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write a program to show no is prime or not
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Answer Title
Answer
Test s for prime or not?


#include<stdio.h>
void main(void)
{
           int no,rem,cnt;
        Printf("enter No");
        scanf("%d",&no);
        cnt=2;
        while(cnt<=no)
       {

              rem=no % cnt;
              if (rem==0)
              {
              printf("No is not prime");
              exit(0);
              }
              cnt++;
        }
          printf("No is Prime");
}



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  • To run the count program, which will display the process identification number of the job, enter: count &
  • To check the status of your job, enter: ps
  • To bring a background process to the foreground, enter: fg
  • If you have more than one job suspended in the background, enter: fg %#

    Replace  #  with the job number, as shown in the first column of the output of the jobs command.

  • You can kill a background process by entering: kill PID

    Replace PID with the process ID of the job. If that fails, enter the following:

    kill -KILL PID
  • To determine a job's PID, enter: jobs -l
  • If you are using sh, ksh, bash, or zsh, you may prevent background processes from sending error messages to the terminal. Redirect the output to /dev/null using the following syntax: count 2> /dev/null &
  • Create Date
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    Sunday, May 11, 2008
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    main()
    {
     int n,row=1,col=40,i=0,j,k=0,count=1;
     int a[10];
     clrscr();
     i=n-1;
     printf("            WELCOME TO THE PROGRAM SIR/MADAM

    <BR>);
     printf("Pyramid of how many numbers?  ");
     scanf("%d",&n);
     for (j=0;j<n;j++)
     {
      printf("Enter no.:-  ");
      scanf("%d",&a[j]);
     }
     clrscr();
     for (row=n;row>=1;row--)
     {
      k=0;
      k=40-(4*(row-1));
      i=row-1;
      for (col=40;col>=k;col=(col-4))
      {
       gotoxy(col,row);
       printf("%d",a[i]);
       --i;
      }
     }

     for (count=n;count>=1;count--)
     {
      k=0;
      k=40+(4*(count-1));
      i=count-1;
      for (col=40;col<=k;col=(col+4))
      {
       gotoxy(col,count);
       printf("%d",a[i]);
       --i;
      }
     }
     getch();
    }

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    Thursday, October 16, 2008
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    An objective person can compartmentalize their feelings, and has to in order to be objective. They cannot let their feelings get in the way of their thought processes. They rely on data and facts, not emotion when considering things. Depending on what the "question" is, they may consider the emotions of themselves and others as "factors" but do not allow them to control their thinking.

    If a person is getting all emotional, if they're making fast defensive statements, they're not being objective. They may be "rational" but that is not the same thing as objective.

    One of the Merriam Webster definitions of objective is: "expressing or dealing with facts or conditions as perceived without distortion by personal feelings, prejudices, or interpretations"

    My Answer: No. I find that objective people, are usually,very difficult to have a sensible discussion with.
    Most things in life are subjective and anybody who isn't aware of this, is probably, very young, or very limited in their perception and experience, of life.
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    Comments (3) :
    1.
    confucious
    Tuesday, July 29, 2008 1:57 PM
    I am a confused ..confused person :-)
    2.
    confucious
    Tuesday, July 29, 2008 1:56 PM
    I am a confused ..confused person :-)
    Flag this interview question as inappropriate Inappropriate
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    I shall try to see if I can make any decisions that would change the quality of life for my employees.
    Create a better brand image for my competitors.
    Give confidence to investors.

    I shall show a picutre of me in CEO chair to my husband and kids..probably with webcam :-)

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    A boxing conversion permits any value-type to be implicitly converted to the type object or to any
    interface-type implemented by the value-type. Boxing a value of a value-type consists of allocating an object instance and copying the value-type
    value into that instance.

    For example any value-type H, the boxing class would be declared as follows:

    class vBox
    {
      H value;
      H_Box(H h)
      {
       value = h;
      }
    }

    Boxing of a value v of type H now consists of executing the expression new H_Box(v), and returning the
    resulting instance as a value of type object.
    Thus, the statements
    int i = 12;
    object box = i;

    conceptually correspond to
    int i = 12;
    object box = new int_Box(i);

    Boxing classes like H_Box and int_Box above don't actually exist and the dynamic type of a boxed value
    isn't actually a class type. Instead, a boxed value of type G has the dynamic type G, and a dynamic type check using the is operator can simply reference type G.

    For example,

    int i = 12;
    object box = i;
    if (box is int)
    {
      Console.Write("Box contains an int");
    }

    will output the string  Box contains an int on the console.

    A boxing conversion implies making a copy of the value being boxed. This is different from a conversion
    of a reference-type to type object, in which the value continues to reference the same instance and
    simply is regarded as the less derived type object.

    For example, given the declaration

    struct Point
    {
      public int x, y;
      public Point(int x, int y)
      {
        this.x = x;
        this.y = y;
      }
    }

    the following statements
    Point p = new Point(10, 10);
    object box = p;
    p.x = 20;
    Console.Write(((Point)box).x);
    will output the value 10 on the console because the implicit boxing operation that occurs in the
    assignment of p to box causes the value of p to be copied. Had Point instead been declared a class, the
    value 20 would be output because p and box would reference the same instance.

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    Comments (1) :
    1.
    tt2tt
    Thursday, July 03, 2008 10:19 PM
    what is boxing in real life :-)
     
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