Print Number Star Right Triangle Pattern Type5

June 23, 2019

PRINTING PATTERN:

10*9*8*7

6*5*4

3*2

1

PREREQUISITE:

Basic knowledge of C language and use of loops.

ALGORITHM:

  1. Take the number of rows/columns as input from the user and store it in any variable.(‘r‘ in this case).
  2. Run a loop ‘r’ number of times to iterate through each of the rows. From i=0 to i<r The loop should be structured as for( i=0 ; i<r: i++).
  3. Run a nested loop to iterate through the columns. From j=0 to j<r. The loop should be structured as for(j=0 ; j<r; j++).
  4. increment count to calculate the max value.
  5. Run a loop ‘r’ number of times to iterate through each of the rows. From i=0 to i<r The loop should be structured as for( i=0 ; i<r: i++).
  6. Run a nested loop to iterate through the columns. From j=r to j>i. The loop should be structured as for(j=r; j>i; j++).
  7. decrement count and run an if condition if(j!=r).
  8. if true then print star and count else print only count.
  9. outside the nested loop print a newline.

CODE IN C:

#include<stdio.h>
int main()
{
int i,j,r,count; //declaring integer variables i,j for loops , r for number of rows and count for increment in value
count=1; //initialising count
printf("Enter the number of rows :\n"); //Asking user for input
scanf("%d",&r);
for(i=0;i<r;i++)
  { 
    for(j=0;j<=i;j++)
     {
       count+=1;
     }
  } //taking number of rows and saving it in variable r
for(i=0;i<r;i++) // loop for number of rows
  {
    for(j=r;j>i;j--) // loop for digits per each row
      {
        count--;
        if(j!=r) //incrementing count
           printf("*%d",count);
        else
           printf("%d",count); //printing digits
     }
    printf("\n"); // printing newline after each row
  }
}

TAKING INPUT:DISPLAYING OUTPUT:

1 comments on “Print Number Star Right Triangle Pattern Type5”


  • Ajitha

    Python
    n=4
    start1=10
    for i in range(n,0,-1):
    num1=start1
    for j in range(1,i+1):
    print(num1,end=””)
    if j<i:
    print("*",end="")
    num1-=1
    print()
    start1=(start1-i)