Java Program to Print Exponentially Increasing Number Pattern

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Printing Exponentially Increasing Number Pattern

In the previous article, we have discussed Java Program to Print Inverted Right-Angled Triangle with Decreasing Number Pattern

In this article we will see exponentially increasing number pattern.

For example :

Enter the number of lines: 5

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APPROACH:

  • Enter how many lines you have to print and store it in variable noOflines.
  • Using Scanner class Take inputs from User.
  • Take a for loop to iterate the Row.
  • Take an inner for loop to print the values in each row.

Java to Print Exponentially Increasing Number Pattern

// Java program for print the
// given numeric pattern in exponentially order
import java.util.*;
class Main
{
    public static void main(String args[])
    {
    	//Taking input from USER
        Scanner read=new Scanner(System.in);
        System.out.print("Enter the number: ");
    	int num = read.nextInt();
    	System.out.print("Enter the number of lines: ");
    	int noOflines = read.nextInt();
    
    
        {
    	    int n = num, num2 = 0,
    		x = 1, limit = 1;
    
        	// Number of rows to  be printed
        	for (int row = 0;
        			row < noOflines; row++)
        	{
        		
        	// Number of elements to  be printed
        		
        		for (int col = 0; col < limit; col++)
        		{
        			if (col == 0)
        				num2 = num;
        
        			// Print all the element
        			System.out.print(num2++ + " ");
        		}
        		
        		num *= 2;
        		limit = num / n;
        		System.out.println();
        	}
        }
    }
}

Output:

Enter the number: 1
Enter the number of lines: 5

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C++ to Print Exponentially Increasing Number Pattern

// Java program for print the
// given numeric pattern in exponentially order.
#include <bits/stdc++.h>

using namespace std;

int main()
{

    int num = 1;
    int noOflines = 5;

    int n = num, num2, x = 1, limit = 1;

    // Number  of rows to be printed
    for (int row = 0; row < noOflines; row++) {
        
        // Number of elements to be printed in each row
        for (int col = 0; col < limit; col++) {
            if (col == 0)
                num2 = num;

            // Print all the element
            cout << num2++ << " ";
        }
        num *= 2;
        limit = num / n;
        cout << endl;
    }


    return 0;
}

Output:

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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

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