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Number Pattern - 14

1234567
12345
123
1

int main()
{
  int i,j;
  for(i=7;i>=1;i-=2)
  {
    for(j=1;j<=i;j++)
    {
      printf("%d",j);
    }
    printf("\n");
  }
  return 0;
}


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7 comments:

  1. The above program is conceptually incorrect. It will run only for 7.
    Here is a generic program that will run for any number of rows ->

    void numberPatternC(int rows) {
    int i;
    int j;
    int count = 1;
    for(i=1; i<rows; i++) {
    count = count + 2;
    }
    for(i=0; i<rows; i++) {
    for(j=1; j<=count; j++) {
    System.out.print(j);
    }
    System.out.println();
    count = count - 2;
    }
    }

    ReplyDelete
    Replies
    1. you can use variable to store any dynamic value and use that in loop

      Delete
  2. #include
    #include
    void main()
    {
    int i,j,n;
    clrscr();
    for(i=3;i>=0;i--)
    {
    n=2*i+1;
    for(j=1;j<=n;j++)
    {
    printf("%d",j);
    }
    printf("\n");
    }
    getch();
    }
    //////////// sir, i think its simple :)

    ReplyDelete
  3. //////the more simpler program
    #include
    int main()
    {
    int i,j,n=7;
    for(i=1; i<=4; i++)
    {
    for(j=1; j<=n; j++)
    printf("%d",j);
    n-=2;
    printf("\n");
    }
    return 0;
    }

    ReplyDelete
  4. #include
    #include
    void main()

    {
    int i, j;
    for(i=7;i>=1;i--)
    {
    if(i%2 != 0)
    {
    for(j=1;j<=i;j++)
    {
    printf("%d",j);
    }
    printf("\n");
    }

    }

    getch();


    }

    ReplyDelete
  5. this program logic is first for(i=3;i>=1;i--)
    {
    for(j=1;j<=(i*3+1);j++)
    {
    s.o.pln(j);
    }

    ReplyDelete
  6. #include

    int main()
    {
    int r,i,j;

    printf("Enter the number of rows : ");
    scanf("%d",&r);

    for(i=r;i>=1;i--)
    {
    for(j=1;j<=(2*i)-1;j++)
    {
    printf("%d",j);
    }
    printf("\n");
    }

    return 0;
    }

    ReplyDelete