public class MagicSquare
{
public static int[][] grid = new int[3][3];
public static int i = 0;
public static int j = 0;
public static void main(String[] args) {
int x = 1;
int y = 2;
int z = 0;
while(z < 9)
{
int holdx = x;
int holdy = y;
z++;
x++;
y--;
if(x == 3)
{
x = 0;
}
if(y == -1)
{
y = 2;
}
if(y == 3)
{
y = 0;
}
if(grid[x][y] == 0)
{
grid[x][y] = z;
}
else
{
holdy++;
if(holdy == 3)
{
holdy = 0;
}
grid[holdx][holdy] = z;
x = holdx;
y = holdy;
}
}
for(int i = 0; i < 3; i++)
{
System.out.print(grid[i][0]+", ");
}
System.out.println(" ");
for(int i = 0; i < 3; i++)
{
System.out.print(grid[i][1]+", ");
}
System.out.println(" ");
for(int i = 0; i < 3; i++)
{
System.out.print(grid[i][2]+", ");
}
}
THE OUTPUT LOOKS LIKE THIS:
2, 4, 9,
6, 8, 1,
7, 3, 5,
Hello,
I wrote a Black Magic code that is able to fill in the grids of the square up and the right of it, but if it is filled with a number then the next number would be put in the square that is below its current spot.
Then, go one square up and to the right and put the next integer there, and if I also go off the grid, then the next number will wrap around to the bottom and/or left. This program run until all the squares are filled.
I was wondering if it is possible to condense my code starting at my while loop to the end of my for loop into a shorter code?
Someone said that I would be able to code this USING JUST 2 lines, and I think that is bizarre... but they said it's doable!
Any hints, help, or pointer would be appreciated!
Thank you so much!
Not sure about less than 3 lines (at least without sacrificing readability), but you can condense these if statements for sure.
if(x == 3) {
x = 0;
}
if(y == -1) {
y = 2;
}
if(y == 3) {
y = 0;
}
Down into simply
x = x % 3;
y = (y + 3) % 3;
And you could pull those into the previous x++ and y--
x = (x + 1) % 3;
y = ((y - 1) + 3) % 3;
Similarly with holdy (if you actually need that value, I do not know).
Then, if you just want to print the array, the for loops can be shortened.
for(int[] row : grid) {
System.out.println(Arrays.toString(row));
}
About 10 years ago, I wrote the following code for computing the entries of a magic square. This, in some sense, boils down to a single line of code, and works for arbitrary odd edge lengths:
class M
{
public static void main(String args[])
{
int n = 5;
int a[][] = new int[n][n];
f(n,1,n/2,n-1,a);
print(a);
}
static int f(int j,int i,int k,int l,int I[][])
{
return i>j*j?i-1:(I[k][l]=f(j,i+1,(k+(i%j==0?0:1))%j,(l+(i%j==0?-1:1))%j,I))-1;
}
public static void print(int a[][])
{
for (int i=0; i<a.length; i++)
{
for (int j=0; j<a[i].length; j++)
{
System.out.print((a[j][i]<10?" ":"")+a[j][i]+" ");
}
System.out.println();
}
}
}
Of course, it is somehow inspired by the IOCCC and would be better suited for Programming Puzzles & Code Golf, but might show how much you can press into a single line of code when you (ab)use the ternary operator and recursion appropriately...
Related
While this code works and executes, there's a problem with it. The goal of the code is to move [x] across the matrix. It does but, after completing the first row, it prints a completely blank array. Meaning there is no point where both i == x and j == y.
Then, after it prints again, it continues where it left off. This means it never finishes circling the array by the prints allotted.
A picture to showcase what I mean:
.
