I need a project ;P - Page 2

[N3wbi3C0d3r]

cout << "Gradual increase:\n";
    for (int a=1; a <= 5; a++) {
        for (int b=0; b <= a; b++)
        cout << b;
        cout << endl;
    }
    cin.get ();
}

Shows up as :
01
012
0123
01234
012345
I've been asking my friend, and his given me a couple of hints, on how to shorten the program and stablize the rest of it. But this program didnt work out, i cant figure out what exactly i did wrong... Don't answer it for me, just tell or show me the line i messed up on, and ill figure it out form there, wanna work my brain so i emorize how to do all of this.

[Narue]

>show me the line i messed up on
This one:

for (int b=0; b <= a; b++)

[N3wbi3C0d3r]

Thankyou, tried taking 0 and putting in 1, but it was so late last night, and i didnt press compile i guess, but now that part works, and the next drawing will be a peice of cake.

[N3wbi3C0d3r]

for (int a=1; a>=5; a--) {
        for (int b=1; b<=a; b++)

Wich one of these two lines is incorect, i've tried switching things around many times, nothing's worked so far, each time just gives me a bunch of blank space..

[Narue]

>for (int a=1; a>=5; a--) {
Is 1 greater than or equal to 5?

[N3wbi3C0d3r]

No, lol, got it thx, didnt even read it that hard to see it. now comes the pyrimid, and i have NO idea on how to do it. Maybe you can gimme an algorithm of it, and ill translate it into code.

[N3wbi3C0d3r]

Narue, maybe just a hint then? cuz ive tried some things, unlucky so far.

[Narue]

>Narue, maybe just a hint then?
Instead of one loop inside of a loop, use two loops inside of a loop. The outer loop handles rows, the first inner loop handles spaces, and the second inner loop handles the numbers:

for each row
  for each space
    print a space
  for each number
    print a number

The trick is in working out how many spaces to print and how many numbers to print in relation to the row. Start by drawing this:

*
   **
  ***
 ****
*****

Then work out the math required to print another n - 1 fill characters after the center line. That way you break the problem into two steps that are easier to solve than the entire problem all at once.

[N3wbi3C0d3r]

cout << "**********" << endl;
    cout << "Pyrimid:\n";
    for (int a=5; a>0; a--) {
        for (int b=0; b<9; b++)
        cout << " ";    
    cout << "\n";
        for (int c=9; c>0; c--)
        cout << c; 
    cout << "\n";   
    }             
    cin.get ();
}

This shows up as:

987654321

987654321

987654321

987654321

987654321

How do i get it to do the pyrimid? lol, ive tried several things.

[Narue]

>How do i get it to do the pyrimid?
You're going to smack yourself when you see the solution that I was pushing you for:

#include <iostream>

using std::cout;
using std::endl;
using std::cin;

int main()
{
  int h;

  cout<<"Enter the height of your new pyramid: ";
  if ( cin>> h ) {
    for ( int i = 0; i < h; i++ ) {
      for ( int j = 0; j < h - 1 - i; j++ )
        cout<<' ';
      for ( int j = 0; j < i * 2 + 1; j++ )
        cout<<'*';
      cout<<endl;
    }
  }
}

It takes only three minor changes to print digits starting at 1, though you can only print up to a height of 5 before double digit numbers mess up the formatting.

[N3wbi3C0d3r]

#include <iostream>

using std::cout;
using std::endl;
using std::cin;

int main()
{
    cout << "Full box:\n";
    for ( int a=0; a<5; a++ ) {
        for ( int b=5; b>0; b-- ) {
            cout<< b;
        }
        cout << endl;
    }
    cout << "**********" << endl;
    cout << "Empty box:\n";
    for (int a=5;a>0; a--) {
        for (int b=5; b > 0; b-- )
          if ( ((a > 1) && (a < 5)) && (b > 1) && (b < 5))
   cout << " ";
        else cout << b;
      cout << "\n";
    }
    cout << "**********" << endl;
    cout << "Gradual increase:\n";
    for (int a=1; a<=5; a++) {
        for (int b=1; b<=a; b++)
        cout << b;
        cout << endl;
    }
    cout << "**********" << endl;
    cout << "Gradual decrease:\n";
    for (int a=5; a>=1; a--) {
        for (int b=1; b<=a; b++)
        cout << b;
        cout << endl;
    }        
    cout << "**********" << endl;
    cout << "Pyrimid:\n";
    for (int a=0;a<5;a++) {
        for (int b=1;b<10;b++)
        if ( (b<(5-a)) || (b>(5+a)) )
                cout << " ";
        else cout << b;
        cout << endl;
    }
    cin.get ();
}

There it is finished, all done.