ZJ c292

// Judge: https://zerojudge.tw/ShowProblem?problemid=c292

#include <bits/stdc++.h>
using namespace std;

/*
    This problem is simple but a little hard to code for beginners.
    For beginners, they may write four or even more 'if' & 'else' to the for directions respectively.
    However, we have a faster, concise, and powerful method by using arrays.

    Think about the relative position for representing directions:
    Left, Up, Right, and Down!
    Notice that how I designed the order.

    The four directions actualy correspond to what problem says:
        0 -> Left, 1 -> Up, 2 -> Right, 3 -> Down
    Another advantage of this order is that, if your current direction is D, you can do
        'D = (D+1)%4' to turn right,
        'D = (D+3)%4' to turn left.
    See my code below.
*/
//direction:L  U   R  D
int dx[] = {0, -1, 0, 1};
int dy[] = {-1, 0, 1, 0};

int arr[49][49];// input numbers
bool vis[49][49];// vis[i][j] is 1 when arr[i][j] has been printed

int main(){
    // inputs
    int n, dir;
    cin >> n >> dir;
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            cin >> arr[i][j];
        }
    }
    --dir;// that way, '(dir+1)%4' in the line 43 will be the corrent direction
    int x = n/2, y = n/2;
    while(0 <= x && x < n && 0 <= y && y < n){// checking boundaries
        cout << arr[x][y];
        vis[x][y] = 1;
        int tryd = (dir+1) % 4;// the right direction
        if(vis[x + dx[tryd]][y + dy[tryd]] == 0){
            // change direction if available
            dir = tryd;
        }
        // move
        x += dx[dir];
        y += dy[dir];
    }
    cout << '\n';
    return 0;
}