C++ pattern

[jlieu.ww]

Design a C++ program by considering the following conditions;
1. User must keyed-in “start” to start the program and “stop” to 
end the program.
.
2. If the user keyed-in “choice1”, print out PATTERN A.
3. If the user keyed-in “choice2”, print out PATTERN B.
Use if else statement, while loop and nested for loop in your program.
PATTERN A PATTERN B
X X X X X
  X
    X
      X
X X X X X
X X X X X
      X
    X
  X
X X X X X

i really need help for this..

[NathanOliver]

What do you have so far? We only give homework help to those that show effort.

[jlieu.ww]

I'm trying for the whole day and can't even reach to get the pattern. only the pattern, not the whole question.

[NathanOliver]

That still doesnt show us what you have. Even if it doesnt compile show us what you are doing and we can help to get it to work.

[jlieu.ww]

#include <iostream>
#include <string>

using namespace std;

int main() 
{



    for (int i = 1; i <= 5; i++){
        for (int j = 1; j <= 5; j++)
        {
            if ((i >= j) && (i<=j))
            {
                cout << "  ";
            }
            else
            {
                cout << "X ";

            }
        }
        cout << endl;
    }
    cout << endl;
    system("pause");
}

I can't get the Z pattern afterall, i tried editing and get many-many pattern but not Z. And, i'm still not fully understand about nested loop.

[NathanOliver]

Okay, lets break this down a little. If we want to print the backwards Z and we have a nested for loop like you have what are the conditions of i and j that would print an X

   jjjjj
   01234
i0 XXXXX
i1  X      
i2   X    
i3    X  
i4 XXXXX

From the above picture we can see that if i == 0 or i == 4 print X so that coves the top and bottom rows. Now to get the diagonal you can see that i and j equal each other where there is an X so that would be if i == j print X. So to pat that all together we get (this is pusedo code):

for i = 0 to i == 4
    for j = 0 to j == 4
        if i == 0  or i == 4 print "X"
        else if i == j print "X"
        else print " "

For printing the normal Z the first if is still code but wee need to change the second one.

   jjjjj
   01234
i0 XXXXX
i1    X    
i2   X    
i3  X    
i4 XXXXX

From the above we can see a patter amerge with when to print an X. The X's are at (i, j): (1, 3) , (2, 2) , (3, 1). If you add the pairs for each one of those you get 4 so you would have an if like if i + j == 4 print X and the code would look like (this is pusedo code):

for i = 0 to i == 4
    for j = 0 to j == 4
        if i == 0  or i == 4 print "X"
        else if i + j == 4 print X
        else print " "