C Programming Code for this display

C Programming Code for this display

These type of questions are asked to track your analytical skills ,creativity and problem solving skills. good understanding of loop is also needed to solve this question.

For most of these questions there are a lot of solutions in which a few are explained here.

• let us solve this question without loop.
   

algorithm 

1. At first clear the console screen.
2. Then print string " * " 
3. Go to the next line
4. Then pint string " * * "
5. Go to the next line
6. Then pint string " * * * "
7. Go to the next line
8. Then pint string " * * * * "
9. Go to the next line
10. Then pint string " * * * "
11. Go to the next line
12. Then pint string " * * "
13. Go to the next line
14. Then pint string " * "
15. stop the program
16.  

program will be like this

#include <stdio.h>

#include <conio.h>

int main(){

clrscr();

printf(“ * n”);

printf(“ * * n”);

printf(“ * * * n”);

printf(“ * * * * n”);

printf(“ * * * n”);

printf(“ * * n”);

printf(“ * n”);

getch();

return 0;

}

this explain your analytical skills ,creativity and problem solving skills but decrease your technical knowledge rating. if you do like this next they will ask  about loop. you can replace all print statements to a single print statement like this 

int main(){

clrscr();

printf(“ * n * * n * * * n * * * * n * * * n * * n * ”);

algorithm 

1. At first clear the console screen.
2. store this string in an character array " * n * * n * * * n * * * * n * * * n * * n * "
3. display the array using loop
4. stop the program

#include <stdio.h>

#include <conio.h>

int main(){

char a[20]=" * n * * n * * * n * * * * n * * * n * * n * ";

int count;

clrscr();

for (count =0;a[count]!='';count++){

        printf("%c",a[count]);

}

getch();

return 0;

}

• next is using nested for loops

C Programming Code for this display

int I,j,n;……………1

printf(“Enter the number of * in middle line:”);…………….2

scanf(%d, &n);…………….3

for(i=1;i<=n;i++)                       // first for loop

{

                for(j=1;j<=I;j++)

                {

                                printf(“*”);

                }

printf(“n”);

}

for(i=n-1;i>=1;i–)                    //second for loop

{

                for(j=1;j<=I;j++)

                {

                                printf(“*”);

                }

printf(“n”);

}

Let’s understand the logic behind the C code:

If we consider the * layout in the form of table then we have to print 1star in 1^St row, then leave the line, 2 starts in 2^ND row, then again leave the line and so on till the loop reaches to 4 (number that you enter as input).

DRY run of the code:

Line number 1 is the variable declaration

Line number 2 and 3 will take input from the user, suppose 4 in this case. So n=4.

First for loop:

1.       First Iteration:

I=1, check i<=n (1<=4, true) – Enter into the loop

             J=1, check j<=I (1<=1, true) – Enter into the loop

                          Print *

             Now j++ i.e. j=2, check j<=i (2<=1, false) – Exit from the loop

Print(“n”) – leave this line

Now i=2.

2.       Second Iteration:

I=2, check i<=n (2<=4, true) – Enter into the loop

             J=1, check j<=i (1<=2, true) – Enter into the loop

                          Print *

                         Now j++ i. e. j=2, check j<=i (2<=2, true) – be in the loop

                         Print 2^nd *

            Now j++ i. e. j=3, check j<=i (3<=2, false) – Exit from the loop

Print(“n”) – leave this line

Now i=3.

And so on. Like this you can continue your dry run till the condition i<=n become false.

Second for loop: Same logic as there for first for loop.

Try it yourself now and learn the star series

C Programming Code for this display

This program reads input n and gets the desired output.

#include<stdio.h>
main()
{
    int i,j,n;
    printf("nEnter the limit:");
    scanf("%d",&n);
    for(i=1;i<=n;++i)
    {
        for(j=1;j<=i;++j)
            printf("* ");
        printf("n");
    }
    for(i=n-1;i>0;–i)
    {
        for(j=1;j<=i;++j)
            printf("* ");
        printf("n");
    }
}