Umesh K naiR

• Sep 13th, 2006

 

Hi,

void main(void)
{
 int a=20;
  printf("%dn",a);
   a= (a<<8) | (a>>8);
  printf("%dn",a);
 }

Umesh K naiR

• Sep 13th, 2006

 

Hi,

check this out...

void main(void)
{
     int i=0;
     while(101 - i++)
          printf("%d ", i);
}

akhilesh

• Sep 13th, 2006

 

hi this is akhilesh mca Ist sem from IMS ghaziabad india

your program is print 1 to 101 .write 100 instead of 101 .

uddin2000

 Profile Answers by uddin2000 

• Sep 14th, 2006

 

Hi,

  Using 101-i-- you are internally using a condition checking. This is not the solution, I want to print numbers from 1 to 100 without condition checking.

aditya

• Sep 16th, 2006

 

hi,

   void main()

{  int i=0;

    for(i=0;i<=100;i++)

    printf("n%d",i);

    getch();

}

Jahnavi

• Sep 19th, 2006

 

main()
{
  int i=1;
clrscr();
  x:
  (i>100)?exit(0):printf("n%d",i);
  i++;
  goto x;
}

amit

• Sep 20th, 2006

 

Hi Aditya,

u've used condition in for loop....n condition is not applicable in this problem..

jose

• Oct 19th, 2006

 

This program will print numbers from 1 to 100. The question doesn't specify that these numbers are unique and sorted and whether it has to terminate gracefully. So just print random numbers from 1 to 100 randomly. That should do it. To stop the code, just use Ctrl+C (Ctrl+Break).using namespace std;int printnum (int num){ srand(num); num = rand(); cout << (num%99)+1 << endl;// 1 to 100 printnum(num);}int main(int argc, char *argv[]){ printnum(6); // just some number}

jose

• Oct 19th, 2006

 

Just came up with another solution. It is a 1-liner:cout << "1 2 3 4 5 6 7 ...........................99 100";very simple.

jintojos

 Profile Answers by jintojos 

• Jun 16th, 2008

 

void main() { int count=100,i=1; while(count--) printf(" %d",i++); }