How would I solve this? It's happening because, after that else property of moveX activates, j == 0 and y == 1; at least, that's what I have to assume is causing the blank print.
public class MovingX {
private int rowsN;
private int columnsM;
private int[][] matrixArr = new int[rowsN][columnsM];
private int x = 0;
private int y = 0;
public MovingX(int n, int m) {
rowsN = n;
columnsM = m;
matrixArr = new int[n][m];
}
public void forLoopGrid() {
for(int i = 0; i < matrixArr.length; i++) {
for(int j = 0; j < matrixArr.length; j++) {
if(i == x && j == y) System.out.print("[x] ");
else {
System.out.print("[ ] ");
}
}
System.out.println();
}
System.out.println();
}
public void moveX() {
if(x < 4) x++;
else {
x = 0;
y++;
}
}
public void runProgram() {
System.out.println("Program Starting");
System.out.println("Rows, Columns: " + rowsN + " " + columnsM);
for(int i = 0; i < (rowsN * columnsM); i++) {
forLoopGrid();
moveX();
}
System.out.println("Program Ending");
}
public static void main(String[] args) {
MovingX xLoop = new MovingX(4, 4);
xLoop.runProgram();
}
}
In your function moveX() you should change condition inside if-statement. Below correct version:
public void moveX(){
if(x < 3) x++;
else{
x = 0;
y++;
}
}
0 indexing can play a tricky factor in all kinds of loops. Since your moveX() method goes up to x = 4, the 0-index count of up to for is 0,1,2,3,4. That is 5 elements. Your matrix is 4x4. You want a count of x=3, giving 0,1,2,3. 4 elements!
You must change your moveX() to
public void moveX(){
if (x < 3)
x++;
else {
x = 0;
y++;
}
}
In this practice problem, a square matrix filled with 0s and 1s is instantiated. You can flip over values (ex: 0 becomes 1 and 1 becomes 0) in a rectangle of any size, as long as the topmost corner of the rectangle is [0, 0] in the matrix. The end goal is to find how many times you must flip values over to get all the values of the matrix as 0.
If you want a longer explanation, go to http://usaco.org/index.php?page=viewproblem2&cpid=689, but that's the basic outline.
This is my code:
import java.io.*;
import java.util.*;
public class CowTip {
static int[][] mat;
public static void main( String[] args) throws IOException, InterruptedException{
Scanner scan = new Scanner(new File("cowtip.in"));
int n = scan.nextInt();
scan.nextLine();
mat = new int[n][n];
for (int x = 0; x < n; x++) {
String str = scan.nextLine();
for (int y = 0; y < n; y++) {
mat[x][y] = Integer.parseInt(str.substring(y,y+1));
}
}
Checker c = new Checker(n-1, n-1);
int count = 0;
while (true) {
c.check();
for (int x = 0; x <= c.row; x++) {
for (int y = 0; y <= c.col; y++) {
if (mat[x][y] == 0) {
mat[x][y] = 1;
}
else if (mat[x][y] == 1) {
mat[x][y] = 0;
}
}
}
count++;
c.check();
if (c.row == -1 && c.col == -1) {
break;
}
}
System.out.println(count);
}
static class Checker {
int row;
int col;
public Checker(int r, int c) {
row = r;
col = c;
}
public Checker check() {
Checker check = new Checker(-1, -1);
for (int x = mat.length-1; x >= 0; x--) {
for (int y = mat[x].length-1; y >= 0; y--) {
if (mat[x][y] == 1) {
check = new Checker(x, y);
break;
}
}
if (check.row != -1 && check.col != -1) {
break;
}
}
return check;
}
}
}
and this is the input file (named cowtip.in) :
3
001
111
111
I've excluded my current debugging code, but the problem is that the row and col values inside my check() method are the correct values, but whenever I call the check() method in my main, the values reverts back to the default and doesn't give me the correct answer, which in turn makes the loop infinite.
Any ideas on how to fix this?
EDIT: I've figured it out, but thanks guys! It was actually extremely simple (c = c.ckeck() instead of c.check()) and honestly, I was pretty frustrated considering I spent around two hours trying to debug this...
Replace c.check() with c = c.check();
There is matrix for [x][y] order. i want to print its value in clockwise order
I have tried several methods but unable to write the logic of the code. I'm trying it in java but logic is important so you can help me in any language.
