just some fun little programming puzzles I found around online...



Write a "Hello World" program in 'C' without using a semicolon.
(Note: #include in C doesn't need a semicolon but import does)



Problem #1 Write a "Hello World" program in D with only a semicolon on import statement.

Problem #2 Test if an int is even or odd without looping or if statement (Cant use: do, while, for, foreach, if).

Problem #3 Write a program without using any loop (if, for, while etc) to print numbers from 1 to 100 and 100 to 1;

Problem #4 Find if the given number is a power of 2.



Post Solutions to this root, comments to someones solution in that thread.

Wyverex wrote:
> just some fun little programming puzzles I found around online...
> 
> 
> 
> Write a "Hello World" program in 'C' without using a semicolon.
> (Note: #include in C doesn't need a semicolon but import does)
> 
> 
> 
> Problem #1 Write a "Hello World" program in D with only a semicolon on import statement.
> 
> Problem #2 Test if an int is even or odd without looping or if statement (Cant use: do, while, for, foreach, if).
> 
> Problem #3 Write a program without using any loop (if, for, while etc) to print numbers from 1 to 100 and 100 to 1;
> 
> Problem #4 Find if the given number is a power of 2.
> 
> 
> 
> Post Solutions to this root, comments to someones solution in that thread.

Solutions, Don't Peek if you havn't tried them yet!


Problem #1********************************************
import tango.io.Stdout;

void main()
{
  if(Stdout("Hello World")) {}
}


Problem #2*********************************************
import tango.io.Stdout;

void main()
{
  int x = 2;

  char[][2] ans = ["Even", "Odd"];
  Stdout( ans[ x & 1 ] ).newline;
}

Problem #3*******************************************
import tango.io.Stdout;

void countUP( int i )()
{
  Stdout(i)(" ");
  countUP!(i+1);
}

void countUP( int i : 101 )() {}

void countDN( int i )()
{
  Stdout(i)(" ");
  countDN!(i-1);
}

void countDN( int i : 0 )() {}

void main()
{
  countUP!(1);
  Stdout.newline;
  countDN!(100);
}

Problem#4******************************************
import tango.io.Stdout;

bool test( in uint i )
{
  int c = 0;
  while( i > 0 )
  {
    c += i & 1;
    i = i >> 1;
  }
  return ( c == 1? true : false );
}

void main()
{
   for(uint x = 0; x < uint.max ; ++x)
      if(test(x)) Stdout(x).newline;
}

Reply to wyverex,

> just some fun little programming puzzles I found around online...
> 
> Problem #2 Test if an int is even or odd without looping or if
> statement (Cant use: do, while, for, foreach, if).

bool isEven(int i) {return !(i & 0x01);}

> 
> Problem #3 Write a program without using any loop (if, for, while etc)
> to print numbers from 1 to 100 and 100 to 1;

void DoIt(int i = 1)
{
   writef("%d\n", i);
   i==100 || DoIt(i+1);
   writef("%d\n", i);
}

> 
> Problem #4 Find if the given number is a power of 2.
> 

bool isPow(int i)
{
 if(!i) return true;
 while(!(i & 0x01)) i>>=1;
 return !(i ^ 0x01);
}

Wyverex wrote:

> Problem #1******************************************** import tango.io.Stdout;
> 
> void main()
> {
>    if(Stdout("Hello World")) {}
> }

this works, no semicolon needed with phobos:

void main()
{
    if( printf("Hello World"), true )
    {
    }
}

Sweet! Didn't know printf was accessible without an import!
10pts!

Lutger wrote:
> Wyverex wrote:
>  
>> Problem #1********************************************
>> import tango.io.Stdout;
>>
>> void main()
>> {
>>    if(Stdout("Hello World")) {}
>> }
> 
> this works, no semicolon needed with phobos:
> 
> void main()
> {
>     if( printf("Hello World"), true )
>     {
>     }
> }

Through another one in....



(Multiply x by 7 without using multiplication (*) operator.)

Bonus Problem!
Multiply X by Y without using multiplication (*) operator..  Faster the better!!!

Wyverex wrote:
> Through another one in....
> 
> 
> 
> (Multiply x by 7 without using multiplication (*) operator.)
> 
> Bonus Problem!
> Multiply X by Y without using multiplication (*) operator..  Faster the better!!!


import tango.io.Stdout;

uint mult ( in uint a, in uint b )
{
   uint shift, ans;

   while( b )
   {
      ans += ( b & 1 ? a << shift : 0 );
      shift ++;
      b >>= 1;
   }
   return ans;
}

void main()
{
  uint x = 92;
  uint y = 478;

  Stdout( x * y ).newline;  //here to prove value is right!!
  Stdout( mult(x, y) ).newline;
}

Reply to wyverex,

> Through another one in....
> 
> (Multiply x by 7 without using multiplication (*) operator.)
> 
> Bonus Problem!
> Multiply X by Y without using multiplication (*) operator..  Faster
> the
> better!!!

int Mul(int i, int j)
{
  int ret = 0;
  while(j)
  {
     if(j & 0x01) ret += i;
     j >>= 1;
     i <<= 1;
  }
  return ret;
}


now do div :)

char a=0,b=0,c=0,d=0;
while(a<5 || b<9 || c<5 || d<9)
{
  writef("%d%d:%d%d\n", a,b,c,d);

  // add a single /expression/ here
}

make that count from "00:00" to "59:59" (minutes and seconds)

I seem to remember I figured this out once, but I don't remember how I did it.

On Thu, 07 Aug 2008 01:50:56 +0400, Wyverex <wyverex.cypher@gmail.com> wrote:

> just some fun little programming puzzles I found around online...
>
>
>
> Write a "Hello World" program in 'C' without using a semicolon.
> (Note: #include in C doesn't need a semicolon but import does)
>
>
>
> Problem #1 Write a "Hello World" program in D with only a semicolon on import statement.
>
> Problem #2 Test if an int is even or odd without looping or if statement (Cant use: do, while, for, foreach, if).
>
> Problem #3 Write a program without using any loop (if, for, while etc) to print numbers from 1 to 100 and 100 to 1;
>
> Problem #4 Find if the given number is a power of 2.
>

// 0 is considered a power of two here
bool isPowerOfTwo(int i) {
	return (i & (i-1) == 0);
}