CK1602-Editorial

The answer is the resultant matrix after r anti-clockwise rotation. This can be achieved by taking out each outermost spiral present in the matrix and storing it in
another array. Now the elements of each spiral will be in a separate array which will be in another array. Then we just have to shift the elements and store them in 
their right position as determined by r each time.


Python code:

 
from copy import deepcopy
m, n, r = map(int, raw_input().split())
matrix = []
for i in xrange(m):
    matrix.append(map(int, raw_input().split()))
k = min(m, n) / 2
rows = []
for ii in xrange(k):
    row = []
    for i in xrange(ii, m - 1 - ii):
        row.append(matrix[i][ii])
    for i in xrange(ii, n - 1 - ii):
        row.append(matrix[m - 1 - ii][i])
    for i in xrange(m - 1 - ii, ii, -1):
        row.append(matrix[i][n - 1 - ii])
    for i in xrange(n - 1 - ii, ii, -1):
        row.append(matrix[ii][i])
    rows.append(row)
 
result = deepcopy(matrix)
 
for ii in xrange(k):
    row = rows[ii]
    shift = r % len(row)
    idx = -shift
    for i in xrange(ii, m - 1 - ii):
        result[i][ii] = row[idx]
        idx += 1
        idx %= len(row)
    for i in xrange(ii, n - 1 - ii):
        result[m - 1 - ii][i] = row[idx]
        idx += 1
        idx %= len(row)
    for i in xrange(m - 1 - ii, ii, -1):
        result[i][n - 1 - ii] = row[idx]
        idx += 1
        idx %= len(row)
    for i in xrange(n - 1 - ii, ii, -1):
        result[ii][i] = row[idx]
        idx += 1
        idx %= len(row)
for i in result:
    print " ".join(map(str, i))
//