1097 : [기초-종합+배열] 바둑알 십자 뒤집기

문제 설명   

 

 

아버지를 기다리던 경곽이는 검정/흰 색 바둑알을 바둑판에 꽉 채워 깔아 놓고 놀다가...

"음... 이거... 십(+)자 뒤집기를 해볼까?"하고 생각하였다.

바둑판(19×19$19\times 19$)에 흰돌(1$1$)/검정돌(0$0$) 모두 꽉 채워놓여있을 때,

n$n$개의 좌표를 입력받아 십(+)자 뒤집기한 결과를 출력하는 프로그램을 작성해보자.

입력

바둑알이 깔려 있는 상황이 19×19$19\times 19$ 로 입력된다.
십자바꾸기 횟수(n$n$)가 입력된다.
십자바꾸기 좌표가 횟수(n$n$) 만큼 입력된다. 
단, n$n$은 10$10$이하의 자연수이다.

출력

십자 바꾸기 결과를 출력한다.

입력 예시

0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 2 10 10 12 12

출력 예시

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

#include <iostream>
int main()
{
    using namespace std;
    int a, a1[19][19] = { 0, }, x[19], y[19];
 
    for (int i = 0; i < 19; i++)
        for (int j = 0; j < 19; j++)
            cin >> a1[i][j];
 
    cin >> a;
 
    for (int i = 0; i < a; i++)
        cin >> x[i] >> y[i];
 
     
    for (int k = 0; k < a; k++)
    {
        for (int i = 0; i < 19; i++)
        {
            a1[i][y[k] - 1] = !a1[i][y[k] - 1];
        }
        for (int j = 0; j < 19; j++)
        {
            a1[x[k] - 1][j] = !a1[x[k] - 1][j];
        }
    }
 
    for (int i = 0; i < 19; i++, puts(""))
        for (int j = 0; j < 19; j++)
            printf("%d ", a1[i][j]);
     
    return 0;
}