Patterns(For Loop)

[CY0T3R]

1
1 0
1 0 1
1 0 1 0
1 0 1 0 1

How To Print The Above Pattern ?? (Using For-Loops Only)

[DJSAN10]

1.Take number of rows as input from user
2.Loop until all rows are done
3.According to the pattern, number of elements in a row = row number itself. Loop for printing elements in 1 row. If the element number you are printing is odd ,print 1, if it is even print 0.

Start coding and post back if you have any further problems in your code.

[Ali_2101]

int main() 
{
    for(int i=0;i<5;i++)
    {
      int prev=0,curr=0;
      for(int j=1;j<=i+1;j++)
      {
        curr=prev+pow(double(-1),j+1);
        cout<<curr<<" ";
        prev=curr;
      }
      cout<<endl;
    }
return 0;
}

[StuXYZ]

The most scary thing about the above code is the pow(double(-1),j+1) ...
please think, an alternating pattern is required.

I initial thought that it was homework help, but it is actually nicely formatted obstifacted C++. So am not sure that the code deserves the -1 reputation. :)

[CY0T3R]

Thanks for the help guys i've made the source code for this pattern -

for(int i=1;i<=5;i++)
{
for(int j=1;j<=i;j++)
{
if(j%2!=0)
cout<<" "<<"1";
else cout<<" "<<"0";
}
cout<<"\n";
}

[ravenous]

if you were so inclined, you don't actually need the if... else structure in the inner-loop:

int rows = 5;
for( int i = 0; i < rows; ++i )
{
    for( int j = 1; j <= i; ++j )
        std::cout << " " << j % 2;

    std::cout << "\n";
}