how to cout symbols in specific shapes

[catalin10]

I have to solve a problem which will help me with a more difficult one. I must write an algorithm that displays a triangle made of numbers. More details below:

#include <iostream>
using namespace std;
main()
{
    int n, i, j;                        
    cin>>n; // n is the variable which determines the size of the triangle
    for(i=1;i<=n;i++)  //i represents the number of elements from one line
    {for(j=1; j<=i;j++) //j represents the number of lines which is equal to i

        cout<<2*j-1<<" "<<endl;
    }
}

/* the program should return:
1
1 3
1 3 5
...
1 3 5...2n-1
 
but instead returns:
1
1
3
1
3
5
...
1
3
5
...
2n-1 */  

After doing this I'll have to write an algorithm that displays a rhombus made of stars with specific dimensions. But this one will come only after the one I showed you now.

I realized that I really don't know how to give shape to the data returned by the program. I even tried with a square matrix ( v[i][j]) but the numbers were still displayed one under another like in the example above.

[Berzeger]

That's because you cout endl after each number. You will need to cout endl after each line not each number. Move the cout << endl; part to the outter for loop.

[Smatik]

{
int n, i, j;
cin>>n; // n is the variable which determines the size of the triangle
for(i=1;i<=n;i++) //i represents the number of elements from one line
{for(j=1; j<=i;j++) //j represents the number of lines which is equal to i

cout<<2*j-1<<" ";
}
cout<<\n;
}

try this one...i am a beginner but i think it should work as you want

[Smatik]

And yea i also want to solve the rhombus of asterisks problem. Please tell me if you get it work!

[catalin10]

Thanks. This is how it must be written:

  for(i=1;i<=n;i++)  //i represents the number of elements from one line
    {for(j=1; j<=i;j++) //j represents the number of lines which is equal to i
{


cout<<2*j-1<<" ";
    }
    cout<<endl;
} 

About that rhombus problem, I'll work on it tomorrow, or the day after tomorrow and I'll post my trial.

[Smatik]

COOL

[Berzeger]

It is not nice to mix "\n" and endl. Also you shouldn't have just main () { ... } but for example int main (int argc, char ** argv) { ... return 0; }. Maybe you should do some proper reading and not just hopping into programming hoping all the problems will eventually solve themselves.

[cire]

There is no reason not to use '\n' "\n" or std::endl as you wish. std::endl is functionally the same as a newline followed by a flush of the output stream. "\n" is a c-string, of course, so it requires a (slight) bit more than '\n' for std::cout to process, but it's very unlikely you'd ever notice a difference.

It is true that main always returns int, and the prototype you give is accurate for programs that wish to access parameters the program was invoked with, but the return 0; is the same as letting execution fall off the end of main and isn't necessary. main is the only function for which this is true.