Print Number Star Square Pattern Type2

June 23, 2019

PRINTING PATTERN:

13*14*15*16

9*10*11*12

5*6*7*8

1*2*3*4

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.(‘l‘ in this case).
  2. Run a loop ‘l’ number of times to iterate through each of the rows. From i=0 to i<l. The loop should be structured as for( i=0 ; i<l : i++).
  3. Inside this loop run another nested loop to iterate through the columns. From j=0 to j<l. The loop should be structured as for(j=0 ; j<l ; j++).
  4. And increment count to get the max number required.
  5. Run a loop ‘l’ number of times to iterate through each of the rows. From i=0 to i<l. The loop should be structured as for( i=0 ; i<l : i++).
  6. Inside this loop run another nested loop to iterate through the columns. From j=0 to j<l. The loop should be structured as for(j=0 ; j<l ; j++).
  7. increment count and run an if condition if(j==l-1).
  8. if true then print count else print count and a star after it.
  9. outside the nested loop reinitialize count=count-2*l then print a newline

CODE IN C:

#include<stdio.h>
int main()
{
int i,j,l,count=0; //declaring integers i,j for loops and l for number of rows
printf("Enter the number of rows/columns\n"); //Asking user for input
scanf("%d",&l); //Taking the input for number of rows
for(int i=0;i<l;i++) //Outer loop for number of rows
  {
    for(j=0;j<l-1;j++)
      {
        count++;
      }
  }
    for(i=0;i<l;i++)
      {
        for(j=0;j<l;j++)
          {
            count++;
            if(j==l-1)
              {  
                printf("%d",count);
              }
            else
              {
                printf("%d*",count);
              }

          }
    count=count-2*l;
    printf("\n");
  } 

}

TAKING INPUT:DISPLAYING OUTPUT:

4 comments on “Print Number Star Square Pattern Type2”


  • Ajitha

    n=4
    start1=13
    for i in range(1,n+1):
    num1=start1
    for j in range(1,n+1):
    print(num1,end=””)
    if j<4:
    print("*",end="")
    num1+=1
    print()
    start1=start1-n


  • Bhupendra

    #include
    using namespace std;

    int main()
    {
    int n;
    cin>>n;
    int count=n*n;
    for (int i=1; i<=n; i++){
    count=count-4*i+1;
    for (int j=1; j<=n; j++){
    cout<<count++<<"*";
    }
    count=n*n;
    cout<<endl;
    }

    return 0;
    }


  • Prasoon

    #include
    using namespace std;

    int main() {
    int rows,count=13;
    cout<>rows;
    for(int i=0; i<rows; i++) {
    for(int j=0; j<rows; j++) {
    cout<<count++<<"*";
    }
    count = count-8;
    cout<<"\n";
    }
    return 0;
    }