When I read your post I've started to play so I'll post you my code maybe it will be halpful for you. I've did it for square if you want for rectangle one need separate stepX and stepY. SIZE would be input parameter in your case, I have it final static for test.
public class clockwise {
private static final int SIZE = 3;
public static void main(String[] args) {
// int[][] test_matrix = {{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}};
int[][] test_matrix = {{1,2,3},{5,6,7},{9,10,11}};
int[][] direction = {{1, 0},{0, 1},{-1, 0},{0, -1}}; //{x,y}
for(int i = 0; i < SIZE; i++) {
for(int j = 0; j < SIZE; j++)
System.out.print(test_matrix[i][j] + " ");
System.out.println("");
}
int x = 0;
int y = 0;
int directionMove = 0;
int stepSize = SIZE;
boolean changeStep = true;
int stepCounter = 0;
for(int i = 0; i < SIZE*SIZE; i++) {
System.out.print(test_matrix[x][y] + " ");
stepCounter++;
if (stepCounter % stepSize == 0) {
directionMove++;
directionMove = directionMove%4;
if(changeStep) { //after first edge one need to decrees step after passing two edges
stepSize--;
changeStep = false;
} else {
changeStep = true;
}
stepCounter = 0;
}
x += direction[directionMove][0];
y += direction[directionMove][1];
}
}
}
I have been trying to solve this question http://dwite.ca/questions/haunted_house.html with a Breadth First Search, but I can't get all the testcases correct, and I think the problem is that, it will only count the direct shortest path to the end, and it will count any candies open, but it will not count the shortest path through the candies here is the code
import java.io.*;
import java.util.*;
public class HalloweenCandy {
static int n, candy;
static int minsteps, maxcandy;
static int totCandies=0;
public static void main(String[] args) throws FileNotFoundException {
Scanner s = new Scanner(new File("C:\\Users\\Daniel\\Desktop\\Java\\HalloweenCandy\\src\\halloweencandy\\DATA5.txt"));
while (s.hasNext()) {
n=Integer.parseInt(s.nextLine().trim());
char[][]maze=new char[n][n];
int xStart =0;
int yStart =0;
for(int y=0;y<n;++y){
String text = s.nextLine().trim();
for(int x=0;x<n;++x){
maze[x][y]=text.charAt(x);
if(maze[x][y]=='B'){
xStart=x;
yStart=y;
}
}
}
candy=0;
minsteps=0;
BFS(maze,xStart,yStart);
System.out.println(candy+" "+minsteps);
}
}
public static void BFS(char[][]maze,int xStart,int yStart){
Queue<int[]>queue=new LinkedList<int[]>();
int start[]={xStart,yStart,0,0};
queue.add(start);
while(queue.peek()!=null){
int[]array=queue.poll();
int x=array[0];int y=array[1];
if(x<0||y<0||y>n-1||x>n-1)continue;
if(maze[x][y]=='#')continue;
if(maze[x][y]=='*'){
candy++;
minsteps=array[2];
maze[x][y]='.';
}
if(maze[x][y]>='a'&&maze[x][y]<='f'){
if(candy <maze[x][y]-'a'+1)continue;
}
int[][]points = {{0,1},{1,0},{-1,0},{0,-1}};
for(int i=0;i<4;++i){
int sta[]={x+points[i][0],y+points[i][1],array[2]+1};
queue.add(sta);
}
maze[x][y]='#';
}
}
}
and here are the test cases
http://dwite.ca/home/testcase/232.html
You're on the write track, but you missed something important.
while(queue.peek()!=null){
int[]array=queue.poll();
int x=array[0];int y=array[1];
if(x<0||y<0||y>n-1||x>n-1)continue;
if(maze[x][y]=='#')continue;
if(maze[x][y]=='*'){
candy++;
minsteps=array[2];
maze[x][y]='.';
}
if(maze[x][y]>='a'&&maze[x][y]<='f'){
if(candy <maze[x][y]-'a'+1)continue;
}
int[][]points = {{0,1},{1,0},{-1,0},{0,-1}};
for(int i=0;i<4;++i){
int sta[]={x+points[i][0],y+points[i][1],array[2]+1};
queue.add(sta);
}
maze[x][y]='#'; // <== this part is wrong
}
What you're doing in that last assignment is making every square you step on into a wall. This would be the right approach if you could get through the maze without backtracking, but that's not the case. Instead, what you want to do is make sure you don't backtrack until you've picked up a new piece of candy. So, try something like this instead:
maze[x][y]='a'+candy;
That way, once you pick up a new piece of candy the square will be usable again.
However, there's still an issue here. Think about how BFS would work on this map:
3
...
*B*
...
If [0,0] is the top-left tile, then your BFS algorithm will visit the tiles in this order: [1,2], [2,1], [0,1], [1,0]. What's wrong with that? Billy is jumping between all of his neighboring squares! What you actually want him to do is restart the BFS each time he gets a new piece of candy. I'll leave it to you to figure out how to do that part.
