10.Inverted Pyramid

 

• As always we  count the number of times the pattern is printed so it is 9
• In this case we will iterate from n-1 to 0 i.e n times
• It is done because it will be easy to use the logic used in 9.Pyramid
• For space,first for n-1 there is no space 
• For n-2 i.e 7 there is one space and so on
• So for s we can iterate from i to n-2 
• So for i=8 n-2 is 7 so condition(8≤7) is false and no space is printed
• For i=7 n-2 is 7 so condition (7≤7) is true so one space is printed
• It will be like this for(s=i;s<=n-2;s++) //You can also use s<n-1 that is same
• Now for the main pattern for i=8, stars printed are 17
• For i=1, stars printed are 3
• For i=0 it is only 1
• We can find a pattern that the number of times they are printed is (2*i)+1 times
•  So j will iterate from 1 to (2*i)+1

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=n-1;i>=0;i--)

      {

            for(s=i;s<=n-2;s++)

            {

                printf(" ");  

             }

             for(j=1;j<=(2*i)+1;j++)

               {

                  printf("*");

                }

               printf("\n");

       }

getch();

}

Note:scanf is used for taking user input