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Monday, 29 August 2016

Chapter 8 Exercise 23, Introduction to Java Programming, Tenth Edition Y. Daniel LiangY.

 
*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)



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;
 
 }
 
 
}

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