I am supposed to create a box based on user input. For example:

Enter the number of rows (max 99): 10

Enter the number of cols (max 99): 20

Enter an interior character: |

Enter an exterior character: +

+ + + + + + + + + + + + + + +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ + + + + + + + + + + + + + +

Could anyone help me get started? I am having a hard time with this for a C++ class! Thanks :)

Enter the number of rows (max 99): 10

Enter the number of cols (max 99): 20

Enter an interior character: |

Enter an exterior character: +

+ + + + + + + + + + + + + + +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ + + + + + + + + + + + + + +

Could anyone help me get started? I am having a hard time with this for a C++ class! Thanks :)

Last edited on

roughly... but what if they enter 1 and 1 ? needs a tweak for that but this should get you started

for(i = ... all the rows)

{

cout <<exterior;

for (j = ...all the cols-2)

cout << interior;

cout << exterior << endl;

}

for(i = ... all the rows)

{

cout <<exterior;

for (j = ...all the cols-2)

cout << interior;

cout << exterior << endl;

}

I would be happy to help you. I would recommend using nested loops to solve the problem. In any problem where you need to output characters in rows and columns nested loops can help you get a resolution. Simply create a loop on the outside that represents the rows and loop on the inside that represents the column. The loop that represents the column should print a character to the console. The loop that represents the row should move to the next line. The catch in this problem is that we need to print a different character depending on the position. I would recommend writing a conditional statement inside the column loop that determines which character to print.

How do we determine which coordinates are the exterior points? If we consider a graph with coordinates x and y or better yet an array with two dimension we can begin to visualize our solution. In the array box[x][y] we can easily determine the exterior coordinates or values. So how do we define the boundaries of our array? We know one set of boundaries just be examining the properties of a two-dimensional array. We know that 0th element of the x coordinate is always a boundary and the 0th element of the y coordinate is always a boundary. The outer boundary will be determined by the user input.

y = 0

+ + + + + + + + + + + + + + + x = userInput

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ + + + + + + + + + + + + + + y = userInput

x = 0

Simply put

every coordinate (x,y) is exterior if x is 0 or x is userInput

every coordinate (x,y) is exterior if y is 0 or y is userInput

You should be able to use a conditional statement to reflect this logic and then output the proper character

How do we determine which coordinates are the exterior points? If we consider a graph with coordinates x and y or better yet an array with two dimension we can begin to visualize our solution. In the array box[x][y] we can easily determine the exterior coordinates or values. So how do we define the boundaries of our array? We know one set of boundaries just be examining the properties of a two-dimensional array. We know that 0th element of the x coordinate is always a boundary and the 0th element of the y coordinate is always a boundary. The outer boundary will be determined by the user input.

y = 0

+ + + + + + + + + + + + + + + x = userInput

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ | | | | | | | | | | | | | | | | | | +

+ + + + + + + + + + + + + + + y = userInput

x = 0

Simply put

every coordinate (x,y) is exterior if x is 0 or x is userInput

every coordinate (x,y) is exterior if y is 0 or y is userInput

You should be able to use a conditional statement to reflect this logic and then output the proper character

Last edited on

I'm working on the same problem and so far I have the following. I can't figure out what I'm missing.

#include <iostream>

#include <string>

using namespace std;

int main () {

int rows, columns, x, y;

string intCharacter, exCharacter;

x = 0;

y = 0;

cout << "Enter the number of rows (max 99): ";

cin >> rows;

cout << rows << endl;

cout << "Enter the number of cols (max 99): ";

cin >> columns;

cout << columns << endl;

cout << "Enter an interior character: ";

cin >> intCharacter;

cout << intCharacter << endl;

cout << "Enter an exterior character: ";

cin >> exCharacter;

cout << exCharacter << endl;

while (x < rows) {

y = 0;

while (y < columns) {

y = y + 1;

if (x == 0) {

cout << exCharacter;

}

if (x == rows -1) {

cout << exCharacter;

}

if (y == 1) {

cout << exCharacter;

}

if (y == columns - 1) {

cout << exCharacter;

}

}

x = x + 1;

cout << endl;

}

return 0;

}

#include <iostream>

#include <string>

using namespace std;

int main () {

int rows, columns, x, y;

string intCharacter, exCharacter;

x = 0;

y = 0;

cout << "Enter the number of rows (max 99): ";

cin >> rows;

cout << rows << endl;

cout << "Enter the number of cols (max 99): ";

cin >> columns;

cout << columns << endl;

cout << "Enter an interior character: ";

cin >> intCharacter;

cout << intCharacter << endl;

cout << "Enter an exterior character: ";

cin >> exCharacter;

cout << exCharacter << endl;

while (x < rows) {

y = 0;

while (y < columns) {

y = y + 1;

if (x == 0) {

cout << exCharacter;

}

if (x == rows -1) {

cout << exCharacter;

}

if (y == 1) {

cout << exCharacter;

}

if (y == columns - 1) {

cout << exCharacter;

}

}

x = x + 1;

cout << endl;

}

return 0;

}

I am also having a hard time with this same problem and cannot figure out what I have done wrong. Can anyone help or did any of you get this to run properly??

#include <iostream>

#include <string>

using namespace std;

int main () {

int rows, cols, x, y;

string intChar, extChar;

cout << "Enter the number of rows (max 99): ";

cin >> rows;

cout << rows << endl;

cout << "Enter the number of cols (max 99): ";

cin >> cols;

cout << cols << endl;

cout << "Enter an interior character: ";

cin >> intChar;

cout << intChar << endl;

cout << "Enter an exterior character: ";

cin >> extChar;

cout << extChar << endl;

while (x < rows) {

while (y < cols) {

if(y == 0 || y == rows - 1){

cout << extChar << " ";

}

else if(x == 0 || x == cols - 1){

cout << extChar << " ";

}

else

cout << intChar << " ";

}

cout << endl;

}

return 0;

}

#include <iostream>

#include <string>

using namespace std;

int main () {

int rows, cols, x, y;

string intChar, extChar;

cout << "Enter the number of rows (max 99): ";

cin >> rows;

cout << rows << endl;

cout << "Enter the number of cols (max 99): ";

cin >> cols;

cout << cols << endl;

cout << "Enter an interior character: ";

cin >> intChar;

cout << intChar << endl;

cout << "Enter an exterior character: ";

cin >> extChar;

cout << extChar << endl;

while (x < rows) {

while (y < cols) {

if(y == 0 || y == rows - 1){

cout << extChar << " ";

}

else if(x == 0 || x == cols - 1){

cout << extChar << " ";

}

else

cout << intChar << " ";

}

cout << endl;

}

return 0;

}

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