Matrix Rotation Anti Clockwise

Challenge

You are given a 2D matrix, a, of dimension MxN and a positive integer R. You have to rotate the matrix R times and print the resultant matrix. Rotation should be in anti-clockwise direction. Note that in one rotation, you have to shift elements by one step only (refer sample tests for more clarity).

It is guaranteed that the minimum of M and N will be even.

Input

First line contains three space separated integers, M, N and R, where M is the number of rows, N is number of columns in matrix, and R is the number of times the matrix has to be rotated. Then M lines follow, where each line contains N space separated positive integers. These M lines represent the matrix.

Output

Print the rotated matrix.

Constraints

2 <= M, N <= 300

1 <= R <= 10^9

min(M, N) % 2 == 0

1 <= a(i,j) <= 10^8, where i is [1..M] & j is [1..N]

Sample Input #00

4 4 1

1 2 3 4

5 6 7 8

9 10 11 12

13 14 15 16

Sample Output #00

2 3 4 8

1 7 11 12

5 6 10 16

9 13 14 15

Sample Input #01

4 4 2

1 2 3 4

5 6 7 8

9 10 11 12

13 14 15 16

Sample Output #01

3 4 8 12

2 11 10 16

1 7 6 15

5 9 13 14

Sample Input #02

5 4 7

1 2 3 4

7 8 9 10

13 14 15 16

19 20 21 22

25 26 27 28

Sample Output #02

28 27 26 25

22 9 15 19

16 8 21 13

10 14 20 7

4 3 2 1

Sample Input #03

2 2 3

1 1

1 1

Sample Output #03

1 1

1 1

Explanation

Sample Case #00: Here is an illustration of what happens when the matrix is rotated once.

1 2 3 4 2 3 4 8

5 6 7 8 1 7 11 12

9 10 11 12 -> 5 6 10 16

13 14 15 16 9 13 14 15

Sample Case #01: Here is what happens when to the matrix after two rotations.

1 2 3 4 2 3 4 8 3 4 8 12

5 6 7 8 1 7 11 12 2 11 10 16

9 10 11 12 -> 5 6 10 16 -> 1 7 6 15

13 14 15 16 9 13 14 15 5 9 13 14

Sample Case #02: Following are the intermediate states.

1 2 3 4 2 3 4 10 3 4 10 16 4 10 16 22

7 8 9 10 1 9 15 16 2 15 21 22 3 21 20 28

13 14 15 16 -> 7 8 21 22 -> 1 9 20 28 -> 2 15 14 27 ->

19 20 21 22 13 14 20 28 7 8 14 27 1 9 8 26

25 26 27 28 19 25 26 27 13 19 25 26 7 13 19 25

10 16 22 28 16 22 28 27 22 28 27 26 28 27 26 25

4 20 14 27 10 14 8 26 16 8 9 25 22 9 15 19

3 21 8 26 -> 4 20 9 25 -> 10 14 15 19 -> 16 8 21 13

2 15 9 25 3 21 15 19 4 20 21 13 10 14 20 7

1 7 13 19 2 1 7 13 3 2 1 7 4 3 2 1

Sample Case #03: As all elements are same, any rotation will reflect the same matrix.

Solution (C)

#include <stdio.h>

#include <string.h>

#include <math.h>

#include <stdlib.h>

int l, m, Row, Col, mat[300][300];

void rotation(int l, int m, int Row, int Col) {

int si, sj, i, j, t, f;

si = l;

sj = m;

t = mat[l][m];

for (i = l + 1; i <= Row; i++) {

f = mat[i][m];

mat[i][m] = t;

t = f;

}

l++;

for (i = m + 1; i <= Col; i++) {

f = mat[Row][i];

mat[Row][i] = t;

t = f;

}

m++;

if (l - 1 < Row) {

for (i = Row - 1; i >= l - 1; i--) {

f = mat[i][Col];

mat[i][Col] = t;

t = f;

}

}

Col--;

if (m - 1 < Col) {

for (i = Col; i >= m; i--) {

f = mat[l - 1][i];

mat[l - 1][i] = t;

t = f;

}

}

Row--;

mat[si][sj] = t;

return;

}

int main() {

int M, N, R, i, j;

scanf("%d%d%d", &M, &N, &R);

int f, K;

for (i = 0; i < M; i++) {

for (j = 0; j < N; j++) {

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

}

}

l = 0;

m = 0;

Row = M - 1;

Col = N - 1;

while (l < Row && m < Col) {

int rot = 2 * (Row - l) + 2 * (Col - m);

f = k % rot;

for (i = 1; i <= f; i++) {

rotation(l, m, Row, Col);

}

l++;

m++;

Row--;

Col--;

}

for (i = 0; i < M; i++) {

for (j = 0; j < N; j++) {

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

}

printf("\n");

}

return 0;

}