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:33 14 42 5Output:*****..**..******..**..******..**..*******************..*..*..*..**..*..*..*..***************..*..*..*..**..*..*..*..***************..*..*..*..**..*..*..*..***************..*..*..*..**..*..*..*..*******************************..*..*..*..*..**..*..*..*..*..******************..*..*..*..*..**..*..*..*..*..*****************

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()