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Star Pattern - 8

    *
   *** 
  *****
 *******
*********
 *******
  *****
   ***   
    *

#include <stdio.h>

int main()
{
    int i, j, k;
    for(i=1;i<=5;i++)
    {
        for(j=i;j<5;j++)
        {
            printf(" ");
        }
        for(k=1;k<(i*2);k++)
        {
                printf("*");
        }
        printf("\n");
    }
    for(i=4;i>=1;i--)
    {
        for(j=5;j>i;j--)
        {
                printf(" ");
        }
        for(k=1;k<(i*2);k++)
        {
                printf("*");
        }
        printf("\n");
    }

    return 0;
}

Related Links:
- More Number Pattern Programs
- Star Pattern Programs in C
- Alphabet Pattern Programs in C
- Series Programs in C

5 comments:

  1. class Dymond
    {
    public static void main(String[] args)
    {
    int x=11;
    int y=x/2; // spaces
    int z=1; // *`s

    boolean b1=true;
    boolean b2= true;
    for(int i=0;i<x;i++)
    {
    for(int j=0;j<y;j++)
    {
    System.out.print(" ");
    }
    for(int k=0;k<z;k++)
    {
    System.out.print("*");
    }
    if(y==0) b1=false;
    if(z==x) b2=false;

    y=b1?y-1:y+1;
    z=b2?z+2:z-2;
    System.out.println();
    }

    }
    }

    ReplyDelete
  2. Generalized version of the code:

    #include
    void main()
    {
    int i,j,k,n;
    printf("Enter the no of rows \n");
    scanf("%d",&n);
    for(i=1;i<=n/2;i++)
    {
    for(j=n/2;j>i;j--)
    {
    printf(" ");
    }
    for(k=1;k<=i*2-1;k++)
    {
    printf("*");
    }
    printf("\n");
    }
    for(i=n;i>n/2+1;i--)
    {
    for(j=n;j>=i;j--)
    {
    printf(" ");
    }
    for(k=n+2;k<i*2-1;k++)
    {
    printf("*");
    }
    printf("\n");
    }
    }

    ReplyDelete
  3. #include
    using namespace std;

    int main(){
    int i,j,k,n;
    cout<<"Enter a odd No for Diamond Star Pattern ";
    cin>>n;
    cout<<endl;
    for(i=1;i<=n;i++){
    if(i<=(n+1)/2){
    for(j=i;j<(n+1)/2;j++){
    cout<<" ";
    }
    }else{
    for(j=1;j<=(i-(n+1)/2);j++){
    cout<<" ";
    }
    }

    if(i<=(n+1)/2){
    for(k=1;k<2*i;k++){
    cout<<"*";
    }
    }else{
    for(k=0;k<(2*n-2*i)+1;k++){
    cout<<"*";
    }
    }
    cout<<endl;
    }
    }


    please review :)

    ReplyDelete
  4. #include

    int main()
    {
    int r,i,j,s,k,m,n;

    printf("Enter the no. of rows (better be odd) : ");
    scanf("%d",&r);

    m = r/2+1;
    n = r-m;

    for(i=1;i<=r;i++)
    {
    if(i<=m)
    {
    for(s=1;s<=m-i;s++)
    {
    printf(" ");
    }

    for(j=1;j<=2*i-1;j++)
    {
    printf("*");
    }
    }
    else
    {
    k = i-m;

    for(s=1;s<=k;s++)
    {
    printf(" ");
    }

    for(j=1;j<=r-2*k;j++)
    {
    printf("*");
    }
    }
    printf("\n");
    }

    return 0;
    }

    ReplyDelete
  5. would you know how i would create this diamond shape in c using the constaint of only 3 printf and 3 n\t\? I would think else if but i cannot figure it out. any help please!!

    ReplyDelete