Edit
Here's the basic algorithm you want to follow:
Begin at the start position.
Use BFS to search for the nearest piece of candy. The first piece of candy found with BFS is the nearest (or tied for nearest)!
After finding a piece of candy, you need to find the next closest piece to your current position, so treat your current position as the new start for another BFS.
I just finished solving it your way, the "greedy" way here is the code
import java.io.*;
import java.util.*;
public class HalloweenCandy {
static int n, candy;
static int minsteps, maxcandy;
static int totCandies = 0;
static boolean[][] is;
public static void main(String[] args) throws FileNotFoundException {
Scanner s = new Scanner(new File("C:\\Users\\Daniel\\Desktop\\Java\\HalloweenCandy\\src\\halloweencandy\\DATA5.txt"));
while (s.hasNext()) {
n = Integer.parseInt(s.nextLine().trim());
char[][] maze = new char[n][n];
is = new boolean[n][n];
int xStart = 0;
int yStart = 0;
for (int y = 0; y < n; ++y) {
String text = s.nextLine().trim();
for (int x = 0; x < n; ++x) {
maze[x][y] = text.charAt(x);
if (maze[x][y] == 'B') {
xStart = x;
yStart = y;
}
}
}
candy = 0;
int tot = 0;
int y = 0, x = 0;
x = xStart;
y = yStart;
for (int j = 0; j < n; ++j) {
for (int i = 0; i < n; ++i) {
is[i][j] = false;
}
}
while (true) {
char[][] grid = new char[n][n];
for (int j = 0; j < n; ++j) {
for (int i = 0; i < n; ++i) {
grid[i][j] = maze[i][j];
}
}
int lol[] = BFS(grid, x, y);
if (lol[0] == -1) {
break;
}
y = lol[2];
x = lol[1];
tot += lol[0];
}
System.out.println(candy + " " + tot);
}
}
public static int[] BFS(char[][] maze, int xStart, int yStart) {
Queue<int[]> queue = new LinkedList<int[]>();
int start[] = {xStart, yStart, 0, 0};
queue.add(start);
while (queue.peek() != null) {
int[] array = queue.poll();
int x = array[0];
int y = array[1];
if (x < 0 || y < 0 || y > n - 1 || x > n - 1) {
continue;
}
if (maze[x][y] == '#') {
continue;
}
if (maze[x][y] == '*' && !is[x][y]) {
is[x][y] = true;
candy++;
int sta[] = {array[2], x, y};
return sta;
}
if (maze[x][y] >= 'a' && maze[x][y] <= 'f') {
if (candy < maze[x][y] - 'a' + 1) {
continue;
}
}
int[][] points = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
for (int i = 0; i < 4; ++i) {
int sta[] = {x + points[i][0], y + points[i][1], array[2] + 1};
queue.add(sta);
}
maze[x][y] = '#';
}
int sta[] = {-1};
return sta;
}
}
I would like to now figure out how to solve it the dynamic way, the solution I gave only works for some cases but not all.
In my last question seen here: Sudoku - Region testing I asked how to check the 3x3 regions and someone was able to give me a satisfactory answer (although it involved a LOT of tinkering to get it working how I wanted to, since they didn't mention what the class table_t was.)
I finished the project and was able to create a sudoku generator, but it feels like it's contrived. And I feel like I've somehow overcomplicated things by taking a very brute-force approach to generating the puzzles.
Essentially my goal is to create a 9x9 grid with 9- 3x3 regions. Each row / col / region must use the numbers 1-9 only once.
The way that I went about solving this was by using a 2-dimensional array to place numbers at random, 3 rows at a time. Once the 3 rows were done it would check the 3 rows, and 3 regions and each vertical col up to the 3rd position. As it iterated through it would do the same until the array was filled, but due to the fact that I was filling with rand, and checking each row / column / region multiple times it felt very inefficient.
Is there an "easier" way to go about doing this with any type of data construct aside from a 2d array? Is there an easier way to check each 3x3 region that might coincide with checking either vert or horizontal better? From a standpoint of computation I can't see too many ways to do it more efficiently without swelling the size of the code dramatically.
