8550. Print the chessboard

You are given a chessboard of size n * n. It is filled with numbers from 1 to n² in the following way: the first ceil(n² / 2) numbers from 1 to ceil(n² / 2) are written in the cells with an even sum of coordinates from left to right from top to bottom. The rest of the n² –  ceil(n² / 2) numbers from ceil(n² / 2) + 1 to n² are written in the cells with an odd sum of coordinates from left to right from top to bottom.

The operation ceil(x / y) means division x by y rounded up.

Input. One integer n (1 ≤ n ≤ 9).

Output. Print the matrix – the chessboard in the given manner. Watch for alignment!

Sample input 1	Sample output 1
5	1 14 2 15 3 16 4 17 5 18 6 19 7 20 8 21 9 22 10 23 11 24 12 25 13
Sample input 2	Sample output 2
6	1 19 2 20 3 21 22 4 23 5 24 6 7 25 8 26 9 27 28 10 29 11 30 12 13 31 14 32 15 33 34 16 35 17 36 18

SOLUTION

array

Algorithm analysis

Enumerate the rows and columns of the chessboard from 1 to n. Set the counter cnt = 1. Iterate over the cells of the resulting two-dimensional table n * n. Assign the value cnt to each cell with an even sum of coordinates, and after each assignment, increase cnt by 1. Then iterate through the cells of the resulting table again and fill the cells with an odd sum of coordinates in the same way.

Example

Consider the first example with n = 5.

The first pass through a two-dimensional array is shown on the left, filling the cells with an even sum of coordinates. The second pass is shown on the right, filling the cells with an odd sum of coordinates.

Algorithm realization

Declare the two-dimensional array.

#define MAX 110

int m[MAX][MAX];

Read the value of n.

scanf("%d", &n);

The first ceil(n² / 2) numbers from 1 to ceil(n² / 2) write to the cells with even sum of coordinates.

cnt = 1;

for (i = 1; i <= n; i++)

for (j = 1; j <= n; j++)

if ((i + j) % 2 == 0) m[i][j] = cnt++;

The rest n² – ceil(n² / 2) numbers from ceil(n² / 2) + 1 to n² write to the cells with odd sum of coordinates.

for (i = 1; i <= n; i++)

for (j = 1; j <= n; j++)

if ((i + j) % 2 == 1) m[i][j] = cnt++;

Print the two dimensional array.

for (i = 1; i <= n; i++)

{

for (j = 1; j <= n; j++)

printf("%2d ", m[i][j]);

printf("\n");

}