C++

Forum

 
How to make pattern of stars (*)

siabapet (15)
How to make pattern of stars (*) from this output
1
2
3
4
5
 
*****
**
* *
*  *
*   *

I know it's right triangle but the bottom side is in top side and I don't get the logic though?

Here's the my right triangle code:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
 
#include<iostream>
using namespace std;

int main()
{
	int i,j,n;
    cout << "input : "; cin >> n;
    for(i=1; i<=n; i++)
    {
        for(j=1; j<=i; j++)
        {
            if(j==1 || j==i || i==n)
            { cout<< "*"; }
            else
            { cout << " "; }
        }
        cout <<endl;
    }
    return 0;
}
Last edited on
salem c (3700)

> I know it's right triangle but the bottom side is in top side and I don't get the logic though?
Here's a hint - for loops can count backwards.
siabapet (15)
is it like this
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
 
#include<iostream>
using namespace std;

int main()
{
	int i,j,n;
    cout << "input : "; cin >> n;
    for(i=1; i<n; i++)
    { cout << "*"; }
    for(i=1; i<=n; i++)
    {
        for(j=1; j<=i; j++)
        {
            if(j==1 || j==i || j==n)
            { cout<< "*"; }
            else
            { cout << " "; }
        }
        cout << endl;
    }
    return 0;
}
Last edited on
seeplus (6598)
Well that seems to produce the required display. However this can be simplified. Consider:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
 
#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
	size_t n {};

	cout << "input : ";
	cin >> n;

	cout << setw(n + 1) << setfill('*') << '\n' << setfill(' ');

	for (size_t i = 2; i <= n; ++i)
		cout << '*' << setw(i) << "*\n";
}



input : 5
*****
**
* *
*  *
*   *

Handy Andy (5051)
Hello siabapet,

If the idea is to produce a right triangle consider this:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
 
#include<iostream>
using namespace std;

int main()
{
    int numRows{};

    cout << "input : "; cin >> numRows;

    for (int i = 0; i < numRows; i++)
    {
        for (int j = 1; j <= i; j++)
        {
            if (j == 1 || j == i || j == numRows)
            {
                cout << "*";
            }
            else
            {
                cout << " ";
            }
        }

        cout << endl;
    }

    for (int i = 0; i < numRows; i++)  // <--- Moved down.
    {
        cout << "*";
    }

    return 0;  // <--- Not required, but makes a good break point.
}

Give your variables a proper name. It makes reading the code easier.

Be careful starting for loops at 1 and using (<=). This will be a problem in the future as most things you will use in a for loop will need to start at zero.

This produced this:

input : 5

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


Andy
zsh (1)
Hi siabapet,

this is my first reply. I look into your code, but I think what you code is the exactly one.
and I run it to get the result.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
 
#include<iostream>
using namespace std;

int main()
{
	int i,j,n;
    cout << "input : "; cin >> n;
    for(i=1; i<=n; i++)
    {
        for(j=1; j<=i; j++)
        {
            if(j==1 || j==i || i==n)
            { cout<< "*"; }
            else
            { cout << " "; }
        }
        cout <<endl;
    }
    return 0;
}




input : 5
*
**
* *
*  *
*****




input : 7
*
**
* *
*  *
*   *
*    *
*******


did I misunderstand? do you want the bottom line to top
Last edited on
seeplus (6598)
For a right angled triangle, then consider:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
 
#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
	size_t numRows {};

	cout << "input : ";
	cin >> numRows;

	for (size_t i = 1; i < numRows; ++i)
		cout << '*' << setw(i - 1) << ((i > 1) ? "*" : "") << '\n';

	cout << setw(numRows + 1) << setfill('*') << '\n';
}



input : 7
*
**
* *
*  *
*   *
*    *
*******

Last edited on
Topic archived. No new replies allowed.