2669. Rotation

An array of size n × m is given. Rotate it 90° clockwise.

Input. The first line contains two positive integers n and m (1 ≤ n, m ≤ 50).

Each of the next n lines contains m nonnegative integers, each not exceeding 10⁹ – the elements of the array.

Output. Print the resulting array in the same format as in the input.

Sample input	Sample output
3 4 1 2 3 4 5 6 7 8 9 10 11 12	4 3 9 5 1 10 6 2 11 7 3 12 8 4

SOLUTION

two-dinensional array

Algorithm analysis

When matrix a is rotated 90 degrees clockwise, the element at position (i, j) moves to the position (j, n – i – 1) in matrix b.

If the original matrix a has dimensions n × m, then the resulting matrix b will have dimensions m × n. The indexing of elements in both matrices starts from 0.

Algorithm implementation

Declare the input array a and the resulting reversed array b.

#define MAX 55

int a[MAX][MAX], b[MAX][MAX];

Read the input data.

scanf("%d %d",&n,&m);

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

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

scanf("%d",&a[i][j]);

Reverse array a and store the result in b.

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

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

b[j][n-i-1] = a[i][j];

Print the reversed array b.

printf("%d %d\n",m,n);

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

{

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

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

printf("\n");

}