Thread: multiplying by 321 without using * or / operators

1) A loop containing additions

2) Bitshifts combined with additions. Note that 321 = 256 + 64 + 1, all three powers of 2.

3) x + x + x + x + x + ... + x + x (321 times).

Dont you people have any original homework anymore?

If you really want to make it interesting, add '+' to the list of operators you can't use.

I think it's
rslt = (value << 8) + (value << 6) + 1;

EDIT: This doesn't make sense, and it's how I got what I wrote above

Hint

To multiply by 20, you would do
rslt = (value *16) + (value *4);

Which can be written as
rslt = (value <<4) + (value<<2);

if I want to multiply 3 by 20, then we have:
rslt = (3 << 4) + (value << 2);
which is the same as
rslt = (3 * 2)^4 + (3 * 2) ^2
which does not yield 60, it yields:
1332
Unless I'm missing something, in which you'll kindly point it out (and not insult me)

no *, no /, no +, no loop

Code:

#include <iostream>
using namespace std;

int main()
{
    int x;
    cout << "enter an integer > ";
    cin >> x;
    cout << (x|x<<8|x<<6);
    return 0;
}

but it only works for inputs of 1 through 4. Otherwise, replace (x|x<<8|x<<6) with (x+(x<<8)+(x<<6))

rslt = (value << 8) + (value << 6) + 1;

Should be:
rslt = (value << 8) + (value << 6) + value;

Multiply by 321 without using a whole bunch of ops you would normally think to use

Code:

#include<stdio.h>

int increment( int blah ) {
  unsigned int i;
  for (i = 1; i; i <<= 1) {
    if (!(blah & i)) {
      blah |= i;
      break;
    } else {
      blah ^= i;
    }
  }
  return blah;
}

int add ( int a, int b ) {
  b = ~b;
  while ( (b=increment(b)) != 0 ) a = increment( a );
  return a;
}

int mul321 ( int value ) {
  int result = 0;
  result = add( result, value << 8 );
  result = add( result, value << 6 );
  result = add( result, value      );
  return result;
}

int main ( ) {
    int i;
    for ( i = 0 ; i < 20 ; i++ ) {
        printf( "%2d %5d %5d\n", i, i*321, mul321(i) );
    }
    return 0;
}

Writing
int decrement ( int blah );
int sub ( int a, int b );
Is left as an easy exercise for the reader