Igor Perić

Problem

A secret team of programmers is plotting to disrupt the programming language landscape and bring punched cards back by introducing a new language called Punched Card Python that lets people code in Python using punched cards! Like good disrupters, they are going to launch a viral campaign to promote their new language before even having the design for a prototype. For the campaign, they want to draw punched cards of different sizes in ASCII art.

The ASCII art of a punched card they want to draw is similar to an R×C matrix without the top-left cell. That means, it has (R⋅C)−1 cells in total. Each cell is drawn in ASCII art as a period (.) surrounded by dashes (-) above and below, pipes (|) to the left and right, and plus signs (+) for each corner. Adjacent cells share the common characters in the border. Periods (.) are used to align the cells in the top row.

For example, the following is a punched card with R=3 rows and C=4 columns:

..+-+-+-+
..|.|.|.|
+-+-+-+-+
|.|.|.|.|
+-+-+-+-+
|.|.|.|.|
+-+-+-+-+

There are more examples with other sizes in the samples below. Given the integers R and C describing the size of a punched card, print the ASCII art drawing of it as described above.

Input

The first line of the input gives the number of test cases, T. T lines follow, each describing a different test case with two integers R and C: the number of rows and columns of the punched card that must be drawn.

Output

For each test case, output one line containing Case #x:, where x is the test case number (starting from 1). Then, output (2⋅R)+1 additional lines with the ASCII art drawing of a punched card with R rows and C columns.

Limits

Time limit: 5 seconds.
Memory limit: 1 GB.

Test Set 1 (Visible Verdict)

1≤T≤81
2≤R≤10
2≤C≤10

Sample input

3
3 4
2 2
2 3

Sample output

Case #1:
..+-+-+-+
..|.|.|.|
+-+-+-+-+
|.|.|.|.|
+-+-+-+-+
|.|.|.|.|
+-+-+-+-+
Case #2:
..+-+
..|.|
+-+-+
|.|.|
+-+-+
Case #3:
..+-+-+
..|.|.|
+-+-+-+
|.|.|.|
+-+-+-+

Sample Case #1 is the one described in the problem statement. Sample Cases #2 and #3 are additional examples. Notice that the output for each case contains exactly R⋅C+3 periods.

Solution

There are multiple approaches to solve this easy task. I chose to compute the actual number of rows and columns in the output as (N-1)*2+3 and (M-1)*2+3 and then for each pair of indices i and j determine the rule which maps the pair (i,j) into a character on that spot.

There was no need to store the output matrix, the characters are printed immediately while iterating.

#include <iostream>
#include <vector>
using namespace std;

int main(){
    int t, q = 0;
    cin >> t;
    while (q++ < t) {
        int n, m, i, j;
        cin >> n >> m;
        // compute the actual number of rows in the output
        n = (n - 1) * 2 + 3;
        m = (m - 1) * 2 + 3;

        cout << "Case #" << q << ":" << endl;
        for (i=0;i<n;i++) {
            for (j=0;j<m;j++) {
                // handle top left corner
                if (i < 2 && j < 2) {
                    cout << ".";
                    continue;
                }
                // handle intersection coordinates
                if (i % 2 == 0 && j % 2 == 0)  {
                    cout << "+";
                    continue;
                }
                // handle pipes
                if (i % 2 == 0 && j % 2 == 1)  cout << "-";
                if (i % 2 == 1 && j % 2 == 0)  cout << "|";
              
                // everything else
                if ((i + j) % 2 == 0) cout << ".";
            }
            cout << endl;
        }
    }
    return 0;
}