*8.23 (Game: find the flipped cell) Suppose you are given a 6-by-6 matrix filled with
0s and 1s. All rows and all columns have an even number of 1s. Let the user flip
one cell (i.e., flip from 1 to 0 or from 0 to 1) and write a program to find which
cell was flipped. Your program should prompt the user to enter a 6-by-6 array
with 0s and 1s and find the first row r and first column c where the even number
of the 1s property is violated (i.e., the number of 1s is not even). The flipped cell
is at (r, c). Here is a sample run:
Enter a 6-by-6 matrix row by row:
1 1 1 0 1 1
1 1 1 1 0 0
0 1 0 1 1 1
1 1 1 1 1 1
0 1 1 1 1 0
1 0 0 0 0 1
The flipped cell is at (0, 1)
Enter a 6-by-6 matrix row by row:
1 1 1 0 1 1
1 1 1 1 0 0
0 1 0 1 1 1
1 1 1 1 1 1
0 1 1 1 1 0
1 0 0 0 0 1
The flipped cell is at (0, 1)
import java.util.Scanner; public class ProgrammingEx8_23 { public static void main(String[] args) { Scanner input = new Scanner(System.in); System.out.println("Enter a 6-by-6 matrix row by row:"); int[][] n = new int[6][6]; for (int i = 0; i < n.length; i++) { for (int j = 0; j < n[i].length; j++) { n[i][j] = input.nextInt(); } } int col = calCol(n); int row = calRow(n); if(col!=-1 && row != -1) { System.out.println("The flipped cell is at (" +row + ","+ col +")"); }else { System.out.println("There is no flipped cell."); } } public static int calCol(int array[][]) { int intSum = 0; for (int i = 0; i < array.length; i++) { for (int j = 0; j < array[i].length; j++) { intSum += array[j][i]; } if (intSum % 2 != 0) { return i; } } return -1; } public static int calRow(int array[][]) { int intSum = 0; for (int i = 0; i < array.length; i++) { for (int j = 0; j < array[i].length; j++) { intSum += array[i][j]; } if (intSum % 2 != 0) { return i; } } return -1; } }
No comments :
Post a Comment