ALTERNATING RECTANGLES WITH 0 AND X

Write a code which inputs two numbers m and n and creates a matrix of size m x n (m rows and n columns) in which every element is either X or 0. The Xs and 0s must be filled alternatively, the matrix should have the outermost rectangle of Xs, then a rectangle of 0s, then a rectangle of Xs, and so on. 

Input Format: 

The first line of the input consists of an integer, m that correspond to the number of rows in the input array. 

The next line of the input consists of an integer, n that corresponds to the number of columns in the input array. 

Sample Input: 

3 

3 

Sample Output: 

X X X 

X 0 X 

X X X 

  

Sample Input: 

4 

5 

Sample Output: 

X X X X X 

X 0 0 0 X 

X 0 0 0 X 

X X X X X 

Program Code:

#include <stdio.h>

void fill0X(int m, int n)

{

int i, k = 0, l = 0;

int r = m, c = n;

char a[m][n];

char x = 'X';

while (k < m && l < n)

{

for (i = l; i < n; ++i)

a[k][i] = x;

k++;

for (i = k; i < m; ++i)

a[i][n-1] = x;

n--;

if (k < m)

{

for (i = n-1; i >= l; --i)

a[m-1][i] = x;

m--;

}

if (l < n)

{

for (i = m-1; i >= k; --i)

a[i][l] = x;

l++;

}

x = (x == '0')? 'X': '0';

}

for (i = 0; i < r; i++)

{

int j;

for (j = 0; j < c; j++)

printf("%c ", a[i][j]);

printf("\n");

}

}

int main()

{

int a,b;

scanf("%d%d",&a,&b);

fill0X(a, b);

return 0;

}