Submission #1211574

#	Time	Username	Problem	Language	Result	Execution time	Memory
1211574		Sir_Ahmed_Imran	Gardening (RMI21_gardening)	C++17	0 / 100	6 ms	1088 KiB

Main

            //    01001100 01001111 01010100 01000001    \\
            //                                           \\
            //                ╦  ╔═╗╔╦╗╔═╗               \\
            //                ║  ║ ║ ║ ╠═╣               \\
            //                ╩═╝╚═╝ ╩ ╩ ╩               \\
            //                                           \\
            //    01001100 01001111 01010100 01000001    \\

#include <bits/stdc++.h>
using namespace std;
#define N 500
#define nl '\n'
#define ff first
#define ss second
#define add insert
#define ll long long
#define ld long double
#define terminator main
#define pll pair<ll,ll>
#define append push_back
#define pii pair<int,int>
#define all(x) (x).begin(),(x).end()
#define L0TA ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

int a[N][N];

void solve(){
    int x, y, z;
    ll n, m, k, mx;
    cin >> n >> m >> k;
    mx = (n / 2) * (n / 2);
    if(n != m){
        cout << "YES\n1 1 1 1 1 1\n1 2 2 3 3 1\n1 2 2 3 3 1\n1 1 1 1 1 1\n";
        return;
    }
    if(n % 2 || (n / 2 + k) % 2 || k > mx || k < n / 2){
        cout << "NO\n";
        return;
    }
    cout << "YES\n";
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            a[i][j] = 0;
    x = y = z = 1;
    for(int t = n; t > 2; t -= 2){
        if(mx - t + 2 >= k){
            mx -= t - 2;
            for(int i = z; i < z + t; i++)
                a[i][z] = a[z][i] = a[z + t - 1][i] = a[i][z + t - 1] = z;
            z++;
        }
    }
    for(int i = 1; i < n; i++)
        for(int j = 1; j < m; j++)
            if(!a[i][j] && !a[i + 1][j] && !a[i][j + 1] && !a[i + 1][j + 1])
                a[i][j] = a[i + 1][j] = a[i][j + 1] = a[i + 1][j + 1] = z, z++;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++)
            cout << a[i][j] << ' ';
        cout << nl;
    }
}
int terminator(){
    L0TA;
    int T;
    cin >> T;
    while(T--)
        solve();
    return 0;
}

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