9.Pyramid

 

• First we count the number of times the pattern is printed i.e 9
• In this we have an exception that we won't iterate from 1to n but from 0 to n-1
• The outer for loop will iterate from 0 to n-1
• Before we used to iterate from 1 to n-i but now initial i value is 0
• So we will iterate from 1 to n-i-1
• Now for the main pattern for i=0 , stars printed is 1
• For i=1 stars printed are 3
• For i=2 stars printed are 5
• We can find a pattern that is (2*i)+1
• You can check by substituting the i value
• So j will iterate from 1 to (2*i)+1
• Remember to print "\n" to make it move to next line

Code

#include<stdio.h>

#include<conio.h>

void main()

{

int i,j,s,n;

clrscr();

printf("Enter number of lines ");

scanf("%d",&n);

for(i=0;i<=n-1;i++)

      {

            for(s=1;s<=n-i-1;s++)

            {

                printf(" ");  

             }

             for(j=1;j<=(2*i)+1;j++)

               {

                  printf("*");

                }

               printf("\n");

       }

getch();

}

Note:scanf is used for taking user input