3rd c++ Problem

Replies

• 

  Manish Goyal

  52 views but no replies...
  Give it a try at least...
  otherwise i will give you a clue
• 

  gaurav.bhorkar

  Please give a clue.
  Maybe the problem should not allow use of Decision Control System and loops but allow use of %, /, *, etc.
• 

  Sahithi Pallavi

  Please give us a clue goyal...!
• 

  Hussanal Faroke

  isdivis(a,b)
  {
  x=a;
  while(x>b)
  x-=b
  a-b?return 0:return 1;
  }
  main:-
  for(int j=2,j isdivis(number,j):{cout<<"prime";break;}:continue;
• 

  Hussanal Faroke

  Hussanal Faroke

  isdivis(a,b)
  {
  x=a;
  while(x>b)
  x-=b
  b-x?return 0:return 1;
  }
  main:-
  for(int j=2,j isdivis(number,j)?{cout<<"not prime";exit(0);}:continue;
  cout<<"prime";

  may be there is some errors in extreme... take ur own modification
• 

  Manish Goyal

  Guys Hussanal Faroke gave you a great clue
  yeah You can use ternary operator
  

  Hussanal Faroke

  isdivis(number,j):{cout<<"prime";break;}:continue;

  Hussanal this statement ...seems to be illogical for me
  however nice attempt you are very close..I mean very much close
  Good
• 

  gaurav.bhorkar

  Use a loop to check the condition and goto to jump.
  
  int main ()
  {
  int num ;
  char * array[] = {"even", "odd"} ;
  printf ("Enter any num") ;
  scanf ("%d", &num) ;
  while (num)
  {
  while (num == 1)
  goto label1;
  num -= 2 ;
  }
  label1 : printf ("The number is %s", array[num]) ;
  return 0;
  }
• 

  Manish Goyal

  gaurav.bhorkar

  Use a loop to check the condition and goto to jump.
  
  int main ()
  {
  int num ;
  char * array[] = {"even", "odd"} ;
  printf ("Enter any num") ;
  scanf ("%d", &num) ;
  while (num)
  {
  while (num == 1)
  goto label1;
  num -= 2 ;
  }
  label1 : printf ("The number is %s", array[num]) ;
  return 0;
  }

  Hey Gaurav My question was to check whether no is prime or not?
  Please read the Question before
• 

  gaurav.bhorkar

  gaurav.bhorkar

  Use a loop to check the condition and goto to jump.
  
  int main ()
  {
  int num ;
  char * array[] = {"even", "odd"} ;
  printf ("Enter any num") ;
  scanf ("%d", &num) ;
  while (num)
  {
  while (num == 1)
  goto label1;
  num -= 2 ;
  }
  label1 : printf ("The number is %s", array[num]) ;
  return 0;
  }

  char * array[] is the array of pointers to strings.
  
  

  printf ("The number is %s", array[num]) ;

  When value of num is 0, it prints "even" because the string "even" is present at position 0 in the array.
• 

  gaurav.bhorkar

  Oh I am sorry!
  
  I didn't read the question properly.
  
  Sorry!
• 

  Manish Goyal

  I agree with you but that is fine for checking whether no is even or odd
  absolutely correct No doubt
  
  But I want to check whether no is prime or not...
  I hope you got it
  
  for eg
  
  Please enter no
  20
  sory not a prime no
  
  Please enter no
  
  3
  Yes it is a prime no
  
  Take it easy dude.....The clue is already given....
• 

  Manish Goyal

  Okk Guys ...I think now i should give you the code ...
  so here it is
  

  #include
  #include
  int main()
  {
  int i,no,a,b;
  cout<<"Enter any no"<>no;
  a=no;
  for(i=2;i0)
      {
          no=no-i;
      }
      b=(no==0)?-1:1;
      (b==1)?(no=a):(i=no);
  }
  (b!=-1&no!=1)?cout<<"Prime no":cout<<"Not a prime no";
  getch();
  return 0;
  }
  

• 

  gaurav.bhorkar

  Oh! That was easy, wasn't that?
  
  I was thinking too much.
• 

  Manish Goyal

  yeah Garauv
  It was easy..just you were to show some crazIness....
  Anyhow..thanks for giving it a try
• 

  Hussanal Faroke

  I don't know how to delete one post that is why i edited like this!!!!!!!!!!!!!!
  
  
• 

  aicbal

  int main()
  {
  int num,i,j,flag=FALSE;
  printf("Enter a number: ");
  scanf("%d",&num);
  for(i=2;i {
  j=num/i;
  j=0?flag=TRUE:;
  }
  flag=TRUE?printf("Not Prime");printf("Prime");
  }
  
  
  #-Link-Snipped-#