spoj CPTTRN5 - Character Patterns (Act 5)

Using two characters: . (dot) and * (asterisk) print a grid-like pattern. The grid will have l lines, c columns, and each square shaped element of the grid will have the height and width equal to s.

Moreover, each of the grid elements will have a diagonal. The diagonal of the first square in the first line of the grid is directed towards down and right corner - use the \ (backslash) character to print it; while the next diagonal will be directed towards upper right corner - use the / (slash) character to print it. Print the successive diagonals alternately (please consult the example below).

Input

You are given t - the number of test cases and for each of the test case three positive integers: l - the number of lines, c - the number of columns in the grid and s - the size of the single square shaped element.

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 1 2 4 4 1 2 5 2 Output:*****\.**.\******./**/.******\.**.\***************\*/*\*/***********/*\*/*\***********\*/*\*/***********/*\*/*\***************************\.*./*\.*./*\.**.\*/.*.\*/.*.\******************./*\.*./*\.*./**/.*.\*/.*.\*/.*****************

题意：给出行，列及方块的大小 ，输出如下的字符，每个方格的对角线为左上->右下，后序的相反

思路：1、row % (s+1) = 0 或者col%(s+1)=0输出*

2、(row+col)%(2*s+2) = 0输出\

3、(row-col)%(2*s+2)=0输出/

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])    s = int(a[2])    row = l * (s + 1) + 1    col = c * (s + 1) + 1    for i in range(row):        for j in range(col):            if (0 == i % (s + 1) or 0 == j % (s + 1)):                print('*', sep = '', end = '')            elif (0 == (i + j) % (2 * s + 2)):                print('/', sep = '', end = '')            elif (0 == (i - j) % (2 * s + 2)):                print('\\', sep = '', end = '')            else:                print('.', sep = '', end = '')        print()