spoj CPTTRN3 - Character Patterns (Act 3)

Using two characters: . (dot) and * (asterisk) print a grid-like pattern.

Input

You are given t - the number of test cases and for each of the test cases two positive integers: l - the number of lines and c - the number of columns in the grid. Each square of the grid is
of the same size and filled with 4 dots (see the example below).

Output

For each of the test cases output the requested pattern (please have a look at the example). Use one line break in between successive patterns.

Example

Input:
3
3 1
4 4
2 5

Output:
****
*..*
*..*
****
*..*
*..*
****
*..*
*..*
****

*************
*..*..*..*..*
*..*..*..*..*
*************
*..*..*..*..*
*..*..*..*..*
*************
*..*..*..*..*
*..*..*..*..*
*************
*..*..*..*..*
*..*..*..*..*
*************

****************
*..*..*..*..*..*
*..*..*..*..*..*
****************
*..*..*..*..*..*
*..*..*..*..*..*
****************

题意：给出一些例子，每个例子的输入为l,c，输出l行,c列，每块是由4个点组成，外围是星号
思路：实现上就是(3*l+1)行，(3*c+1)列，在遍历过程中，如果行或者列被3整除输出*，否则输出.（用python实现会超时，用c/c++就是ok）

代码如下：
t = int(input())
for cas in range(t):
line = input();
a = line.split(' ');
l = int(a[0])
c = int(a[1])
row = 3 * l + 1
col = 3 * c + 1
for i in range(row):
for j in range(col):
if (0 == i % 3 or 0 == j % 3):
print('*', sep='', end='')
else:
print('.', sep='', end='')
print()