = M - 1;
            while(k){
                if(k == min(N / 2,M / 2)){
                    f(j,ly,ry + 1){
                       ans[lx][j] = k;
                       ans[rx][j] = k;
                    }
                    f(i,lx,rx + 1){
                        ans[i][ly] = k;
                        ans[i][ry] = k;
                    }
                    lx++;
                    rx--;
                    ly++;
                    ry--;
                    N -= 2;
                    M -= 2;
                    k--;
                }
                else{
                    if(N >= M){
                        f(j,ly,ry + 1){
                            ans[lx][j] = ans[lx + 1][j] = ans[lx][j+1] = ans[lx+1][j+1] = k;
                            k--;
                            j++;
                        }
                        N -= 2;
                        lx += 2;
                    }
                    else{
                        f(i,lx,rx + 1){
                            ans[i][ly] = ans[i+1][ly] = ans[i][ly+1] = ans[i+1][ly+1] = k;
                            k--;
                            i++;
                        }
                        M -= 2;
                        ly += 2;
                    }
                }
            }
            f(i,0,n){
                f(j,0,m){
                    ll posa = 0;
                    f(k,0,4){
                        ll x = i + dx[k],y = j + dy[k];
                        if(in(x,y,n,m)){
                            posa += ans[i][j] == ans[x][y];
                        }
                    }
                    ok &= posa == 2;
                }
            }
            if(ok){
                cout<<"YES\n";
                f(i,0,n){
                    f(j,0,m){
                        cout<<ans[i][j]<<" ";
                    }
                    cout<<"\n";
                }
            }
            else{
                assert(0);
                cout<<"NO\n";
            }
        }
    }
}

• Subtask 1: 5 / 5
• Subtask 2: 0 / 6
• Subtask 3: 0 / 10
• Subtask 4: 0 / 18
• Subtask 5: 0 / 39
• Subtask 6: 0 / 22