Print Downward Arrow Mark Symbol Number Pattern
In this article we are going to see how to print the downward arrow mark symbol number pattern.
Exampe-1 When size value= 5 3 3 1 3 5 234 3
Example-2
When size value= 9
5
5
5
5
1 5 9
2 5 8
3 5 7
456
5
Now, let’s see the actual program to print it.
Are you wondering how to seek help from subject matter experts and learn the Java language? Go with these Basic Java Programming Examples and try to code all of them on your own then check with the exact code provided by expert programmers.
Approach:
- Enter size of the pattern and store it in an integer variable
size. - Take one outer for loop to iterate the rows.
- Take one inner for loops, to iterate the columns.
- After each iteration print a new line.
JAVA CODE:
import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
int size, r, c;
//prefer odd number
//Taking size as input from user
System.out.print("Size : ");
Scanner scan = new Scanner(System.in);
size = scan.nextInt();
//Taking middle of the pattern in negative
int mid = -size / 2 + 1;
//Outer Loop
for (r = 1; r <= size; r++)
{
//Inner loop
for (c = 1; c <= size; c++)
{
if (c == size / 2 + 1 || c == mid || c == size - mid + 1)
System.out.print(c);
else
System.out.print(" ");
}
//Prints a newline
System.out.println();
//Incrementing the mid value
mid++;
}
}
}
Output:
Size : 9
5
5
5
5
1 5 9
2 5 8
3 5 7
456
5
C CODE:
#include <stdio.h>
int main()
{
int size, r, c;
//Taking size as input from user
printf("Size : ");
scanf("%d", &size);
//Taking middle of the pattern in negative
int mid = -size / 2 + 1;
//Outer Loop
for (r = 1; r <= size; r++)
{
//Inner loop
for (c = 1; c <= size; c++)
{
if (c == size / 2 + 1 || c == mid || c == size - mid + 1)
printf("%d",c);
else
printf(" ");
}
//Prints a newline
printf("\n");
//incrementing the mid value
mid++;
}
return 0;
}
Output:
Size : 9
5
5
5
5
1 5 9
2 5 8
3 5 7
456
5
C++ CODE:
#include <iostream>
using namespace std;
int main(int argc, char const *argv[])
{
int size, r, c;
//Taking size as input from user
cout << "Size : ";
cin >> size;
//Taking middle of the pattern in negative
int mid = -size / 2 + 1;
//Outer Loop
for (r = 1; r <= size; r++)
{
//Inner loop
for (c = 1; c <= size; c++)
{
if (c == size / 2 + 1 || c == mid || c == size - mid + 1)
cout << c;
else
cout << " ";
}
//Prints a newline
cout << endl;
//Incrementing the mid value
mid++;
}
return 0;
}
Output:
Size : 9
5
5
5
5
1 5 9
2 5 8
3 5 7
456
5