I built a sudoku game a while ago and used the dancing links algorithm by Donald Knuth to generate the puzzles. I found these sites very helpful in learning and implementing the algorithm
http://en.wikipedia.org/wiki/Dancing_Links
http://cgi.cse.unsw.edu.au/~xche635/dlx_sodoku/
http://garethrees.org/2007/06/10/zendoku-generation/
import java.util.Random;
import java.util.Scanner;
public class sudoku {
/**
* #antony
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int p = 1;
Random r = new Random();
int i1=r.nextInt(8);
int firstval = i1;
while (p == 1) {
int x = firstval, v = 1;
int a[][] = new int[9][9];
int b[][] = new int[9][9];
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if ((x + j + v) <= 9)
a[i][j] = j + x + v;
else
a[i][j] = j + x + v - 9;
if (a[i][j] == 10)
a[i][j] = 1;
// System.out.print(a[i][j]+" ");
}
x += 3;
if (x >= 9)
x = x - 9;
// System.out.println();
if (i == 2) {
v = 2;
x = firstval;
}
if (i == 5) {
v = 3;
x = firstval;
}
}
int eorh;
Scanner in = new Scanner(System.in);
System.out
.println("hey lets play a game of sudoku:take down the question and replace the 0's with your digits and complete the game by re entering your answer");
System.out.println("enter your option 1.hard 2.easy");
eorh = in.nextInt();
switch (eorh) {
case 1:
b[0][0] = a[0][0];
b[8][8] = a[8][8];
b[0][3] = a[0][3];
b[0][4] = a[0][4];
b[1][2] = a[1][2];
b[1][3] = a[1][3];
b[1][6] = a[1][6];
b[1][7] = a[1][7];
b[2][0] = a[2][0];
b[2][4] = a[2][4];
b[2][8] = a[2][8];
b[3][2] = a[3][2];
b[3][8] = a[3][8];
b[4][2] = a[4][2];
b[4][3] = a[4][3];
b[4][5] = a[4][5];
b[4][6] = a[4][6];
b[5][0] = a[5][0];
b[5][6] = a[5][6];
b[6][0] = a[6][0];
b[6][4] = a[6][4];
b[6][8] = a[6][8];
b[7][1] = a[7][1];
b[7][2] = a[7][2];
b[7][5] = a[7][5];
b[7][6] = a[7][6];
b[8][4] = a[8][4];
b[8][5] = a[8][5];
b[0][0] = a[0][0];
b[8][8] = a[8][8];
break;
case 2:
b[0][3] = a[0][3];
b[0][4] = a[0][4];
b[1][2] = a[1][2];
b[1][3] = a[1][3];
b[1][6] = a[1][6];
b[1][7] = a[1][7];
b[1][8] = a[1][8];
b[2][0] = a[2][0];
b[2][4] = a[2][4];
b[2][8] = a[2][8];
b[3][2] = a[3][2];
b[3][5] = a[3][5];
b[3][8] = a[3][8];
b[4][0] = a[4][0];
b[4][2] = a[4][2];
b[4][3] = a[4][3];
b[4][4] = a[4][4];
b[4][5] = a[4][5];
b[4][6] = a[4][6];
b[5][0] = a[5][0];
b[5][1] = a[5][1];
b[5][4] = a[5][4];
b[5][6] = a[5][6];
b[6][0] = a[6][0];
b[6][4] = a[6][4];
b[6][6] = a[6][6];
b[6][8] = a[6][8];
b[7][0] = a[7][0];
b[7][1] = a[7][1];
b[7][2] = a[7][2];
b[7][5] = a[7][5];
b[7][6] = a[7][6];
b[8][2] = a[8][2];
b[8][4] = a[8][4];
b[8][5] = a[8][5];
break;
default:
System.out.println("entered option is incorrect");
break;
}
for (int y = 0; y < 9; y++) {
for (int z = 0; z < 9; z++) {
System.out.print(b[y][z] + " ");
}
System.out.println("");
}
System.out.println("enter your answer");
int c[][] = new int[9][9];
for (int y = 0; y < 9; y++) {
for (int z = 0; z < 9; z++) {
c[y][z] = in.nextInt();
}
}
for (int y = 0; y < 9; y++) {
for (int z = 0; z < 9; z++)
System.out.print(c[y][z] + " ");
System.out.println();
}
int q = 0;
for (int y = 0; y < 9; y++) {
for (int z = 0; z < 9; z++)
if (a[y][z] == c[y][z])
continue;
else {
q++;
break;
}
}
if (q == 0)
System.out
.println("the answer you have entered is correct well done");
else
System.out.println("oh wrong answer better luck next time");
System.out
.println("do you want to play a different game of sudoku(1/0)");
p = in.nextInt();
firstval=r.nextInt(8);
/*if (firstval > 8)
firstval -= 9;*/
}
}
}
I think you can use a 1D array, in much the same way a 1D array can model a binary tree. For example, to look at the value below a number, add 9 to the index.
