Today I’ve solved bit algorithm problems.

I’ll write my solution one of those problems.

So Let’s begin.

### Problem

```
Print the "*" by inferring example's rule.
```

### Input

```
Value N(1 ≤ N ≤ 100) will be given.
```

### Output

```
Print from the first line to N line.
```

### Example Input 1

```
1
```

### Example Output 1

```
*
```

### Example Input 2

```
2
```

### Example Output 2

```
*
* *
```

### Example Input 3

```
3
```

### Example Output 3

```
*
* *
* * *
```

### Example Input 4

```
4
```

### Example Output 4

```
*
* *
* * *
* * * *
```

From here

Let’s think about it!

It’s can be solved easily if I use many **conditional expression and output expression**.

But I think It can be solved just a single output function(except newline) with loops and the ternary operator.

So Let’s find the rule of this problem.

For finding the rule, I’ve drawn a table.

We can find some rule of this.

First, We can get the count of ‘i’ raw.
Second, If the N is even value, We can print ‘*’ when the sum of ‘i’ and ‘j’ value is an odd value. If not, We can print ‘*’ when the sum of ‘i’ and ‘j’ is even value.

The first rule can be used for a single output function(except newline).

But the second rule makes the condition. So We need to remove the condition.

I take ‘i’+’j’-‘N’.

So, ‘*’ can be printed when that value is odd.

And,

Third, We can print ‘ ‘ when that value is less than 1.

So Let’s write this to code.

```
#include <iostream>
using namespace std;
int main()
{
int N=0;
cin>>N;
for(int i=1; i<=N; i++)
{
for(int j=1; j<=(N+i-1); j++)
cout<< ((((i+j-N) > 0)&&(i+j-N) % 2 != 0)? "*" : " ");
cout<<endl;
}
return 0;
}
```

So this problem can be solved like this single output function(except newline).