답안 #237495

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
237495	2020-06-07T06:15:19 Z	kartel	Spirale (COCI18_spirale)	C++14	72 / 80	16 ms	396 KB

spirale

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define in(x) freopen(x, "r", stdin)
#define out(x) freopen(x, "w", stdout)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-O3")
#define F first
#define S second
#define pb push_back
#define N +55
#define MaxS N * N
#define M ll(1e9 + 7)
#define sz(x) (int)x.size()
//#define re return
#define oo ll(1e18)
#define el '\n'
#define pii pair <int, int>
using namespace std;
//using namespace __gnu_pbds;
//typedef tree <int, null_type, less_equal <int> , rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef long long ll;
typedef long double ld;

int a[N][N], i, j, k, kol, q, x, y, cmd, n, m;

int main()
{
    srand(time(0));
    ios_base::sync_with_stdio(0);
    iostream::sync_with_stdio(0);
    ios::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);

//    in("input.txt");
//    out("output.txt");

    cin >> n >> m >> q;
    for (i = 1; i <= n; i++)
        for (j = 1; j <= m; j++) a[i][j] = 1e9;
    while (q--)
    {
        cin >> x >> y >> cmd;
        if (cmd == 0)
        {
            a[x][y] = 1;
            kol = 1;
            k = 1;
            while (kol < 50 * 50)
            {
                i = 1;
                while (kol < 50 * 50 && i <= k)
                {
                    x--;
                    i++;
                    kol++;
                    if (x >= 1 && y >= 1 && x <= n && y <= m) a[x][y] = min(a[x][y], kol);
                }
                i = 1;
                while (kol < 50 * 50 && i <= k)
                {
                    y++;
                    i++;
                    kol++;
                    if (x >= 1 && y >= 1 && x <= n && y <= m) a[x][y] = min(a[x][y], kol);
                }
                k++;

                i = 1;
                while (kol < 50 * 50 && i <= k)
                {
                    x++;
                    i++;
                    kol++;
                    if (x >= 1 && y >= 1 && x <= n && y <= m) a[x][y] = min(a[x][y], kol);
                }
                i = 1;
                while (kol < 50 * 50 && i <= k)
                {
                    y--;
                    i++;
                    kol++;
                    if (x >= 1 && y >= 1 && x <= n && y <= m) a[x][y] = min(a[x][y], kol);
                }
                k++;
            }
        }
        else
        {
            a[x][y] = 1;
            kol = 1;
            k = 1;
            while (kol < 50 * 50)
            {
                i = 1;
                while (kol < 50 * 50 && i <= k)
                {
                    x--;
                    i++;
                    kol++;
                    if (x >= 1 && y >= 1 && x <= n && y <= m) a[x][y] = min(a[x][y], kol);
                }
                i = 1;
                while (kol < 50 * 50 && i <= k)
                {
                    y--;
                    i++;
                    kol++;
                    if (x >= 1 && y >= 1 && x <= n && y <= m) a[x][y] = min(a[x][y], kol);
                }
                k++;

                i = 1;
                while (kol < 50 * 50 && i <= k)
                {
                    x++;
                    i++;
                    kol++;
                    if (x >= 1 && y >= 1 && x <= n && y <= m) a[x][y] = min(a[x][y], kol);
                }
                i = 1;
                while (kol < 50 * 50 && i <= k)
                {
                    y++;
                    i++;
                    kol++;
                    if (x >= 1 && y >= 1 && x <= n && y <= m) a[x][y] = min(a[x][y], kol);
                }
                k++;
            }
        }
    }
    for (i = 1; i <= n; i++, cout << el)
      for (j = 1; j <= m; j++) cout << a[i][j] << " ";
}

//  x ^ 2 + y ^ 2 = 1
//  x * a_i + y * b_i
//  a_i = -b_i * tan(alpha)
//  a_i / -b_i = tan(alpha)
//  alpha = atan(a_i / (-b_i))

Subtask #1

72.0 / 80.0

#	결과	실행 시간	메모리	Grader output
1	Correct	5 ms	396 KB	Output is correct
2	Correct	5 ms	384 KB	Output is correct
3	Correct	4 ms	384 KB	Output is correct
4	Correct	5 ms	384 KB	Output is correct
5	Correct	4 ms	384 KB	Output is correct
6	Correct	5 ms	384 KB	Output is correct
7	Correct	4 ms	384 KB	Output is correct
8	Correct	5 ms	384 KB	Output is correct
9	Correct	16 ms	384 KB	Output is correct
10	Incorrect	5 ms	384 KB	Output isn't correct