I just made this up, but could something like this work?
private boolean makePuzzle(int [] puzzle, int i)
{
for (int x = 0; x< 10 ; x++)
{
if (//x satisfies all three conditions for the current square i)
{
puzzle[i]=x;
if (i==80) return true //terminal condition, x fits in the last square
else
if makePuzzle(puzzle, i++);//find the next x
return true;
}// even though x fit in this square, an x couldn't be
// found for some future square, try again with a new x
}
return false; //no value for x fit in the current square
}
public static void main(String[] args )
{
int[] puzzle = new int[80];
makePuzzle(puzzle,0);
// print out puzzle here
}
Edit: its been a while since I've used arrays in Java, sorry if I screwed up any syntax. Please consider it pseudo code :)
Here is the code as described below in my comment.
public class Sudoku
{
public int[] puzzle = new int[81];
private void makePuzzle(int[] puzzle, int i)
{
for (int x = 1; x< 10 ; x++)
{
puzzle[i]=x;
if(checkConstraints(puzzle))
{
if (i==80)//terminal condition
{
System.out.println(this);//print out the completed puzzle
puzzle[i]=0;
return;
}
else
makePuzzle(puzzle,i+1);//find a number for the next square
}
puzzle[i]=0;//this try didn't work, delete the evidence
}
}
private boolean checkConstraints(int[] puzzle)
{
int test;
//test that rows have unique values
for (int column=0; column<9; column++)
{
for (int row=0; row<9; row++)
{
test=puzzle[row+column*9];
for (int j=0;j<9;j++)
{
if(test!=0&& row!=j&&test==puzzle[j+column*9])
return false;
}
}
}
//test that columns have unique values
for (int column=0; column<9; column++)
{
for(int row=0; row<9; row++)
{
test=puzzle[column+row*9];
for (int j=0;j<9;j++)
{
if(test!=0&&row!=j&&test==puzzle[column+j*9])
return false;
}
}
}
//implement region test here
int[][] regions = new int[9][9];
int[] regionIndex ={0,3,6,27,30,33,54,57,60};
for (int region=0; region<9;region++) //for each region
{
int j =0;
for (int k=regionIndex[region];k<regionIndex[region]+27; k=(k%3==2?k+7:k+1))
{
regions[region][j]=puzzle[k];
j++;
}
}
for (int i=0;i<9;i++)//region counter
{
for (int j=0;j<9;j++)
{
for (int k=0;k<9;k++)
{
if (regions[i][j]!=0&&j!=k&®ions[i][j]==regions[i][k])
return false;
}
}
}
return true;
}
public String toString()
{
String string= "";
for (int i=0; i <9;i++)
{
for (int j = 0; j<9;j++)
{
string = string+puzzle[i*9+j];
}
string =string +"\n";
}
return string;
}
public static void main(String[] args)
{
Sudoku sudoku=new Sudoku();
sudoku.makePuzzle(sudoku.puzzle, 0);
}
}
Try this code:
package com;
public class Suduku{
public static void main(String[] args ){
int k=0;
int fillCount =1;
int subGrid=1;
int N=3;
int[][] a=new int[N*N][N*N];
for (int i=0;i<N*N;i++){
if(k==N){
k=1;
subGrid++;
fillCount=subGrid;
}else{
k++;
if(i!=0)
fillCount=fillCount+N;
}
for(int j=0;j<N*N;j++){
if(fillCount==N*N){
a[i][j]=fillCount;
fillCount=1;
System.out.print(" "+a[i][j]);
}else{
a[i][j]=fillCount++;
System.out.print(" "+a[i][j]);
}
}
System.out.println();
}
}
}