Print Number Pattern

Printing Number Pattern

In this program we’re going to code a c program for the number pattern program. We will be using the approach of Basic incrementing Squared Number-Star Pattern + Basic incrementing inverted Squared Number-Star Pattern (alternate)

For any input Number N print the following code – for below code N=5.

Then output –

```1 * 2 * 3 * 4 * 5
11 * 12 * 13 * 14 * 15
21 * 22 * 23 * 24 * 25
16 * 17 * 18 * 19 * 20
6 * 7 * 8 * 9 * 10```

Algorithm:

• Start the program.
• Take number of rows input from the user and store it in any variable (‘n’ in this case).
• Check if ‘n’ is within the range of 1 to 100 (inclusive).
• Iterate over the rows from 1 to ‘n’ with a step of 2 (odd rows)
• Calculate the value of ‘k’ using the formula: (i-1)*n + 1.
• Print the numbers in the row, followed by an asterisk and a space.
• If ‘n’ is odd, If it is, Update ‘p’ to ‘n-1’.
• Iterate over the rows from ‘p’ to 1 with a step of 2 (even rows).
• Calculate the value of ‘k’ using the formula: (i-1)*n + 1.
• Print the numbers in the row, followed by an asterisk and a space.
• End the program.

Code in C++:

Run
```#include<iostream>
using namespace std;
int main()
{
int n;
cout<<"Enter the Number of Rows : ";
cin >> n;
cout << endl << endl;
int p = n;
if(n >= 1 && n <= 100)
{
for(int i = 1; i <= n; i += 2)
{
cout<<"\t";
int k=(i-1)*n+1;
for(int j = 0; j < n-1; j++)
{
cout << k << " * ";
k++;
}
cout << k << " ";
cout << endl;
}
if(n % 2 != 0)
{
p = n-1;
}
for(int i = p; i > 0; i -= 2)
{
cout << "\t";
int k = (i-1) * n + 1;
for(int j = 0; j < n-1; j++)
{
cout << k << " * ";
k++;
}
cout << k << " ";
cout << endl;
}
}
else
{
cout << "Enter a Valid Input(1-100)!";
}
return 0;
}
```

One comment on “Print Number Pattern”

• singhji012002

Simple c++;
#include
using namespace std;

int main()
{
int n;
cin>>n;
int cnt = 1;
for(int i=1; i<=n; i++)
{
if(i%2 != 0)
{
for(int j=1; j<=n; j++)
{
if(j != n)
{
cout<<cnt<<"*";
cnt++;
}
else
{
cout<<cnt;
cnt++;
}
}
}
else
{
int temp = cnt+n;
for(int j=1; j<=n; j++)
{
if(j != n)
{
cout<<temp<<"*";
temp++;
}
else
{
cout<<temp;
temp++;
}
}
}
cout<<endl;
}
}